Wave Optics

Class 12 Physics Chapter 10 Important Extra Questions Wave Optics

Very Short Answer

Question 1.
How does the fringe width of interference fringes change, when the whole apparatus of Young’s experiment is kept in a liquid of refractive index 1.3? (Delhi 2008)
Answer:

Fringe width becomes yL times of its initial value.

Question 2.
How does the angular separation of interference fringes change in Young’s experiment, if the distance between the slits is increased? (Delhi 2008)
Answer:

When separation between two slits is increased, angular separation decreases.

Question 3.
State the reason, why two independent sources of light cannot be considered as coherent sources. (Delhi 2008)
Answer:
Two independent sources of light cannot be coherent. This is because light is emitted by individual atoms, when they return to ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.

Question 4.
How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced with red light? (All India 2008)
Answer:

When incident violet light is replaced with red light, the angle of minimum deviation of a glass decreases.

Question 5.
If the angle between the pass axis of polarizer and the analyser is 45°, write the ratio of the intensities of original light and the transmitted light after passing through the analyser. (Delhi 2008)
Answer:

Question 6.
What type of wavefront will emerge from a
(i) point source, and
(ii) distant light source? (Delhi 2008)
Answer:
(i) Point source – Spherical wavefront
(ii) Distant light source – Plane wavefront.

Question 7.
Unpolarized light is incident on a plane surface of glass of refractive index µ at angle i. If the reflected light gets totally polarized, write the relation between the angle i and refractive index µ. (Delhi 2008)
Answer:
µ = tan i_p.

Question 8.
Draw a diagram to show refraction of a plane wave front incident in a convex lens and hence draw the refracted wave front. (Delhi 2008)
Answer:

Question 9.
At what angle of incidence should a light beam strike a glass slab of refractive index \(\sqrt{3}\), such that the reflected and the refracted rays are perpendicular to each other? (Delhi 2008)
Answer:

Question 10.
Differentiate between a ray and a wave front. (Delhi 2008)
Answer:
Ray defines the path of light.
Wave front is the locus of points in the light wave’ having the same phase of oscillation at any instant.

Question 11.
How would the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen is doubled? (All India 2008)
Answer:
Angular separation \(\theta=\frac{\lambda}{d}\) and is independent of slit-screen separation
∴ There will be no change

Question 12.
How does the angular separation between fringes in single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled? (All India 2012)
Answer:

When the distance D of seperation between the slits and the screen is doubled, the angular seperation θ remains unchanged.

Question 13.
In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band. (All India 2012)
Answer:
If the width of the diffraction slit is doubled, the size of the central diffraction band will become half and its intensity will become four times of its original value.

Question 14.
How does the fringe width, in Young’s double-slit experiment, change when the distance of separation between the slits and screen is doubled? (All India 2012)
Answer:
If the distance between slits and screen (D) is doubled, the fringe width in double slit

Question 15.
In what way is plane polarized light different from an unpolarized light? (Comptt. All India 2012)
Answer:
In case of polarized light, the directions of electric field vector are restricted to only a particular / plane whereas in an unpolarized light the direction of \(\overrightarrow{\mathrm{E}}\) is in all possible directions in a plane perpendicular to the direction of propagation.

Question 16.
In a single slit diffraction experiment, the width of the slit is reduced to half its original width. How would this affect the size and intensity of the central maximum? (Comptt. Delhi 2012)
Answer:

Question 17.
Which of the following waves can be polarized
(i) Heat waves
(ii) Sound waves? Give reason to support your answer. (Delhi 2013)
Answer:
Heat waves can be polarized as they are transverse in nature.

Question 18.
Define the term ‘coherent sources’ which are required to produce interference pattern in Young’s double slit experiment. (Comptt. Delhi 2014)
Answer:
Two monochromatic sources, which produce light waves, having a constant phase difference are defined as coherent sources.

Question 19.
Define the term ‘wavefront’.(Comptt. All India 2013)
Answer:
The wavefront is defined as the locus of all particles of a medium, which are vibrating in the same phase.

Question 20.
Draw the shape of the wavefront coming out of a convex lens when a plane wave is incident on it. (Comptt. All India 2013)
Answer:

Question 21.
Draw the shape of the wavefront coming out of a concave mirror when a plane wave is incident on it. (Comptt. All India 2013)
Answer:

Question 22.
Why does Sun appear red at sunrise and sunset? (All India 2016)
Answer:
It is due to least scattering of red light as it has the longest wavelength.
[As per Rayleigh’s scattering, the amount of light

Question 23.
A beam of unpolarised light is incident, on the boundary between two transparent media, at an angle of incidence = i_B, the Brewester’s angle. At what angle does the reflected light get polarised? (Comptt. All India 2016)
Answer:
At an angle of incidence = iB, the reflected light gets polarised.

Question 24.
Sketch the refracted wavefront emerging from convex tens, If a plane wavefront is an incident normally on it.
Answer:
The figure is as shown.

Question 25.
How would you explain the propagation of light on the basis of Huygen’s wave theory?
Answer:
To explain the propagation of light we have to draw a wavefront at a later instant when a wavefront at an earlier instant is known. This can be drawn by the use of Huygen’s principle.

Question 26.
Draw the shape of the reflected wavefront when a plane wavefront is an incident on a concave mirror.
Answer:
The reflected wavefront is as shown.

Question 27.
Draw the shape of the refracted wavefront when a plane wavefront is an incident on a prism.
Answer:
The shape of the wavefront is as shown.

Question 28.
Draw the type of wavefront that corresponds to a beam of light diverging from a point source.
Answer:
The wavefront formed by the light coming from a very far off source is a plane and for a beam of light diverging from a point, a wavefront is a number of concentric circles.

Question 29.
Draw the type of wavefront that corresponds to a beam of light coming from a very far off source.
Answer:
The wavefront is as shown.

Question 30.
Name two phenomena that establish the wave nature of light.
Answer:
Interference and diffraction of light.

Question 31.
State the conditions which must be satisfied for two light sources to be coherent.
Answer:
(a) Two sources must emit light of the same wavelength (or frequency).
(b) The two light sources must be either in-phase or have a constant phase difference.

Question 32.
Draw an intensity distribution graph for diffraction due to a single-slit.
Answer:
The intensity distribution for a single-slit diffraction pattern is as shown.

Question 33.
Name one device for producing plane polarised light. Draw the graph showing the variation of intensity of polarised light transmitted by an analyser.
Answer:
Nicol prism can be used to produce plane polarised light. The graph is as shown.

Question 34.
State Huygens’ principle of diffraction of light. (CBSE AI 2011C)
Answer:
Huygens principle states that
(a) Each point on a wavefront is a source of secondary waves which travel out with the same velocity as the original waves.
(b) The new wavefront is given by the forward locus of the secondary wavelets.

Question 35.
In what way is a plane polarised tight different from an unpolarised light? (CBSE AI 2012C)
Answer:
Plane polarized light vibrates 1n only one plane.

Question 36.
Which of the following waves can be polarised: (i) Heatwaves (ii) Sound waves? Give a reason to support your answer. (CBSE Delhi 2013)
Answer:
Heatwaves are transverse In nature.

Question 37.
Define the term ‘wavefront’. (CBSE AI 2014C)
Answer:
It Is defined as the locus of all points In a medium vibrating in the same phase.

Question 38.
Define the term ‘coherent sources’ which are required to produce interference pattern in Young’s double-slit experiment. (CBSE Delhi 2014C)
Answer:
Two sources that are In phase or have a constant phase difference are called coherent sources.

Question 39.
What change would you expect if the whole of Young’s double-slit apparatus were dipped into the water?
Answer:
The wavelength λ, of light In water, is less than that in air. Since the fringe width β is directly proportional to the wavelength of light, therefore, the fringe width will decrease.

Question 40.
When light travels from a rarer to a denser medium, it loses some speed. Does the reduction in speed Imply a reduction in the energy carried by the light wave?
Answer:
No, the energy carried by a wave depends upon the amplitude of the wave and not on Its speed of propagation.

Question 41.
If one of the slits say S₁, is covered then what changes occur in the Intensity of light at the centre of the screen?
Answer:
The intensity 1s decreased four times because l ∝ 4a² where a is the amplitude of each wave.

Question 42.
How does the angular separation between fringes in a single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled? (CBSE AI 2012)
Answer:
No change.

Question 43.
What is the effect on the interference fringes in Young’s double-slit experiment If the separation between the screen and slits Is Increased?
Answer:
The fringe width Increases.

Question 44.
How does the Intensity of the central maximum change If the width of the slit Is halved in a single-slit diffraction experiment?
Answer:
The width of the central maxima is doubled and the intensity is reduced to one-fourth of Its original value.

Question 45.
The polarising angle of a medium Is 60°, what is the refractive index of the medium? (CBSE Delhi 2019)
Answer:
Using the expression
μ = tan i_p = tan 60° = 1.732.

Question 46.
In the wave picture of light intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light?
Answer:
For a given frequency intensity of light in the photon, the picture is determined by the number of photons crossing a unit area per unit time.

Question 47.
How does the Intensity of the central maximum change if the width of the slit is halved in the single-slit diffraction experiment?
Answer:
The width of the central maximum is doubled and the intensity is reduced to one-fourth of its original value.

Question 48.
What would happen If the path difference between the interfering beams that is S₂P – S₁P became very large?
Answer:
If the path difference becomes very large it may exceed the coherent length. Thus the coherence of the waves reaching P is lost and no interference takes place.

Question 49.
In Young’s double-slit experiment, what would happen to the intensity of the maxima and the minima if the size of the hole illuminating the two coherent holes were gradually Increased?
Answer:
The fringe width will decrease and finally, there will be general illumination on the screen.

Question 50.
What is the Brewster angle for air to glass transition? (Refractive index of glass =1.5.) (NCERT)
Answer:
Given μ = 1.5, i_P = ?
Using the relation μ = tan i_P we have
l_p = tan^-1 (1.5) = 56.3°

Question 51.
Is Huygen’s principle valid for longitudinal sound waves? (NCERT Exemplar)
Answer:
Yes

Question 52.
Consider a point at the focal point of a convergent lens. Another convergent lens of the short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image? (NCERT Exemplar)
Answer:
Spherical.

Question 53.
What is the shape of the wavefront on earth for sunlight? (NCERTExemplar)
Answer:
Spherical with a huge radius as compared to the earth’s radius so that it is almost a plane.

Question 54.
Draw a graph showing the intensity distribution of fringes due to diffraction at a single-slit. (CBSE 2018C)
Answer:

Short Answer Type

Question 1.
How can one distinguish between an unpolarised and linearly polarised light beam using polaroid? (CBSE Delhi 2019)
Answer:
The two lights will be allowed to pass through a polariser. When the polarizer is rotated in the path of these two light beams, the intensity of light remains the same in all the orientations of the polariser, then the light is unpolarised. But if the intensity of light varies from maximum to minimum then the light beam is a polarised light beam.

Question 2.
What is meant by plane polarised light? What type of waves shows the property of polarisation? Describe a method of producing a beam of plane polarised light?
Answer:

1. The light that has its vibrations restricted in only one plane is called plane polarised light.
2. Transverse waves show the phenomenon of polarization.
3. Light is allowed to pass through a polaroid. The polaroid absorbs those vibrations which are not parallel to its axis and allows only those vibrations to pass which are parallel to its axis.

Question 3.
Write the Important characteristic features by which the Interference can be distinguished from the observed diffraction pattern. (CBSE AI 2015)
Answer:
(a) In the interference pattern the bright fringes are of the same width, whereas in the diffraction pattern they are not of the same width.
(b) In interference all bright fringes are equally bright while in diffraction they are not equally bright.

Question 4.
State Brewster’s law. The value of Brewster’s angle for the transparent medium is different for the light of different colours. Give reason. (CBSE Delhi 2016)
Answer:
When the reflected ray and the refracted ray are perpendicular then μ = tani_p where i_p is the polarising angle or Brewster angle.

Brewster’s angle depends upon the refractive index of the two media in contact. The refractive index in turn depends upon the wavelength of light used (different colours) hence Brewster’s angle is different for different colours.

Question 5.
State one feature by which the phenomenon of interference can be distinguished from that of diffraction.
A parallel beam of light of wavelength 600 nm is incident normally on a slit of width ‘a’. If the distance between the slits and the screen is 0.8 m and the distance of 2^nd order maximum from the centre of the screen is. 15 mm, calculate the width of the slit. (All India 2008)
Answer:
(i) In interference all the maxima are of equal intensity.
In diffraction pattern central fringe is of maximum intensity while intensity of secondary maxima falls rapidly.

Question 6.
Define the term ‘linearly polarised light’. When does the intensity of transmitted light become maximum, when a polaroid sheet is rotated between two crossed polaroids? (All India 2008)
Answer:
Linearly polarised light is one in which the
vibration of light is present in one line only.

Intensity is maximum.

Question 7.
(i) State the principle on which the working of an optical fiber is based.
(ii) What are the necessary conditions for this phenomenon to occur? (All India 2009)
Answer:
(i) Working of an optical fibre is based on the principle of total internal reflection.
(ii) (a) Light should travel from a denser to rarer medium.
(b) Angle of incidence should be more than

Question 8.
(a) Why are coherent sources necessary to produce a sustained interference pattern?
(b) In Young’s double slit experiment using mono-chromatic light of wavelength X, the intensity of light at a point on the screen where path difference is X, is K units. Find out the intensity of light at a point where path difference is \(\frac{2 \lambda}{3}\). (Delhi 2012)
Answer:
(a) Coherent sources have a constant phase difference and, therefore, produce a sustained interference pattern.
These sources are needed to ensure that the position of maxima and minima do not change with time.

Question 9.
State two conditions required for obtaining co-herent sources.
In Young’s arrangement to produce interference pattern, show that dark and bright fringes appearing on the screen are equally spaced. (Comptt. Delhi 2009)
Answer:
Two conditions for obtaining coherent sources: (0 Two sources should give monochromatic light.
(ii) Coherent sources of light should be obtained from a single source by some device.
The fringe width (dark and bright) is given by

Hence, it is same for both dark and bright fringes So they are equally spaced on the screen.

Question 10.
Laser light of wavelength 640 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of light which produces interference fringes separated by 8.1 mm using same arrangement. Also find the minimum value of the order ‘n’ of bright fringe of shorter wavelength which coincides with that of the longer wavelength. (Comptt. All India 2012)
Answer:
Distance between two bright fringes = Fringe width

Calculation of minimum value of order: for n to be minimum
(n + 1)^th maxima of shorter wavelength should coincide with n^th maxima of longer wavelength

Question 11.
Yellow light (λ = 6000Å) illuminates a single slit of width 1 x 10-4 m. Calculate
(i) the distance between the two dark lines on either side of the central maximum, when the diffraction pattern is viewed on a screen kept 1.5 m away from the slit;
(ii) the angular spread of the first diffraction minimum. (Comptt. All India 2012)
Answer:
(i) Distance between two dark lines, on either

(ii) Angular spread of the first diffraction minimum (on either side)

Question 12.
A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Calculate the width of the slit. (All India 2013)
Answer:

Question 13.
A parallel beam of light of 600 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1.2 m away. It is observed that the first minimum is at a distance of 3 mm from the centre of the screen. Calculate the width of the slit. (All India 2013)
Answer:

Question 14.
Write the distinguishing features between a diffraction pattern due to a single slit and the interference fringes produced in Young’s double slit experiment?
Answer:
Difference between interference and diffraction of light

	Interference	Diffraction
1.	Interference is due to superposition of two distinct waves com­ing from two coherent sources.	Diffraction is due to super­position of the secondary wavelets coming from dif­ferent parts of the same wavefront.
2.	Interference fringes may or may not be of the same width.	Diffraction fringes are not to be of the same width.
3.	The intensity of minima is generally zero.	The intensity of minima is never zero.
4.	All bright fringes are of uniform intensity.	All bright fringes are not of uniform intensity.

Question 15.
Answer the following questions :
(i) In what way is diffraction from each slit related to the interference pattern in a double slit experiment?
(ii) When a tiny circular obstacle is placed in the path of light from a distance source, a bright spot is seen at the centre of the shadow of the obstacle. Explain, why. (Comptt. All India 2013)
Answer:
(i) Diffraction from each slit is related to interference pattern in a double slit experiment in the following ways :

• The intensity of minima for diffraction is never zero, while for interference it is generally zero.
• All bright fringes for diffraction are not of uniform intensity, while for interference, these are of uniform intensity

(ii) Waves from the distant source are diffracted by the edge of the circular obstacle and these diffracted waves interfere constructively at the centre of the obstacle’s shadow produc¬ing a bright spot.

Question 16.
(a) Write the conditions under which light sources can be said to be coherent.
(b) Why is it necessary to have coherent sources in order to produce an interference pattern? (Comptt. All India 2013)
Answer:
(a) Coherent sources of light. The sources of light, which emit continuously light waves of the same wavelength, same frequency and in same phase are called Coherent sources of light.
Interference pattern is not obtained. This is because phase difference between the light waves emitted from two different sodium lamps will change continuously.

(b) Conditions for interference. The important conditions for obtaining interference of light are :

1. The two sources of light must be coherent. i.e. they should exist continuous waves of same wavelength or frequency.
2. The two sources should be monochromatic.
3. The phase difference of waves from two sources should be constant.
4. The amplitude of waves from two sources should be equal.
5. The coherent sources must be very close to each other.

Question 17.
(i) Monochromatic light of frequency 6.0 × 10¹⁴ Hz is produced by a laser. The power emitted is 2.0 × 10^-3 W. Estimate the number of photons emitted per second on an average by the source.
(ii) Draw a plot showing the variation of photoelectric current versus the intensity of incident radiation on a given photosensitive surface. (Delhi 2014)
Answer:

Question 18.
For a single slit of width “a”, the first minimum of the interference pattern of a monochromatic light of wavelength λ occurs at an angle of \(\frac{\lambda}{a}\) .
At the same angle of \(\frac{\lambda}{a}\), we get a maximum for a two narrow slits separated by a distance “a”. Explain (Delhi 2014)
Answer:
For a single slit of width ‘a’,
the n^th minimum, \(\sin \theta_{n}=\frac{n \lambda}{a}\)

Question 19.
Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the Polaroid sheet will the transmitted intensity be maximum? (Delhi 2015)
Answer:
Let the rotating Polaroid sheet make an angle θ with the first polaroid.
∴ angle with the other polaroid will be (90 – θ)

Question 20.
Distinguish between unpolarised and a linearly polarised light. Describe, with the help of a diagram, how unpolarised light gets linearly polarised by scattering. (Comptt. Delhi 2016)
Answer:
Unpolarized light : A light wave, in which the electric vector oscillates in all possible directions in a plane perpendicular to the direction of propagation is known as unpolarized light.
Linearly polarized light : If the oscillations of the electric vectors are restricted to just one direction, in a plane perpendicular to the direction of propagation, the corresponding light is known as linearly polarized light.

It is due to scattering of light by molecules of earth’s atmosphere.

Under the influence of the electric field of the incident (unpolarized) wave, the electrons in the molecules acquire components of motion in both these directions. Charges, accelerating parallel to the double arrows, do not radiate energy towards the observer since their acceleration has no transverse component.
The radiation scattered by the molecules is therefore represented by dots, i.e., it is polarized perpendicular to plane of figure.

Question 21.
State Brewster’s law. The value of Brewster angle for a transparent medium is different for light of different colours. Give reason. (Delhi 2016)
Answer:
(i) Brewster’s law : When unpolarised light is incident on the surface separating two media, the reflected light gets (completely) polarized only when the reflected light and refracted light become perpendicular to each other.
The refractive index of denser medium, with respect to rarer medium, is given by

(ii) Since refractive index (µ) of a transparent medium is different for different colours, hence Brewster angle also is different for different colours.

Question 22.
Draw the intensity pattern for single slit diffraction and double slit interference. Hence, state two differences between interference and diffraction patterns. (All India 2017)
Answer:
(i) Intensity distribution in the diffraction due to single slit

(ii) Intensity pattern for double slit interference.

(iii) Difference between Interference and Difference Patterns:
Interference pattern and Diffraction pattern :
The diagram, given here, shows several fringes, due to double slit interference, ‘contained’ in a broad diffraction peak. When the separation between the slits is large compared to their width, the diffraction pattern becomes very flat and we observe the two slit interference pattern.

Basic features of distinction between interference and diffraction patterns :
(i) The interference pattern has a number of equally spaced bright and dark bands while diffraction pattern has a central bright maximum which is twice as wide as the other maxima.
(ii) Interference pattern is the superimposition of two waves slits originating from two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.
(iii) For a single slit of width ‘a’ the first null of diffraction pattern occurs at an angle of \(\frac{\lambda}{a}\). At the same angle of \(\frac{\lambda}{a}\), we get a maxima for two narrow slits separated by a distance ‘a’.

Question 23.
Unpolarised light is passed through a polaroid P₁ When this polarised beam passes through another polaroid P₂ and if the pass axis of P₂ makes angle 6 with the pass axis of P₁, then write the expression for the polarised beam passing through P₂. Draw a plot showing the variation of intensity when θ varies from 0 to 2π. (All India 2017)
Answer:

Question 24.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids.
Answer:
Let l_o be the intensity of polarised light after passing through the first polarizer P₁. Then the intensity of light after passing through the second polarizer P2 will be l = l_ocos 2θ, where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be
l = l_o cos² θ cos² (90° – θ) = l_o cos² θ sin² θ = (l_o /4)sin² 2θ

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 25.
Is energy conserved in interference? Explain.
Answer:
Yes, energy is conserved in interference. Energy from the dark fringes is accumulated in the bright fringes. If we take
l = 4a²cos²\(\frac{\phi}{2}\), then intensity at bright points is l_max = 4a² and intensity at the minima l_min = 0. Hence average intensity in the pattern of the fringes produced due to interference is given by
Ī = \(\frac{I_{\max }+I_{\min }}{2}=\frac{4 a^{2}+0}{2}\) = 2a²

But if there is no interference then total intensity at every point on the screen will be l = a² + a² = 2a², which is the same as the average intensity in the interference pattern.

Question 26.
An incident beam of light of intensity lo is made to fall on a polaroid A. Another polaroid B is so oriented with respect to A that there is no light emerging out of B. A third polaroid C is now introduced midway between A and B and is so oriented that its axis bisects the angle between the axes of A and B. What is the intensity of light now between (i) A and C (ii) C and B? Give reasons for your answers.
Answer:
Polaroids A and B are oriented at an angle of 90°, so no light is emerging out of B. On placing polaroid C between A and B such that its axis bisects the angle between axes of A and B, then the angle between axes of polaroids A and B is 45° and that of C and B also 45°.
(a) Intensity of light on passing through Polaroid A or between A and C is l₁ = \(\frac{l_{0}}{2}\)
(b) On passing through polaroid C, intensity of light between C and B becomes
l₂ = l₁ cos² θ = \(\frac{l_{0}}{2}\) × cos² 45° = \(\frac{l_{0}}{4}\)

Question 27.
One of the slits of Young’s double-slit experiment is covered with a semi¬transparent paper so that it transmits lesser light. What will be the effect on the interference pattern?
Answer:
There will be an interference pattern whose fringe width is the same as that of the original. But there will be a decrease in the contrast between the maxima and the minima, i.e. the maxima will become less bright and the minima will become brighter.

Question 28.
Light from a sodium lamp is passed through two polaroid sheets P₁ and P₂ kept one after the other. Keeping P₁, fixed, P₂ is rotated so that its ‘pass axis can be at different angles, θ, with respect to the pass-axis of P₁.
An experimentalist records the following data for the intensity of light coming out of P₂ as a function of the angle θ.

I = Intensity of beam falling on P₁
(a) One of these observations is not in agreement with the expected theoretical variation of I, identify this observation and write the correct expression.
(b) Define the Brewster angle and write the expression for It in terms of the refractive index of the medium.
Answer:
(a) The observation \(\frac{1}{\sqrt{2}}\) is not correct. It should be 1/2.
(b) It is the angle of incidence at which the refracted and the reflected rays are perpendicular to each other. It is related to the refractive index as tan i_p = μ

Question 29.
How will the interference pattern in Young’s double-slit experiment get affected, when
(a) distance between the slits S₁, and S₂ reduced and
(b) the entire set-up is immersed in water? Justify your answer in each case. (CBSE Delhi 2011C)
Answer:
We know that fringe width β of the dark or bright fringes is given by β = \(\frac{D \lambda}{d}\) where d is the distance between the slits.
(a) When the distance between the slits, i. e. d is reduced then p will increase. The interference pattern will thus become broader.
(b) When the entire set up is immersed in water, the pattern will become narrow due to the decrease in the wavelength of light. The new wavelength λ’ = λ/n, hence β’= β/n

Question 30.
Discuss the intensity of transmitted light when a Polaroid sheet is rotated between two crossed polaroids? (NCERT)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁. Then the intensity of light after passing through the second polarizer P₂ will be l = l_o cos² θ

where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 31.
A Polaroid (I) is placed In front of a monochromatic source. Another Polaroid (II) is placed in front of this Polaroid (I) and rotated till no light passes. A third Polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II)? Explain. (NCERT Exemplar)
Answer:
Only in the special case when the pass axis of (III) is parallel to fill or (II) there shall be no light emerging. In all other cases, there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

Question 32.
How is a wavefront defined? Using Huygen’s construction draw a figure showing the propagation of a plane wave refracting at a plane surface separating two media. Hence verify Snell’s law of refraction. (Delhi 2008)
Answer:
(i) Wavefront : Wavefront is defined as the continuous locus of all such particles of the medium which are vibrating in the same phase at any instant.

(ii) We take a plane wavefront AB incident at a plane surface XY. We use secondary wavelets starting at different times. We get refracted wavefront only when the time taken by light to travel along different rays from one wavefront to another is same. We take any arbitrary ray starting from point ‘P’ on incident wavefront to refracted wavefront at point ‘O’. Let total time be ‘t’.

As time should be independent of the ray to be considered
The coefficient of AO in the above equation should be zero

Where ’µ₂ is called refractive index of medium 2 w.r.t. medium 1. This is Snell’s law of refraction.

Question 33.
How is a wavefront defined? Using Huygen’s construction draw a figure showing the propagation of a plane wave reflecting at the interface of the two media. Shpw that the angle of incidence is equal to the angle of reflection. (Delhi 2008)
Answer:
Wavefront : Wavefront is defined as the continuous locus of all such particles of the medium which are vibrating in the same phase at any instant.

We take any point Q on the incident wavefront. When disturbance from point B on the incident wavefront reaches point B’, the disturbance from point Q reaches Q’ via point K on the reflecting surface. Since B’A’ represents the reflected wavefront, time by light to travel from any point on incident wavefront to the corresponding point on the reflected wavefront should always be same. Let total time be ‘t’

As time of the ray to be considered should be independent, the coefficient of AK in the above equation should be zero.
That is, sin i = sin r or i = r
Hence, angle of incidence is equal to angle of reflection.

Question 34.
Distinguish between unpolarised and plane polarised light. An unpolarised light is incident on the boundry between two transparent media. State the condition when the reflected wave is totally plane polarised. Find out the expression for the angle of incidence in this case. (All India 2008)
Answer:
(a) Unpolarised light: A beam of light in which electrical vector oscillates in all possible planes, in a direction normal to the direction of propagation of wave.

(b) Polarised light : A beam of light in which electrical vector oscillates in a direction normal to the direction of propagation of wave on a single plane only.
From Snell’s law.

Question 35.
In a single slit diffraction experiment, when a tiny circular obstacle is placed in the path of light from a distance source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
State two points of difference between the interference pattern obtained in Young’s double slit experiment and the diffraction pattern due to a single slit. (Delhi 2009)
Answer:
Wave diffracted from the edge of any circular obstacle undergoes constructive interference to form a bright spot at the centre of shadow.

	Young’s double slit experiment	Single slit experiment
1.	Light      originating from two coherent sources.	Light      originating from single source.
2.	Fringes are of equal width.	Fringe width decreases with order.
3.	Intensity of all the bright fringes is the brightness is the same.	Intensity falls with increasing order. The brigtness of succ­essive bright fringes goes on decreasing.

Question 36.
In Young’s double slit experiment, monochromatic light of wavelength 630 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 8.1 mm. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes are separated by 7.2 mm. Find the wavelength of light from the second source. What is the effect on the interference fringes if the monochromatic source is replaced by a source of white light? (All India)
Answer:
Position of the n^th bright fringe is given by \(\frac{n \lambda \mathrm{D}}{d}\) from the central bright,
So the separation between two consecutive bright


When the monochromatic light is replaced by a white light:

1. the central bright remains white and
2. all the other colours will form individual maximas with the least wavelength violet forming its bright close to the central bright.

Question 37.
(a) In a single slit diffraction experiment, a slit of width ‘d’ is illuminated by red light of wavelength 650 nm. For what value of ‘d’ will
(i) the first minimum fall at an angle of diffraction of 30°, and
(ii) the first maximum fall at an angle of diffraction of 30°?
(b) Why does the intensity of the secondary maximum become less as compared to the central maximum? (All India 2009)
Answer:

(b) As the order increases only 1/n^th (where n is an odd number) of the slit, will contribute in producing brightness at a point in diffraction. So the higher order maxima are not so bright as the central.

Question 38.
In Young’s double slit experiment, mono-chromatic light of wavelength 600 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 10 mm. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes are separated by 8 mm. Find the wavelength of light from the second source. What is the effect on the interference fringes if the monochromatic source is replaced by a source of white light? (All India 2009)
Answer:

Effect: When the monochromatic light is replaced by a white light:
(i) the central bright remains white and
(ii) all the other colours will form individual maximas with the least wavelength violet forming its bright close to the central bright.

Question 39.
What is an unpolarized light? Explain with the help of suitable ray diagram how an unpolarized light can be polarized by reflection from a transparent medium. Write the expression for Brewster angle in terms of the refractive index of denser medium. (Delhi 2009)
Answer:
Unpolarized light: A light which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarized light.

When unpolarised light is incident on the boundary of two transparent media, the reflected light is polarised with electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other.
Relation between Brewster angle i and refractive index (µ) is :

Question 40.
In Young’s double slit experiment, the two slits 0. 15 mm apart are illuminated by monochromatic light of wavelength 450 nm. The screen is 1.0 m away from the slits.
(a) Find the distance of the second
(i) bright fringe,
(ii) dark fringe from the central maximum.
(b) How will the fringe pattern change if the screen is moved away from the slits? (All India 2009)
Answer:

Question 41.
(a) How does an unpolarised light get polarised when passed through a polaroid?
(b) Two polaroids are set in crossed positions. A third polaroid is placed between the two making an angle 0 with the pass axis of the first polaroid. Write the expression for the intensity of light transmitted from the second polaroid. In what orientations will the transmitted intensity by
(i) minimum and
(ii) maximum? (All India 2009)
Answer:
(a) In a polaroid a long chain of molecules is aligned in a particular direction. The electric vectors (of light waves) along the direction of . aligned molecules gets absorbed.
An unpolarised light wave incident on such a polaroid gets linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules. This direction is called pass-axis of the polaroid.
(b) Not in syllabus.

Question 42.
In Young’s double slit experiment, the two slits 0.12 mm apart are illuminated by monochromatic light of wavelength 420 nm. The screeen is 1.0 m away from the slits.
(a) Find the distance of the second
(i) bright fringe,
(ii) dark fringe from the central maximum.
(b) How will the fringe pattern change if the screen is moved away from the slits? (All India 2009)
Answer:


∴ With increase in the value of D linear width will increase, while the angular width will remain the same.

Question 43.
Describe Young’s double slit experiment to produce interference pattern due to a monochromatic source of light. Deduce the expression for the fringe width. (Delhi 2011)
Answer:
Consider two coherent sources S₁ and S₂ separated by a distance d. Let D be the distance between the screen and the plane of slits S₁ and S₂.
Light waves emitted from S₁ and S₂ reach point O on the screen after travelling equal distances. So path difference and hence phase difference between these waves is zero. Therefore, they meet at O in phase and hence constructive interference

takes place at O. Thus O is the position of the central bright fringe.
Let the waves emitted by S₁ and S₂ meet at point P and the screen at a distance y from the central bright fringe.
The path difference between these waves at P is given by


For constructive interference/maxima:
If path difference is an integral multiple of λ, then bright fringe will be formed at P

…where [m = 1, 2, 3 …
which is the position of mth bright fringe from the central bright fringe.
Fringe width (β) : The distance between any two successive bright fringes (or successive fringes) is called fringe width.

Destructive interferencelMinima: If path difference is odd multiple of \(\lambda / 2\), then dark fringe is formed at P

Which is position of m^th dark fringe from the central bright fringe.
β(fringe width) = y₁ – y₀

Question 44.
Use Huygen’s principle to verify the laws of refraction. (Delhi 2011)
Answer:
(i) Wavefront : Wavefront is defined as the continuous locus of all such particles of the medium which are vibrating in the same phase at any instant.

(ii) We take a plane wavefront AB incident at a plane surface XY. We use secondary wavelets starting at different times. We get refracted wavefront only when the time taken by light to travel along different rays from one wavefront to another is same. We take any arbitrary ray starting from point ‘P’ on incident wavefront to refracted wavefront at point ‘O’. Let total time be ‘t’.

As time should be independent of the ray to be considered
The coefficient of AO in the above equation should be zero

Where ’µ₂ is called refractive index of medium 2 w.r.t. medium 1. This is Snell’s law of refraction.

Question 45.
(a) Describe briefly, with the help of suitable diagram, how the transverse nature of light can be demonstrated by the phenomenon of polarization.
(b) When unpolarized light passes from air to a transparent medium, under what condition does the reflected light get polarized? (Delhi 2011)
Answer:
(a) Light from a source S is allowed to fall normally on the flat surface of a thin plate of a tourmaline crystal, cut parallel to its axis. Only a part of this light is transmitted through A. If now the plate A is rotated, the character of transmitted light remains unchanged. Now another similar plate B is placed at some distance from A such that the axis of B is parallel to that of A. If the light transmitted through A is passed through B, the light is almost completely transmitted through B and no change is observed in the light coming out of B.


If now the crystal A is kept fixed and B is gradually rotated in its own plane, the intensity of light emerging out of B decreases and becomes zero when the axis of B is perpendicular to that of A. If B is further rotated, the intensity begins to increase and becomes maximum when the axes of A and B are again parallel.

Thus, we see that the intensity of light transmitted through B is maximum when axes of A and B are parallel and minimum when they are at right angles.

From this experiment, it is obvious that light waves are transverse and not longitudinal; because, if they were longitudinal, the rotation of crystal B would not produce any change in the intensity of light.

(b) After falling on a transparent medium, unpolarised light will get polarised after reflection only if refracted and reflected rays make a right angle to each other.

Question 46.
(a) Why are coherent sources necessary to produce a sustained interference pattern?
(b) In Young’s double slit experiment using monochromatic light of wavelength A, the intensity of light at a point on the screen where path difference is A, is K units. Find out the intensity of light at a point where path difference is λ/3. (Delhi 2011)
Answer:
(a) Need of coherent sources for the production of interference pattern. When two monochromatic ‘ waves of intensity I₁ I₂ and phase difference ϕ meet at a point, the resultant intensity is given

The last term \(2 \sqrt{I_{1} I_{2}}\) cos ϕ is called interference term.
There are two possibilities :
(i) If cos ϕ remains constant with time, the total intensity at any point will be constant. The intensity will be maximum \(\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}\) at points when cos ϕ is +1 and minimum \(\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}\) at points
where cos ϕ = -1. The sources in this case are coherent.
(ii) If cos ϕ varies continuously with time assuming both positive and negative values, then the average value of cos § will be zero over time interval of measurement. Then interference term averages to zero. There will be same intensity, I = I₁ + I₂ at every point i.e. there will be general illumination on the observation screen. The two sources in two cases are incoherent.

Hence to observe interference, we need to have two sources with same frequency and with a constant phase difference. Such a pair of sources is called coherent source.

Question 47.
Use Huygens’s principle to explain the formation of diffraction pattern due to a single slit
illuminated by a monochromatic source of light. When the width of the slit is made double the
original width, how would this affect the size and intensity of the central diffraction band?
(Delhi 2011)
Answer:
(a) According to Huygen’s principle “The net effect at any point due to a number of wavelets is equal to sum total of contribution of all wavelets with proper phase difference.”

The point O is maxima because contribution from each half of the slit S₁S₂ is in phase, i.e., the path difference is zero.

At point P
(i) If S₂P – S₁P = nλ ⇒ the point P would be minima.
(ii) If S₂P – S₁P = (2n + 1)\(\frac{\lambda}{2} \Rightarrow\) the point would be maxima but with decreasing intensity.

When widths of slits are doubled, contrast between maxima and minima decreases due to the overlapping of interference patterns formed by various narrow pairs of the two slits.

∴ Size of central maxima will be reduced to half and intensity of central maxima will be four times.

Question 48.
(a) Why are coherent sources necessary to produce a sustained interference pattern?
(b) In Young’s double slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3. (Delhi 2011)
Answer:
(a) Coherent sources are needed to ensure that the positions of maxima and minima do not change with time. These have a constant phase difference and, therefore produce sustained interference pattern.

Question 49.
Explain briefly, giving a suitable diagram, how an unpolarised light incident on the interface separating two transparent media gets polarised on reflection. Deduce the necessary condition for it. (Comptt. Delhi 2011)
Answer:
Unpolarized light: A light which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarized light.

When unpolarised light is incident on the boundary of two transparent media, the reflected light is polarised with electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other.
Relation between Brewster angle i and refractive index (µ) is :

Question 50.
(a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment?
(b) Two wavelengths of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction taking place at a single slit of aperture 2 × 10^-4 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases. (Delhi 2011)
Answer:
(a) When there are only a few sources, say two interfering sources, then the result is usually called interference, but if there is a large number of them, it seems that the word diffraction is more often used. In the double slit experiment, we must note that the pattern on the screen is actually a superposition of single slit diffraction from each slit and double slit interference pattern. As a result, there appears a broader diffraction peak in which there occur several fringes of smaller widths due to double slit interference.

Question 51.
(a) Write two characteristic features distinguishing the diffraction pattern from the interference fringes obtained in Young’s double slit experiment.
(b) Two wavelengths of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction taking place due to a single slit of aperture 1 × 10^-4 m. The distance between the slit and the screen is 1.8 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases. (Delhi 2011)
Answer:

Diffraction	Interference
1. Width of principal maxima is twice the width of other fringes.	1. Width of all fringes are the same.
2. Intensity goes on decreasing as order of the diffraction bands increases.	2. All fringes are of same intensity.

Question 52.
(a) What is linearly polarized light ? Describe briefly using a diagram how sunlight is polarised.
(b) Unpolarised light is incident on a polaroid. How would the intensity of transmitted light change when the polaroid is rotated? (All India 2011)
Answer:
(a) Linearly polarised light : Linearly or plane polarised light is that light in which vibrations of electric field vector are taking place only in one particular plane perpendicular to the direction of propagation of light wave. Sun-light is unpolarised light having electric field vector oscillating in all planes, when it passes through a polariser which can be a nicol prism or tourmaline crystal, only those vibrations of light pass through crystal which are parallel to axis of polarizer and hence we get a plane polarised light having vibrations in one plane. Plane polarized light can be observed by using analyzer.

(b) When the polaroid is rotated, the intensity of the transmitted light changes. When the angle of rotation is 45°, the intensity become I₀/2. When the polaroid is further rotated and at an angle of 90°, the intensity is minimum. Thus the intensity of polarised light changes with the angle of rotation of polaroid. It is shown graphically.

Question 53.
In a modified set-up of Young’s double slit experiment, it is given that SS₂ – SS₁ = λ/4, i.e. the source ‘S’ is not equidistant from the slits S₁ and S₂.
(a) Obtain the conditions for constructive and destructive interference at any point P on the screen in terms of the path difference δ = S₂P-S₁P.

(b) Does the observed central bright fringe lie above or below ‘O’? Give reason to support your answerP₃ (Comptt. All India 2011)
Answer:
(a) Conditions for interferance

Question 54.
(a) Using the phenomenon of polarisation, show how transverse nature of light can be demonstrated?
(b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity I₀ is incident on P₁. A third polaroid P₃ is kept in between P₁ and P₂ such that its pass axis makes an angle of 30° with that of P₁. Determine the intensity of light transmitted through P₁, P₂ and P₃. (All India 2011)
Answer:
(a) Light from a source S is allowed to fall normally on the flat surface of a thin plate of a tourmaline crystal, cut parallel to its axis. Only a part of this light is transmitted through A. If now the plate A is rotated, the character of transmitted light remains unchanged. Now another similar plate B is placed at some distance from A such that the axis of B is parallel to that of A. If the light transmitted through A is passed through B, the light is almost completely transmitted through B and no change is observed in the light coming out of B.


If now the crystal A is kept fixed and B is gradually rotated in its own plane, the intensity of light emerging out of B decreases and becomes zero when the axis of B is perpendicular to that of A. If B is further rotated, the intensity begins to increase and becomes maximum when the axes of A and B are again parallel.

Thus, we see that the intensity of light transmitted through B is maximum when axes of A and B are parallel and minimum when they are at right angles.

From this experiment, it is obvious that light waves are transverse and not longitudinal; because, if they were longitudinal, the rotation of crystal B would not produce any change in the intensity of light.

Question 55.
(a) Show, with the help of a diagram, how unpolarised sunlight gets polarised due to scattering.
(b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other.
Unpolarised light of intensity I₀ is incident on P₁. A third polaroid P₃ is kept in between P₁ and P₂ such that its pass axis makes an angle of 45° with that of P₁. Determine the intensity of light transmitted through P₁, P₂ and P₃. (All India 2011)
Answer:
(a) Polarisation by scattering. When a beam of white light is passed through a medium, then the beam gets scattered. When the scattered light is seen in a direction perpendicular to the direction of incidence, it is found to be plane polarised. This phenomenon is called polarisation by scattering

A beam of unpolarised light is incident along Z-axis on a particle. When it is taken along X-axis or Y-axis, then only the vibrations which are parallel to Y-axis or
X-axis, are seen. Hence, light scattered in a direction perpendicular to the incident light is always plane polarised.

Question 56.
(a) Show, giving a suitable diagram, how unpolarised light can be polarised by reflection.
(b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other.
Unpolarised light of intensity I₀ is incident on P₁. A third polaroid P₃ is kept in between P₁ and P₂ such that its pass axis makes an angle of 60° with that of P₁. Determine the intensity of light transmitted through P₁, P₂ and P₃. (All India 2011)
Answer:
(a) Polarisation of light by reflection. The simplest method to produce plane polarised light is by reflection.

When unpolarised light is reflected from a surface, the reflected light may be completely polarised, partially polarised or unpolarised depending on the incident angle.
The angle of incidence at which the reflected light is completely polarised, is called polarising angle. It is represented by i_β.
The value of i_β depends on the wavelength of light used. Therefore, complete polarisation is possible only for monochromatic light. The reflected light along OB is completely plane polarised. The light refracted along OC is unpolarised.
Hence, the reflected light is completely plane polarised in the plane of incidence.

Question 57.
A parallel beam of monochromatic light falls normally on a narrow slit of width ‘a’ to produce a diffraction pattern on the screen placed parallel to the plane of the slit. Use Huygens’ principle to explain that
(i) the central bright maxima is twice as wide as the other maxima.
(ii) the intensity falls as we move to successive maxima away from the centre on either side.
(Comptt. Delhi 2011)
Answer:
(i) In diffraction pattern, intensity will be minimum at an angle θ = nλ/a
∴ There will be a first minimum at an angle
θ = λ/a, on either side of central maximum
∴ Width of central maxima = 2λ/a
∴ The central bright maxima is twice as wide as the other maxima.

(ii) The intensity of maxima decreases as the order (n) or diffraction maxima increases. This is because, on dividing the slit into odd number of parts, the contributions of the corresponding (outermost) pairs cancel each other, leaving behind the contribution of only the innermost segment.

For example, for first maximum, dividing slit into three parts out of these three parts of the slit, the contributions from first two parts cancel each other; only 1/3^rd portion of the slit contributes to the maxima of intensity. Similarly for second maxima, dividing slit into five parts, contribution of first four parts will be zero (as they cancel each other). The remaining 1/5^th portion only will contribute for maxima and so on.

Question 58.
(a) Unpolarised light of intensity I₀ passes through two polaroids P₀ and P₂ such that pass axis of P₂ makes an angle θ with the pass axis of P₁ Plot a graph showing the variation of intensity of light transmitted through P₂ as the angle θ varies from zero to 180°.
(b) A third polaroid P₃ is placed between P₁ and P₂ with pass axis of P₃ making an angle β with that of P₁. If I₁ I₂ and I₃ represent the intensities of light transmitted by P₁, P₂ and P₃, determine the values of angle θ and β for which I₁ = I₂ = I₃. (Comptt. All India 2011)
Answer:

Question 59.
(a) Two monochromatic waves emanating from two coherent sources have the displacements represented by
y₁ = a cos ωt and y₂ = a cos (ωt + ϕ ), where ϕ is the phase difference between the two displacements. Show that the resultant intensity at a point due to their superposition is given by I = 4I₀ cos² ϕ/2, where I₀ = a₂.
(b) Hence obtain the conditions for constructive and destructive interference. (Comptt. All India 2011)
Answer:
(a) (i) Two independent monochromatic
sources of light cannot produce a sustained interference pattern. The phase difference between these two sources will continuously vary; and the positions of maxima and minima will change with time.

(b) (i) For constructive interference :

(ii) For destructive interference:

Question 60.
Answer the following questions:
(i) In a double slit experiment using light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.
(ii) Light of wavelength 5000 A propagating in air gets partly reflected from the surface of water. How will the wavelengths and frequencies of the reflected and refracted light be affected? (Delhi 2015)
Answer:

(ii) For reflected light : Wavelength remains same and frequency remains the same.
For refracted light : Wavelength decreases, but frequency remains the same.

Question 61.
State clearly how an unpolarised light gets linearly polarised when passed through a Polaroid.
(i) Unpolarised light of intensity I₀ is incident on a polaroid P₁ which is kept near another Polaroid P₂ whose pass axis is parallel to that of P₁. How will the intensities of light, I₁ and I₂, transmitted by the polaroids P₁ and P₂ respectively, change on rotating Px without disturbing P₂?
(ii) Write the relation between the intensities I₁ and I₂. (All India 2015)
Answer:
When unpolarised light is passed through a polaroid, the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules.

I₁ remains unchanged on rotating Pj; while, according to Malus’ law,

when P₁ is rotated.

Question 62.
Use Huygens’ principle to show how a plane wavefront propagates from a denser to rarer medium. Hence verify Snell’s law of refraction. (All India 2015)
Answer:
Huygens’ geometrical construction for a plane wave propagation. Let AB be a section of primary wavefront at any instant t. Take points 1, 2, 3, 4, … on the wavefront AB. Taking each point as centre, draw spheres of radius r = ct, where c is the velocity of light in the medium.
Draw a surface A₁B₁ touching tangentially at the secondary wavelets in the forward direction. The surface A₁B₁ is the secondary wavefront after time t.

A surface A₂B₂ touching tangentially all the secondary wavelets in the backward direction can be drawn to give a backward secondary wavefront.
Laws of refraction using Huygen’s principle. Let XY be the refracting surface separating two mediums 1^st and 2^nd. Let c₁ and c₂ be the velocities of light in these mediums.

Let AB be a plane wavefront moving through the surface XY meeting at the point B.
The time taken ‘t’ of A to reach at A’B then

To obtain new front, draw a circle with point B as centre and BD as radius in 2nd medium. Draw a tangent A’D from point A’. Then A’D represents the refracted wavefront.
Since PB be incident ray and BD be refracted ray

This constant V is called the refractive index of material which proves Snell’s law of refraction.

Question 63.
Explain by drawing a suitable diagram that the interference pattern in a double slit is actually a superposition of single slit diffraction from each slit.
Write two basic features which distinguish the interference pattern from those seen in a coherently illuminated single slit. (Comptt. Delhi 2015)
Answer:
Interference pattern and Diffraction pattern :
The diagram, given here, shows several fringes, due to double slit interference, ‘contained’ in a broad diffraction peak. When the separation between the slits is large compared to their width, the diffraction pattern becomes very flat and we observe the two slit interference pattern.

Basic features of distinction between interference and diffraction patterns :
(i) The interference pattern has a number of equally spaced bright and dark bands while diffraction pattern has a central bright maximum which is twice as wide as the other maxima.
(ii) Interference pattern is the superimposition of two waves slits originating from two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.
(iii) For a single slit of width ‘a’ the first null of diffraction pattern occurs at an angle of \(\frac{\lambda}{a}\). At the same angle of \(\frac{\lambda}{a}\), we get a maxima for two narrow slits separated by a distance ‘a’.

Question 64.
(a) The ratio of the widths of two slits in Young’s double slit experiment is 4 : 1. Evaluate the ratio of intensities at maxima and minima in the interference pattern.
(b) Does the appearance of bright and dark fringes in the interference pattern violate, in any way, conservation of energy? (Comptt. All India 2015)
Answer:

(b) The appearance of bright and dark fringes in the interference pattern does not violate the principle of conservation of energy, because the light energy is distributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy.

Question 65.
(a) Good quality sun-glasses made of polaroids are preferred over ordinary coloured glasses. Justify your answer.
(b) Two polaroids P₁ and P₂ are placed in crossed positions. A third polaroid P₃ is kept between P₃ and P₂ such that pass axis of P₃ is parallel to that of P₁. How would the intensity of light (I₂) transmitted through P₂ vary as P₃ is rotated? Draw a plot of intensity I₂‘ vs the angle ‘θ’ between pass axes of P₁ and P₃. (Comptt. All India 2015)
Answer:
(a) Such good quality sun-glasses made of polaroids are preferred ’ over ordinary coloured glasses because they protect the eyes from the glare.
(b)
Let the rotating Polaroid sheet make an angle θ with the first polaroid.
∴ angle with the other polaroid will be (90 – θ)

Question 66.
(i) State law of Malus.
(ii) Draw a graph showing the variation of intensity (I) of polarised light transmitted by an analyser with angle (θ) between polariser and analyser.
(iii) What is the value of refractive index of a medium of polarising angle 60°?(All India 2015) Answer:
(i) Law of Malus : When the pass axis of a poloroid makes an angle 0 with the plane of polarisation of polorised light of intensity I0 incident on it, then the intensity of the transmitted emergent light is given by
I = I₀ cos²θ
(ii) Variation of intensity with θ

Question 67.
Define the term wave front. State Huygen’s principle.
Consider a plane wave front incident on a thin convex lens. Draw a proper diagram to show how the incident wave front traverses through the lens and after refraction focusses on the focal point of the lens, giving the shape of the emergent wave front. (All India 2015)
Answer:
(i) Wave front : It is defined as the locus of all points which oscillate in phase.
(ii) Huygen’s Principle :
1 Each point of the wave front is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions. These travel with the same velocity as that of the original wave front.
2. The shape and position of the wave front, after time Y, is given by the tangential envelope to the secondary wavelets.
(iii) Refraction of a plane wave-front from a convex lens

Question 68.
The figure, drawn here, shows the geometry of path differences for diffraction by a single slit of width a.
Give appropriate ‘reasoning’ to explain why the intensity of light is

(i) maximum of the central point C on the screen.
(ii) (nearly) zero for point P on the screen when

Hence write an expression for the total linear width of the central maxima on a screen kept at a distance D from the plane of the slit. (Comptt. Delhi 2015)
Answer:
(i) At central point C, angle θ is zero, all path differences are zero. Hence, all the parts of the slit contribute in same phase. This gives maximum intensity at point C.
(ii) From the diagram :

It implies that the contribution from corresponding points in two halves of the slit have a phase difference of π. Therefore, contributions from two halves cancel each other in pairs, resulting in a zero net intensity at point P on the screen.

Question 69.
Two polaroids, P₁ and P₂, are ‘set-up’ so that their ‘pass-axis’ are ‘crossed’ with respect to each other. A third polaroid, P₃, is now introduced between these two so that its ‘pass-axis’ makes an angle θ with the ‘pass-axis’ of P₁.
A beam of unpolarised light, of intensity I, is incident on P₁ If the intensity of light, that gets transmitted through this combination of three polaroids, is I’, find the ratio \(\left(\frac{\mathbf{I}^{\prime}}{\mathbf{I}}\right)\) when θ equals :
(i) 30°,
(ii) 45° (Comptt. Delhi 2015)
Answer:
A beam of unpolarised light of intensity I is incident on P₁.

Question 70.
A plane wavefront is incident at an angle of incidence i on a reflecting surface. Draw a diagram showing incident wavefront, reflected wavefront and verify the laws of reflection. (Comptt. Delhi 2015)
Answer:
Diagram of a plane wave front for Reflection :

Question 71.
(a) In a Young’s double slit experiment, the two slits are illuminated by two different lamps having same wavelength of light. Explain with reason, whether interference pattern will be observed on the screen or not.
(b) Light waves from two coherent sources arrive at two points on a screen with path differences of 0 and A/2. Find the ratio of intensities at the points. (Comptt. All India 2015)
Answer:
(a) Interference pattern in Young’s double slit experiment will not be observed, because two independent lamps do not constitute ‘coherent sources’.
(b) (i) When path difference is zero, corresponding phase difference will also be zero.

Question 72.
(a) Explain how the intensity of diffraction pattern changes as the order (n) of the diffraction band varies.
(b) Two wavelengths of sodium light 590 nm and 596 nm are used in turn to study the diffraction at a single slit of size 4 mm. The distance between the slit and screen is 2 m. Calculate the separation between the positions of the first maximum of the diffraction pattern obtained in the two cases. (Comptt. All India 2015)
Answer:
(a) Intensity of diffraction pattern drops rapidly with order n, because every higher order
maxima gets intensity only from \(\left(\frac{1}{2 n+1}\right)\) part of the slit. The central maxima gets intensity from the whole slit (n = 0).
1^st secondary maxima gets its intensity only from 1/3 of slit.
2^nd secondary maxima gets its intensity only from 1/5 of slit and so on.

Question 73.
Why are coherent sources necessary to produce interference in Young’s double slit experiment? Light waves from two coherent sources have intensities in the ratio of 4 : 9. Find the ratio of intensities of maxima and minima in the interference pattern. (Comptt. Delhi 2015)
Answer:
If sources are not coherent, the superposition pattern (intensity pattern) is not stable. It keeps on changing with time and hence it is necessary to have coherent sources to observe interference in Young’s double slit experiment.

Question 74.
State the two features to distinguish between interference and diffraction phenomena. Two wavelengths of light 600 nm and 610 nm are used in turn, to study the diffraction at a single slit of size 2 mm. The distance between the slits and screen is 2 m. Calculate the separation between the positions of the second order maximum of the diffraction pattern obtained in the two cases. (Comptt. All India 2015)
Answer:
For distinction between interference and diffraction :
Interference pattern and Diffraction pattern :
The diagram, given here, shows several fringes, due to double slit interference, ‘contained’ in a broad diffraction peak. When the separation
between the slits is large compared to their width, the diffraction pattern becomes very flat and we observe the two slit interference pattern.

Basic features of distinction between interference and diffraction patterns :
(i) The interference pattern has a number of equally spaced bright and dark bands while diffraction pattern has a central bright maximum which is twice as wide as the other maxima.

(ii) Interference pattern is the superimposition of two waves slits originating from two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.

(iii) For a single slit of width ‘a’ the first null of diffraction pattern occurs at an angle of \(\frac{\lambda}{a}\). At the same angle of \(\frac{\lambda}{a}\), we get a maxima for two narrow slits separated by a distance ‘a’.
Given : λ₁ = 600 nm = 600 × 10^-9 m,

Question 75.
(a) Good quality sunglasses made of polaroids are preferred over ordinary coloured glasses. Explain why.
(CBSE AI2019)
Answer:
Polaroid sunglasses are preferred over coloured sunglasses because they reduce the intensity of light.

(b) How is plane polarised light defined?
Answer:
Plane polarised light: Light in which vibrations of electric field vector are restricted to one plane containing the direction of propagation.

(c) A beam of plane polarised light is passed through a polaroid. Show graphically, a variation of the intensity of the transmitted light with the angle of rotation of the Polaroid.
Answer:
The graphical variation is as shown.

Long Answer Type

Question 1.
Define the term wavefront. Using Huygen’s wave theory, verify the law of reflection.
Or
Define the term, “refractive index” of a medium. Verify Snell’s law of refraction when a plane wavefront is propagating from a denser to a rarer medium. (CBSE Delhi 2019)
Answer:
The wavefront is a locus of points that oscillate in the same phase.

Consider a plane wavefront AB incident obliquely on a plane reflecting surface MM^–. Let us consider the situation when one end A of was front strikes the mirror at an angle i but the other end B has still to cover distance BC. The time required for this will be t = BC/c.

According to Huygen’s principle, point A starts emitting secondary wavelets and in time t, these will cover a distance c t = BC and spread. Hence, with point A as centre and BC as radius, draw a circular arc. Draw tangent CD on this arc from point C. Obviously, the CD is the reflected wavefront inclined at an angle ‘r’. As incident wavefront and reflected wavefront, both are in the plane of the paper, the 1^st law of reflection is proved.

To prove the second law of reflection, consider ΔABC and ΔADC. BC = AD (by construction),
∠ABC = ∠ADC = 90° and AC is common.

Therefore, the two triangles are congruent and, hence, ∠BAC = ∠DCA or ∠i = ∠r, i.e.the angle of reflection is equal to the angle of incidence, which is the second law of reflection.
Or
The refractive index of medium 2, w.r.t. medium 1 equals the ratio of the sine of the angle of incidence (in medium 1) to the sine of the angle of refraction (in medium 2), The diagram is as shown.

From the diagram

Question 2.
(a) Sketch the refracted wavefront for the incident plane wavefront of the light from a distant object passing through a convex lens.
Answer:

(b) Using Huygens’s principle, verify the laws of refraction when light from a denser medium is incident on a rarer medium.
Answer:
Refraction from denser to the rarer medium: Let XY be plane refracting surface separating two media of refractive index μ₁ and μ₂ (μ₁ > μ₂)

Let a plane wavefront AB incident at an angle i. According to Huygen’s principle, each point on the wavefront becomes a source of secondary wavelets and

Time is taken by wavelets from B to C = Time taken by wavelets from A to D


(c) For yellow light of wavelength 590 nm incident on a glass slab, the refractive index of glass Is 1.5. Estimate the speed and wavelength of yellow light Inside the glass slab. (CBSE 2019C)
Answer:
Given λ = 590 nm, μ = 1.5
Velocity of light inside glass slab.
∴ v = \(\frac{C}{\mu}=\frac{3 \times 10^{8}}{1.5}\) = 2 × 10⁸ ms^-1

Wavelength of yellow light inside the glass slab.
λ₁ = \(\frac{\lambda}{\mu}=\frac{290}{1.5}\) = 393.33 nm

Question 3.
(a) State the postulates of Huygen’s wave theory.
Answer:
The postulates are
All points on a given wavefront are taken as point sources for the production of spherical secondary waves, called wavelets, which propagate outward with speed characteristic of waves in that medium.

(b) Draw the type of wavefront that corresponds to a beam of light (i) coming from a very far off the source and (ii) diverging from a point source.
Answer:
After some time has elapsed, the new position of the wavefront is the surface tangent to the wavelets or the envelope of the wavelets in the forward direction.


Question 4.
What is meant by the diffraction of light? Obtain an expression for the first minimum of diffraction.
Answer:
The divergence of light from its initial line of travel when it passes through an opening or an obstacle is called diffraction or the phenomenon of bending of light around the sharp corners and spreading into the regions of the geometrical shadow is called diffraction.

Consider that a monochromatic source of light S, emitting light waves of wavelength λ, is placed at the principal focus of the convex lens L₁. A parallel beam of light, i.e. a plane wavefront, gets incident on a narrow slit AB of width ‘a’ as shown in the figure.

The diffraction pattern is obtained on a screen Lying at a distance D from the slit and at the focal plane of the convex lens L₂.

Consider a point P on the screen at which wavelets travelLing in a direction making angle O with CO are brought to focus by the lens. The wavelets from different parts of the slit do not reach point P in phase, although they are initially in phase. It is because they cover unequal distances in reaching point R The waveLets from points A and B will have a path difference equal to BN.

From the right angLed ANB, we have
BN = AB sin θ or BN = a sin θ …(1)

Suppose that the point P on the screen is at such a distance from the centre of the screen that BN = λ. and the angle θ = θ₁.

Then, equation 1 gives
λ = a sin θ₁ or sin θ₁ = \(\frac{λ}{a}\)

Such a point one screen will be the position of the first secondary minimum.

Question 6.
(a) What is plane polarised light? Two polaroids are placed at 90° to each other and the transmitted intensity is zero. What happens when one more polaroid is placed between these two, bisecting the angle between them? How will the intensity of transmitted light vary on further rotating the third polaroid?
(b) If a light beam shows no intensity variation when transmitted through a polaroid which is rotated, does it mean that the light is unpolarised? Explain briefly. (Delhi 2015)
Answer:
(a) Plane polarised light is the light wave in which optical vector (electric vector) oscillates only in a single plane in a direction perpendicular to the direction of propagation of wave.
Obviously, the axis of P₃ is inclined at 45° to the axis of P₁ and P₂.
Let the maximum intensity transmitted by polarised,

Thus the final intensity becomes one-fourth of the maximum intensity transmitted by the first polariod. On further rotating the third polaroid, the intensity of transmitted light decreases till it becomes zero.

(b) Yes, the light beam is unpolarised. The polaroid cuts down the intensity to half due to polarisation of light. When polaroid is rotated, only the plane of polarisation rotates but the intensity of light does not show any change.

Question 7.
(a) What are coherent sources of light? Two slits in Young’s double slit experiment are illuminated by two different sodium lamps emitting light of the same wavelength. Why is no interference pattern observed?
(b) Obtain the condition for getting dark and bright fringes in Young’s experiment. Hence write the expression for the fringe width.
(c) If s is the size of the source and its distance is a from the plane of the two slits, what should be the criterion for the interference fringes to be seen? (All India 2015)
Answer:
(a) Two sources of light having same frequency and a constant or a zero phase difference are said to be coherent.
Light wave emitted from an ordinary source (like a sodium lamp) undergoes abrupt phase changes in times of the order of 10^-10 seconds. Thus two independent sources of light will not have a fixed phase relationship and would be incoherent.

Question 8.
State Huygens’s principle. Show, with the help of a suitable diagram, how this principle is used to obtain the diffraction pattern by a single slit. Draw a plot of intensity distribution and explain clearly why the secondary maxima become weaker with increasing order (n) of the secondary maxima. (Delhi 2010)
Answer:
Huygen’s principle :
“(i) Each point on a wavefront acts as a fresh source of new disturbance called secondary waves or wavelets;
(ii) The secondary wavelets spread out in all directions with the speed of wave in the given medium;
(iii) The new wavefront at any later time is given by the forward envelope (tangential surface in the forward direction) of the secondary wavelets at that time.

The reason is that the intensity of the central maxima is due to the constructive interference of wavelets from all parts of the slit, the first secondary maxima is due to the contribution of wavelets from one third part of the slit (wavelets from remaining two parts interfere destructively), the second secondary maxima is due to the contribution of wavelets from the one fifth part only (the remaining four parts interfere destructively) and so on. Hence, the intensity’ of secondary maxima decreases with the increase in the order n of the maxima.

Question 9.
Explain briefly how the phenomenon of total internal reflection is used in fibre optics. (Delhi 2011)
Answer:
Each optical fibre consists of a core and cladding. Refractive index of the material of the core is higher than that of cladding. When a video signal is directed into an optical fibre at a suitable angle, it undergoes internal reflections repeatedly along the length of the optical fibre and comes out of it with almost neglible loss of intensity.

Question 10.
(i) State the importance of coherent sources in the phenomenon of interference.
(ii) In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence deduce the expression for the fringe width.
(iii) How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water? (All India 2010)
Answer:
(i) Importance of coherent source : Coherent sources are necessary to produce sustained interference pattern. Otherwise the phase difference between the two interfering waves will change rapidly and the interference pattern will be lost. .
(ii)
(a) For constructive interference : We will have constructive interference resulting in a bright fringe when path difference is equal to nλ.

Since the separation between the centres of two consecutive bright fringes is called fringe width. It is denoted by β

Hence all bright and dark fringes are of equal width.
Observations :
(i) Fringe width is directly proportional to the wavelength of light i.e. p X.
(ii) Fringe width is inversely proportional to the distance between two sources

(iii) Fringe width is directly proportional to the distance between screen and two sources i.e. \(\beta \propto \frac{1}{d}\).

Suppose at any distance x from the central maximum,

The bright fringes will coincide at the least distance x,


Then fringe width becomes \(\frac{\mathbf{1}}{\mu}\) times the original fringe width i.e., it will decrease in water.

Question 11.
(a) State Huygen’s principle. Using this principle explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a narrow beam coming from a monochromatic source of light is incident normally.
(b) Show that the angular width of the first diffraction fringe is half of that of the central fringe.
(c) If a monochromatic source of light is replaced by white light, what change would you observe in the diffraction pattern? (All India 2011)
Answer:
(a)
Huygen’s principle :
“(i) Each point on a wavefront acts as a fresh source of new disturbance called secondary waves or wavelets;
(ii) The secondary wavelets spread out in all directions with the speed of wave in the given medium;
(iii) The new wavefront at any later time is given by the forward envelope (tangential surface in the forward direction) of the secondary wavelets at that time.

The reason is that the intensity of the central maxima is due to the constructive interference of wavelets from all parts of the slit, the first secondary maxima is due to the contribution of wavelets from one third part of the slit (wavelets from remaining two parts interfere destructively), the second secondary maxima is due to the contribution of wavelets from the one fifth part only (the remaining four parts interfere destructively) and so on. Hence, the intensity’ of secondary maxima decreases with the increase in the order n of the maxima.

(c) If monochromatic source of light is replaced by a source of white light, instead of white fringes we obtain few coloured fringes and then uniform illumination.

Question 12.
(a) In Young’s double slit experiment, derive the condition for
(i) constructive interference and
(ii) destructive interference at a point on the screen.
(b) A beam of light consisting of two wavelenths, 800 nm and 600 nm is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide. (All India 2011)
Answer:
(a) For constructive interference : We will have constructive interference resulting in a bright fringe when path difference is equal to nλ.

Since the separation between the centres of two consecutive bright fringes is called fringe width. It is denoted by β

Hence all bright and dark fringes are of equal width.
Observations :
(i) Fringe width is directly proportional to the wavelength of light i.e. p X.
(ii) Fringe width is inversely proportional to the distance between two sources

(iii) Fringe width is directly proportional to the distance between screen and two sources i.e. \(\beta \propto \frac{1}{d}\).

Suppose at any distance x from the central maximum,

The bright fringes will coincide at the least distance x,

Question 13.
(a) How does an unpolarized light incident on a polaroid get polarized?
Describe briefly, with the help of a necessary diagram, the polarization of light by reflection from a transparent medium.
(b) Two polaroids ‘A’ and ‘B’ are kept in crossed position. How should a third polaroid ‘C’ be placed between them so that the intensity of polarized light transmitted by polaroid B reduces to 1/8* of the intensity of unpolarized ‘ light incident on A? (All India 2010)
Answer:
(a) In a polaroid a long chain of molecules is aligned in a particular direction. The electric vectors (of light waves) along the direction of . aligned molecules gets absorbed.

Diagram of polarization of light by reflection.

Polarisation of light by reflection. The simplest method to produce plane polarised light is by reflection.

When unpolarised light is reflected from a surface, the reflected light may be completely polarised, partially polarised or unpolarised depending on the incident angle.
The angle of incidence at which the reflected light is completely polarised, is called polarising angle. It is represented by i_β.
The value of i_β depends on the wavelength of light used. Therefore, complete polarisation is possible only for monochromatic light. The reflected light along OB is completely plane polarised. The light refracted along OC is unpolarised.
Hence, the reflected light is completely plane polarised in the plane of incidence.(b) Let the intensity of unpolarized light = I₀

Hence, the angle between polaroid A and C should be 45°.

Question 14.
(a) State Huygen’s principle. Using this principle draw a diagram to show how a plane wave front incident at the interface of the two media gets refracted when it propagates from a rarer to a denser medium. Hence verify Snell’s law of refraction.
(b) When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons :
(i) Is the frequency of reflected and refracted light same as the frequency of incident light?
(ii) Does the decrease in speed imply a reduction in the energy carried by light wave? (Delhi 2010)
Answer:
(a) Huygen’s principle ; Huygen’s principle is based on two assumptions :
(i) Each point on the primary wavefront is a source of a new disturbance called secondary wavelets which travel in all directions with same velocity as that of original waves.
(ii) A surface tangential to the secondary . wavelets gives the position and shape of new wavefront at any instant. This is called secondary wavefront.

The principle leads to the well known laws of reflection and refraction.
Verification of Snell’s Law, From the figure

(b) (i) Yes, frequency is the property of source. Hence, frequency does not change when light is reflected or refracted.
(ii) No, decrease in speed does not imply reduction in energy carried by light wave.

This is because the frequency does not change and according to the formula E = hv, energy will be independent of speed. Energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.

Question 15.
(a) Describe briefly how an unpolarized light gets linearly polarized when it passes through a polaroid.
(b) Three identical polaroid sheets P3, P2 and P3 are oriented so that the pass axis of P₁, P₂ and P₃ are inclined at angles of 60° and 90° respectively with respect to the pass axis of P₁. A monochromatic source S of unpolarized light of intensity I₀ is kept in front of the polaroid sheet P₁ as shown in the figure. Determine the intensities of light as observed by the observers O₁ O₂ and O₃ as shown. (Comptt. Delhi 2010)

Answer:
(a) A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules.

Question 16.
(a) Use Huygen’s geometrical construction to show how a plane wave-front at t = 0 propagates and produces a wave-front at a later time.
(b) Verify, using Huygen’s principle, Snell’s law of refraction of a plane wave propagating from a denser to a rarer medium.
(c) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency. Explain why. (Comptt. Delhi)
Answer:
(a) Huygens’ geometrical construction for a plane wave propagation.
Huygens’ geometrical construction for a plane wave propagation. Let AB be a section of primary wavefront at any instant t. Take points 1, 2, 3, 4, … on the wavefront AB. Taking each point as centre, draw spheres of radius r = ct, where c is the velocity of light in the medium.
Draw a surface A₁B₁ touching tangentially at the secondary wavelets in the forward direction. The surface A₁B₁ is the secondary wavefront after time t.

A surface A₂B₂ touching tangentially all the secondary wavelets in the backward direction can be drawn to give a backward secondary wavefront.
Laws of refraction using Huygen’s principle. Let XY be the refracting surface separating two mediums 1^st and 2^nd. Let c₁ and c₂ be the velocities of light in these mediums.

Let AB be a plane wavefront moving through the surface XY meeting at the point B.
The time taken ‘t’ of A to reach at A’B then

To obtain new front, draw a circle with point B as centre and BD as radius in 2nd medium. Draw a tangent A’D from point A’. Then A’D represents the refracted wavefront.
Since PB be incident ray and BD be refracted ray

This constant V is called the refractive index of material which proves Snell’s law of refraction.

(b) Laws of refraction using Huygen’s principle.
Huygens’ geometrical construction for a plane wave propagation. Let AB be a section of primary wavefront at any instant t. Take points 1, 2, 3, 4, … on the wavefront AB. Taking each point as centre, draw spheres of radius r = ct, where c is the velocity of light in the medium.
Draw a surface A₁B₁ touching tangentially at the secondary wavelets in the forward direction. The surface A₁B₁ is the secondary wavefront after time t.

A surface A₂B₂ touching tangentially all the secondary wavelets in the backward direction can be drawn to give a backward secondary wavefront.
Laws of refraction using Huygen’s principle. Let XY be the refracting surface separating two mediums 1^st and 2^nd. Let c₁ and c₂ be the velocities of light in these mediums.

Let AB be a plane wavefront moving through the surface XY meeting at the point B.
The time taken ‘t’ of A to reach at A’B then

To obtain new front, draw a circle with point B as centre and BD as radius in 2nd medium. Draw a tangent A’D from point A’. Then A’D represents the refracted wavefront.
Since PB be incident ray and BD be refracted ray

This constant V is called the refractive index of material which proves Snell’s law of refraction.

(c) When monochromatic light is incident on a surface separating two media, the redacted and refracted lights both have the same frequency.
In case of refraction, due to change in medium, only wavelength changes by λ/µ factor.
In case of reflection, both wavelength and frequency remain unchanged.

Question 17.
(a) (i) “Two independent monochromatic sources of light cannot produce a sustained interference pattern”. Give reason.
(ii) Light waves each of amplitude “a” and frequency “ω”, emanating from two coherent light sources superpose at a point. If the displacements due to these waves is given by y₁ = a cos ωt and y₂ = a cos(ωt + ϕ ) where ϕ is the phase difference between the two, obtain the expression for the resultant intensity at the point.
(b) In Young’s double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3. (Delhi 2010)
Answer:
(a) (i) Two independent monochromatic
sources of light cannot produce a sustained interference pattern. The phase difference between these two sources will continuously vary; and the positions of maxima and minima will change with time.

Question 18.
(a) How does one demonstrate, using a suitable diagram, that unpolarised light when passed through a Polaroid gets polarised?
(b) A beam of unpolarised light is incident on a glass-air interface. Show, using a suitable ray diagram, that light reflected from the interface is totally polarised, when μ = tan i_B, where μ is the refractive index of glass with respect to air and i_B is the Brewster’s angle. (Delhi 2014)
Answer:
(a) Polarisation of light. Ordinary light is unpolarised light having vibrations distributed in all directions in a plane perpendicular to direction of propagation of light.
Therefore, the phenomenon of restricting the vibrations of light in a particular direction is called polarisation of light. The tourmaline crystal acts as a polariser.

When unpolarised light (ordinary light) is passed through a tourmaline crystal, only those vibrations of light pass through the crystal which are parallel to crystallographic axis AB. All other vibrations are obstructed. The entering light from the crystal is said to be plane polarised light.

(b) When unpolarised light is reflected from a surface, the reflected light may be completely polarised, partially polarised or unpolarised depending on the incident angle.

The angle of incidence at which the reflected light is completely polarised, is called polarising angle. It is represented by i_β.
The value of i_β depends on the wavelength of light used. Therefore, complete polarisation is possible only for monochromatic light. The reflected light along OB is completely plane polarised. The light refracted along OC is unpolarised.
Hence, the reflected light is completely plane polarised in the plane of incidence. Brewster’s law. It states that “when light is incident at polarising angle, the reflected and refracted rays are perpendicular to each other”. When light is incident at polarising angle, the reflected component along OB and refracted components along OC are mutually perpendicular.

Hence, the tangent of the polarising angle is equal to the refractive index of a medium

Question 19.
(a) In Young’s double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.
(b) The ratio of the intensities at minima to the maxima in the Young’s double slit experiment is 9 : 25. Find the ratio of the widths of the two slits. (All India 2014)
Answer:
(a) Young’s double slit experiment. The path difference between two rays coming from holes S₁ and S₂ is (S₂P – S₁P). If point P corresponds to a maximum,

If x, d << D, then negligible error will be introduced if (S₂P + S₁P) in the denominator is replaced by 2D.

Thus, bright and dark bands appear on the screen. Such bands are called ‘fringes’.
These dark and bright fringes are equally spaced.

Question 20.
(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence obtain the conditions for the angular width of secondary maxima and secondary minima.
(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 × 10^-6 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases. (All India 2014)
Answer:
(a) Diffraction at a single slit. Suppose a parallel beam of monochromatic light of wavelength X falls normally on a slit AB of width d (of the order of the wavelength of light). The diffraction occurs on passing through the slit. The diffraction pattern is focussed on to the screen by a convex lens. The diffraction pattern consists of a central bright fringe (or band), having alternate dark and bright fringes of decreasing intensity on both sides.
1. Position of central maximum. Let C be the centre of the slit AB. According to Huygen’s principle, “when.light falls on the slit, it becomes a source of secondary wavelets.” All the wavelets originating from slit AB are in same phase. These secondary waves reinforce each other resulting the central maximum intensity at O.

2. Position of secondary maxima and minima. Consider a point P on the screen. All the secondary waves travelling in a direction making angle θ with CO, reach at a point P. The intensity at P depends on the path difference between secondary waves.
∴ Path difference between the secondary waves reaching P from points A and B is, BN = d sin θ
(i) The point P will be the position of n^th secondary maxima

(ii) The point P will be the position of n^th secondary minima,

Hence, the diffraction pattern due to single slit consists of a centre bright maximum at O alongwith secondary maxima and minima on either side. Intensity distribution curve. The intensity distribution on the screen.is represented as shown in the figure.

Width of central maximum. If D is the distance between slit and screen and yn is the distance of nth minimum from point O,

Question 21.
(a) Distinguish between linearly polarized and unpolarized light.
(b) Show that the light waves are transverse in nature.
(c) Why does light from a clear blue portion of the sky show a rise and fall of intensity when viewed through a polaroid which is rotated? Explain by drawing the necessary diagram. (Comptt. Delhi 2014)
Answer:
(a) Unpolarized light : A light wave, in which the electric vector oscillates in all possible directions in a plane perpendicular to the direction of propagation is known as unpolarized light.
Linearly polarized light : If the oscillations of the electric vectors are restricted to just one direction, in a plane perpendicular to the direction of propagation, the corresponding light is known as linearly polarized light.

Unpolarized light passing through polaroid P₁ gets linearly polarized.
[As the electric field vector components parallel to the pass axis of P₁ are transmitted whereas the others are blocked.]
When this polarized light is incident on a polaroid P₂, kept crossed with respect to P₁, then these components also get blocked and no light is transmitted beyond P₂ .

(c) It is due to scattering of light by molecules of earth’s atmosphere.

Under the influence of the electric field of the incident (unpolarized) wave, the electrons in the molecules acquire components of motion in both these directions. Charges, accelerating parallel to the double arrows, do not radiate energy towards the observer since their acceleration has no transverse component.
The radiation scattered by the molecules is therefore represented by dots, i.e., it is polarized perpendicular to plane of figure.

Question 22.
(a) Using Huygen’s construction of secondary wavelets explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a monochromatic beam of light is incident normally.
(b) Show that the angular width of the first diffraction fringe is half that of the central fringe.
(c) Explain why the maxima at \(\theta=\left(n+\frac{1}{2}\right) \frac{\lambda}{a}\) become weaker and weaker with increasing n. (Delhi 2014)
Answer:
(a) Diffraction Pattern : We can regard the total contribution of the wavefront LN at some point P on the screen, as the resultant effect of the superposition of its wavelets like LM, MM₂, M₂N. These have to be superposed taking into account their proper phase differences. We, therefore, get maxima and minima, i.e. a diffraction pattern, on the screen.

Hence angular width of first fringe is half of the angular width of central fringe.
(c) Maxima become weaker and weaker with increasing n. This is because the effective part of the wavefront, contributing to the maxima, becomes smaller and smaller, with increasing n.

Question 23.
(a) A point object ‘O’ is kept in a medium of refractive index n₁ in front of a convex spherical surface of radius of curvature R which separates the second medium of refractive index n₂ from the first one, as show in the figure.

Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of n₁, n₂ and R.
(b) When the image formed above acts as a virtual object for a concave spherical surface separating the medium n₂ from n₂ (n₂ > n₁), draw this ray diagram and write the similar [similar to (a)] relation. Hence obtain the expression for the lens maker’s formula. (Delhi 2014)
Answer:


This equation gives us a relation between object and image distance in terms of refractive index of the medium and the radius of the curvature of the curved spherical surface. It holds for any curved spherical surface.

(b) Lens maker’s formula : Consider a thin double convex lens of refractive index n2 placed in a medium of refractive index n₂ Here n₁ < n₂. Let B and D be the poles, C₁ and C₂ be the centres of curvature and R₁ and R₂ be the radii of curvature of the two lens surfaces ABC and ADC, respectively. For refraction at surface ABC, we can write the relation between the object distance v₁, image distance v and radius of curvature R₁ as

Question 24.
(a) Consider two coherent sources S₁ and S₂ producing monochromatic waves to produce interference pattern. Let the displacement of the wave produced by S₁ be given by
Y₁ = a cos ωt
and the displacement by S₂ be Y₂ = a cos (ωt + ϕ )
Find out the expression for the amplitude of the resultant displacement at a point and show that the intensity at that point will be

Hence establish the conditions for constructive and destructive interference,
(b) What is the effect on the interference fringes in Young’s double slit experiment when
(i) the width of the source slit is increased;
(ii) the monochromatic source is replaced by a source of white light? (All India 2014)
Answer:

(b) (i) When width of the source slit is increased, then the angular fringe width remains unchanged, but fringes become less and less sharp, so visibility of fringes decreases. If the condition, \(\frac{s}{S}<\frac{\lambda}{d}\) is not satisfied, the interference pattern disappears.

(ii) When the monochromatic source is replaced by a source of white light, the interference pattern due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are at the same position. Therefore, the central fringe is

Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.

Question 25.
(a) Define a wavefront. How is it different from a ray?
(b) Depict the shape of a wavefront in each of the following cases.
(i) Light diverging from point source.
(ii) Light emerging out of a convex lens when a point source is placed at its focus.
(iii) Using Huygen’s construction of secondary wavelets, draw a diagram showing the passage of a plane wavefront from a denser into a rarer medium. (Comptt. All India 2014)
Answer:
(a) Wavefront is the locus of all points, which oscillate in phase or it is the surface of constant phase.
A ray is defined as the path of energy propagation in the limit of wavelength tending to zero.
(b) Shapes of wave-front

Question 26.
(i) In Young’s double slit experiment, deduce the condition for
(a) constructive,’ and
(b) destructive interference at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position ‘x’ on the screen.
(ii) Compare the interference pattern observed in Young’s double slit experiment with single slit diffraction pattern, pointing out three distinguishing features. (Delhi 2016)
Answer:
(i)
(a) For constructive interference : We will have constructive interference resulting in a bright fringe when path difference is equal to nλ.

Since the separation between the centres of two consecutive bright fringes is called fringe width. It is denoted by β

Hence all bright and dark fringes are of equal width.
Observations :
(i) Fringe width is directly proportional to the wavelength of light i.e. p X.
(ii) Fringe width is inversely proportional to the distance between two sources

(iii) Fringe width is directly proportional to the distance between screen and two sources i.e. \(\beta \propto \frac{1}{d}\).

Suppose at any distance x from the central maximum,

The bright fringes will coincide at the least distance x,


(ii)
Interference pattern and Diffraction pattern :
The diagram, given here, shows several fringes, due to double slit interference, ‘contained’ in a broad diffraction peak. When the separation between the slits is large compared to their width, the diffraction pattern becomes very flat and we observe the two slit interference pattern.

Basic features of distinction between interference and diffraction patterns :
(i) The interference pattern has a number of equally spaced bright and dark bands while diffraction pattern has a central bright maximum which is twice as wide as the other maxima.
(ii) Interference pattern is the superimposition of two waves slits originating from two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.
(iii) For a single slit of width ‘a’ the first null of diffraction pattern occurs at an angle of \(\frac{\lambda}{a}\). At the same angle of \(\frac{\lambda}{a}\), we get a maxima for two narrow slits separated by a distance ‘a’.

Question 27.
(a) Draw a diagram showing the ‘Young’s arrangement’ for producing ‘a sustained interference pattern’. Hence obtain the expression for the width of the interference fringes obtained in this pattern.
(b) If the principal source point S were to be moved a little upwards, towards the slit S₁ from its usual symmetrical position, with respect to the two slits S₁ and S₂, discuss how the interference pattern, obtained on the screen, would get affected. (Compt. Delhi 2016)
Answer:
(a)
Consider two coherent sources S₁ and S₂ separated by a distance d. Let D be the distance between the screen and the plane of slits S₁ and S₂.
Light waves emitted from S₁ and S₂ reach point O on the screen after travelling equal distances. So path difference and hence phase difference between these waves is zero. Therefore, they meet at O in phase and hence constructive interference

takes place at O. Thus O is the position of the central bright fringe.
Let the waves emitted by S₁ and S₂ meet at point P and the screen at a distance y from the central bright fringe.
The path difference between these waves at P is given by


For constructive interference/maxima:
If path difference is an integral multiple of λ, then bright fringe will be formed at P

…where [m = 1, 2, 3 …
which is the position of mth bright fringe from the central bright fringe.
Fringe width (β) : The distance between any two successive bright fringes (or successive fringes) is called fringe width.

Destructive interferencelMinima: If path difference is odd multiple of \(\lambda / 2\), then dark fringe is formed at P

Which is position of m^th dark fringe from the central bright fringe.
β(fringe width) = y₁ – y₀

(b) On shifting principal source point ‘S’ little upwards i.e. towards Sv the position of the central maximum on the screen will shift downwards on the screen, i.e. below its previous position. Hence, whole interference pattern will get shifted little downwards but fringe width will remain the same as that of . the initial arrangement.

Question 28.
Based on Huygen’s construction, draw the shape of a plane wavefront as it gets refracted on passing through a convex lens. (Comptt. All India 2016)
Answer:
The required shape of the wavefront is as shown

Question 29.
When a plane wave front, of light, of wavelength X, is incident on a narrow slit, an intensity distribution pattern, of the form shown is observed on a screen, suitably kept behind the slit. Name the phenomenon observed.

(a) Obtain the conditions for the formation of central maximum and secondary maxima and the minima.
(b) Why is there significant fall in intensity of the secondary maxima compared to the central maximum, where as in double slit experiment all the bright fringes are of the same intensity?
(c) When the width of the slit is made double the original width, how is the size of the central band affected? (Comptt. All India 2016)
Answer:
The phenomenon observed is the phenomenon of ‘diffraction’.
(a) At the central maximum : The contributions due to the secondary wavelets, from all parts of the wave front (at the slit), arrive in phase at the central maxima, At the central maxima θ = 0.
At the secondary maxima : It is only the contributions from (nearly) 1/3 (or 1/5, or 1/ 7, …) of the incident wavefront that do not get cancelled at the locations of the secondary maxima. These occur at points for which

At the secondary minima : The contributions, from ‘corresponding pairs’, of the sub-parts of the incident wavefront, cancel each other and the net contribution, at the location of the minima, is zero. The minima occur at points for which

(b) There is a significant fall in intensity at the secondary maxima because the intensity there, is only due to the contribution of (nearly) (1/3 or 1/5 or 1/7, …) of the incident wavefronts.

(c) The size of the central maximum would get halved when width of the slit is doubled.

Question 30.
(a) Distinguish between unpolarised light and linearly polarised light. How does one get linearly polarised light with the help of a Polaroid?
(b) A narrow beam of unpolarised light of intensity I₀ is incident on a polaroid P₁. The light transmitted by it is then incident on a second polaroid P₂ with its pass axis making angle of 60° relative to the pass axis of P₁. Find the intensity of the light transmitted by P₂. (Delhi 2016)
Answer:
(a) Distinction between unpolarised and polarised light: In an unpolarised light, the oscillations of the electric field, are in random directions, in planes perpendicular to the direction of propagation. For a polarised light, the oscillations are aligned along one particular direction.

Alternatively, Polarised light can be distinguished, from unpolarised light, when it is allowed to pass through a polaroid. Polarised light can show change in its intensity, on passing through a polaroid:
while intensity of transmitted light will remain unchanged in case of unpolarised light.

Polarised light with the help of a polaroid : When unpolarised light wave is incident on a polaroid, the electric vectors along the, direction of its aligned molecules get absorbed; the electric vector, oscillating along a direction perpendicular to the aligned molecules, pass through. This light is called linearly polarised light.

Question 31.
(a) Explain two features to distinguish between the interference pattern in Young’s double slit experiment with the diffraction pattern obtained due to a single slit.
(b) A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen.
Estimate the number of fringes obtained in Young’s double slit experiment with fringe width 0.5 mm, which can be accommodated within the region of total angular spread of the central maximum due to single slit. (Delhi. 2016)
Answer:
(a) Interference pattern and Diffraction pattern :
The diagram, given here, shows several fringes, due to double slit interference, ‘contained’ in a broad diffraction peak. When the separation between the slits is large compared to their width, the diffraction pattern becomes very flat and we observe the two slit interference pattern.

Basic features of distinction between interference and diffraction patterns :
(i) The interference pattern has a number of equally spaced bright and dark bands while diffraction pattern has a central bright maximum which is twice as wide as the other maxima.
(ii) Interference pattern is the superimposition of two waves slits originating from two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.
(iii) For a single slit of width ‘a’ the first null of diffraction pattern occurs at an angle of \(\frac{\lambda}{a}\). At the same angle of \(\frac{\lambda}{a}\), we get a maxima for two narrow slits separated by a distance ‘a’.

Question 32.
(a) Define wavefront. Use Huygens’ principle to verify the laws of refraction.
(b) How is linearly polarised light obtained by the process of scattering of light? Find the Brewster angle for air—glass interface, when the refractive index of glass = 1.5. (All India 2016)
Answer:
(a) Wavefront : The wavefront is the common locus of all points which are in phase (surface of constant phase)
Huygens’ principle to verify laws of refraction :
Huygens’ geometrical construction for a plane wave propagation. Let AB be a section of primary wavefront at any instant t. Take points 1, 2, 3, 4, … on the wavefront AB. Taking each point as centre, draw spheres of radius r = ct, where c is the velocity of light in the medium.
Draw a surface A₁B₁ touching tangentially at the secondary wavelets in the forward direction. The surface A₁B₁ is the secondary wavefront after time t.

A surface A₂B₂ touching tangentially all the secondary wavelets in the backward direction can be drawn to give a backward secondary wavefront.
Laws of refraction using Huygen’s principle. Let XY be the refracting surface separating two mediums 1^st and 2^nd. Let c₁ and c₂ be the velocities of light in these mediums.

Let AB be a plane wavefront moving through the surface XY meeting at the point B.
The time taken ‘t’ of A to reach at A’B then

To obtain new front, draw a circle with point B as centre and BD as radius in 2nd medium. Draw a tangent A’D from point A’. Then A’D represents the refracted wavefront.
Since PB be incident ray and BD be refracted ray

This constant V is called the refractive index of material which proves Snell’s law of refraction.

(b) When unpolarised light gets scattered by molecules of earth’s atmosphere, the scattered light has only one of its two components in it and hence linearly polarised, as shown in the diagram :

Question 33.
Explain with diagram, how plane polarized light can be produced by scattering of sunlight. An incident beam of light of intensity IG is made to fall on a polaroid A. Another polaroid B is so oriented with respect to A that there is no light emerging out of B. A third polaroid C is now introduced mid-way between A and B and is so oriented that its axis bisects the angle between the axes of A and B. Calculate the intensity of light transmitted by A, B and C. (Comptt. Delhi 2016)
Answer:
(i) (c) It is due to scattering of light by molecules of earth’s atmosphere.

Under the influence of the electric field of the incident (unpolarized) wave, the electrons in the molecules acquire components of motion in both these directions. Charges, accelerating parallel to the double arrows, do not radiate energy towards the observer since their acceleration has no transverse component.
The radiation scattered by the molecules is therefore represented by dots, i.e., it is polarized perpendicular to plane of figure.

Question 34.
(a) In Young’s double slit experiment a monochromatic source of light S is kept equidistant from the slits S2 and S2. Explain the formation of dark and bright fringes on the screen.
(b) A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in Young’s double-slit experiment.
(i) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm?
(ii) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Given : the separation between the slits is 4 mm and the distance between the screen and plane of the slits is 1.2 m. (Comptt. Delhi 2016)
Answer:
(a) Young’s double slit experiment. The path difference between two rays coming from holes S₁ and S₂ is (S₂P – S₁P). If point P corresponds to a maximum,

If x, d << D, then negligible error will be introduced if (S₂P + S₁P) in the denominator is replaced by 2D.

Thus, bright and dark bands appear on the screen. Such bands are called ‘fringes’.
These dark and bright fringes are equally spaced.

Question 35.
Describe an experiment to show that light waves are transverse in nature.
Answer:
Light is a transverse wave. This can be shown with the help of this simple experiment. The figure shows an unpolarized light beam incident on the first polarising sheet, called the polariser where the transmission axis is indicated by the straight line on the polariser.

The polariser can be a thin sheet of tourmaline (a complex boro-silicate). The Light, which is passing through this sheet, is polarised vertically as shown, where the transmitted electric vector is E_o. A second polarising sheet called the analyser intercepts this beam with its transmission axis at an angle θ to the axis of the polariser. As the axis of the analyser is rotated slowly, the intensity of Light received beyond It goes on decreasing.

When the transmitting axis of the analyser becomes perpendicular to the transmission axis of the polariser, no beam is obtained beyond the analyser. This means that the anaLyser has further polarised the beam coming from the polariser. Since only transverse waves can be polarised, therefore, this shows that light waves are transverse In nature.

Question 36.
Derive an expression for the width of the central maxima for diffraction of light at a single-slit. How does this width change with an increase in the width of the slit?
Answer:
The width of the central maxima for diffraction of light at a single-slit is the distance from the 1st diffraction minima on one side of the central maxima to the 1st diffraction minima on another side of the central maxima.

If a be the width of slit then for 1st diffraction minima, we have
sin θ = ± λ
or
sin θ = θ = ± \(\frac{λ}{a}\)

Angular width of central maxima
= 2θ = ± \(\frac{2λ}{a}\)

If D be the distance between the slit and the screen, then linear width ‘x’ of the central maxima is given by
x = d × 2θ = D × \(\frac{2λ}{a}\) = \(\frac{2Dλ}{a}\)

As the width (a) of the slit is increased, the linear width of central maxima goes on decreasing because x ∝ \(\frac{1}{a}\)

Question 37.
(a) Sketchthegraphshowingthevariation of the intensity of transmitted light on the angle of rotation between a polarizer and an analyser.
Answer:
The graph is as shown below

(b) A ray of light is incident at an angle of incidence i_p on the surface of separation between air and a medium of refractive index µ, such that the angle between the reflected and refracted ray is 90°. Obtain the relation between i_p and µ.
Answer:
Suppose an unpolarised light beam is an incident on a surface as shown in the figure below. The beam can be described by two electric field components, one parallel to the surface (the dots) and the other perpendicular to the first and to the direction of propagation (the arrows). It is found that the parallel components reflect more strongly than the other component, this results in a partially polarised beam. Furthermore, the refracted ray is also partially polarized.

Now suppose the angle of incidence, i is varied until the angle between the reflected and the refracted beam is 90°. At this particular angle of incidence, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised. The angle of incidence at which this occurs is called the polarising angle i_p.

An expression can be obtained relating the polarising angle to the index of refraction n, of the reflecting surface. From the figure, we see that at the polarising angle
i_p + 90° + r = 180°
or
r = 90° – i_p.

Using Snell’s law we have n = \(\frac{\sin i}{\sin r}=\frac{\sin i_{p}}{\sin r}\)

Now sin r = sin (90° – i_p) = cos i_p, therefore the above expression becomes
n = \(\frac{\sin i_{p}}{\cos i_{p}}\) = tan ip

Question 38.
Describe Young’s double-slit experiment to produce an interference pattern due to a monochromatic source of light. Deduce the expression for the fringe width. (CBSE Delhi 2011)
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r1 and r2 are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,
y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe
y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,
y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, a width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

Question 39.
(a) Explain, with the help of a diagram, how plane polarised light Is obtained by scattering.
Answer:
It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

(b) Between two polaroids placed In crossed position a third Polaroid is introduced. The axis of the third Polaroid makes an angle of 30° with the axis of the first polaroid. Find the intensity of transmitted light from the system assuming / to be the intensity of polarised light obtained from the first polaroid. (CBSE A! 2011C)
Answer:
Let l_o be the intensity of light passing through the first polaroid.

The intensity of light passing through the middle polaroid whose axes are inclined at 30° to the first polaroid by Malus law is
l’ = l_o cos² 30° = l_o × 3/4 = 3l_o/4

The intensity of light passing through the system is, therefore, (for the second crystal θ = 60°)
l” = l’cos² 60° = 3l_o/4 × 1 /4
or
l” = 3l_o/16

Question 40.
(a) Why are coherent sources necessary to produce a sustained interference pattern?
Answer:
Interference will be sustained if there is a constant phase difference between the two interfering waves. This is possible if the two waves are coherent.

(b) In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference Is λ, is K units. Find out the intensity of light at a point where the path difference is λ/3. (CBSE Delhi 2012)
Answer:
Intensity at any point on the screen
l = l₁ + l₂ + 2\(\sqrt{1_{1}1_{2}}\) cos Φ

Let l0 be the intensity of either source, then l₁= l₂ = l_o
When p = λ, Φ = 2π

Question 41.
Use Huygen’s principle to explain the formation of the diffraction pattern due to a single-slit illuminated by a monochromatic source of light.
When the width of the slit is made double the original width, how would this affect the size and intensity of the central diffraction band? (CBSE Delhi 2012)
Answer:
(a) The arrangement is shown below in the figure.

When a plane wavefront WW’ falls on a single-slit AB, each point on the unblocked portion ADB of wavefront sends out secondary, wavelets in all the directions. For secondary waves meeting at point O, the path difference between waves is zero and hence the secondary waves reinforce each other giving rise to the central maximum at the symmetrical point O.

Consider secondary waves travelling in a direction making an angle θ with DO and reaching the screen at point P. Obviously, path difference between the extreme secondary waves reaching point P from A and B.
= BC = AB sin θ = a sin θ.

If this path difference a sin θ = λ, then point P will be minimum Intensity. In this situation, wavefront may be supposed to consist of two equal halves AD and BD and for every point on AD, there will be a corresponding point on DB having a path difference λ/2. Consequently, they nullify the effect of each other and point P behaves as the first secondary minimum. In general, if path difference a sin θ = nλ where n = 1, 2, 3, …. then we have secondary minima corresponding to that angle of diffraction θ_n.

However, if for some point P₁ on the screen secondary waves BP₁ and AP₁ differ In path by 3λ/2 then point P₁ will be the position of the first secondary maxima. Because in this situation, wavefront AB may be divided into three equal parts such that path difference between corresponding points on the first and second will be λ/2 and they will nullify. But secondary waves from the third part remain as such and give rise to the first secondary maxima, whose Intensity will be much less than that of central maxima.

In general, if path difference a sin θ = (2n + 1)θ/2, where n = 1, 2, 3, …… then we have nth secondary maxima corresponding to these angles.

The width of the central maxima is the distance between the first secondary minima on either side of the centre of the screen. The width of the central maxima is twice the angle θ subtended by the first minima on either side of the central maxima. Now sin θ = \(\frac{\lambda}{a}\). Since θ is small a therefore it can be replaced by tan θ, hence sin θ = \(\frac{\lambda}{a}=\frac{y}{L} or y = [latex]\frac{L \lambda}{\mathrm{a}}\).

This gives the distance of the first secondary minima on both sides of the centre of the screen. Therefore, the width of the central maxima is 2y, hence
2y = \(\frac{2L \lambda}{\mathrm{a}}\)

(b) The size reduces by half according to the relation: size = X/d. Intensity becomes twice the original intensity.

Question 42.
(a) Using the phenomenon of polarization, show how the transverse nature of light can be demonstrated.
Answer:
Light is a transverse wave. This can be shown with the help of this simple experiment. The figure shows an unpolarized light beam incident on the first polarising sheet, called the polariser where the transmission axis is indicated by the straight line on the polariser.

The polariser can be a thin sheet of tourmaline (a complex boro-silicate). The Light, which is passing through this sheet, is polarised vertically as shown, where the transmitted electric vector is E_o. A second polarising sheet called the analyser intercepts this beam with its transmission axis at an angle θ to the axis of the polariser. As the axis of the analyser is rotated slowly, the intensity of Light received beyond It goes on decreasing.

When the transmitting axis of the analyser becomes perpendicular to the transmission axis of the polariser, no beam is obtained beyond the analyser. This means that the anaLyser has further polarised the beam coming from the polariser. Since only transverse waves can be polarised, therefore, this shows that light waves are transverse in nature.

(b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity l0 is Incident on P₁. A third polaroid P₃ is kept in between P₁ and P₂ such that its pass axis makes an angle of 30° with that of P₁. Determine the intensity of light transmitted through P₁ P₂ and P₃. (CBSE AI 2014)
Answer:
The intensity of light passing through P₁ is half of the light falling on it. Therefore, the light coming out of P₁ is l0_o/2 Now light coming out of P₃ is

and light coming out of polariser P₂ is

Question 43.
What does a Polaroid consist of? Show using a simple Polaroid that light waves are transverse in nature. The intensity of light coming out of a Polaroid does not change irrespective of the orientation of the pass axis of the polaroid. Explain why. (CBSE AI 2015)
Answer:
A polaroid consists of long-chain molecules aligned in a particular direction.

Let the light from an ordinary source (like a sodium lamps pass through a polaroid sheet P₁ it is observed that its intensity is reduced by half. Now, let an identical piece of polaroid P₂ be placed before P₁ As expected, the light from the lamp is reduced in intensity on passing through P₂2 alone. But now rotating P₁ has a dramatic effect on the light coming from P₂. In one position, the intensity transmitted by P₂ followed by P₁ is nearly zero. When turned by 90° from this position, P₁ transmits nearly the full intensity emerging from P₂ shown in the figure.

Since only transverse waves can be polarised, therefore, this experiment shows that light waves are transverse in nature. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarized light wave is an incident on such a polaroid then the light wave will get linearly polarized with the electric vector oscillating along a direction perpendicular to the aligned molecules; this direction is known as the pass-axis of the Polaroid.

Thus, if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P it is observed that its intensity is reduced by half. Rotating P has no effect on the transmitted beam and transmitted intensity remains constant as there is always an electric vector that oscillates in a direction perpendicular to the direction of the aligned molecules.

Question 44.
Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted Intensity be maximum? (CBSE Delhi 2015)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁. Then the intensity of light after passing through the second polarizer P₂ will be
l = l_o cos² θ

where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be

The transmitted intensity will be maximum when 2θ = π/2 or θ = π/4

Question 45.
Distinguish between unpolarised and linearly polarised light. Describe with the help of a diagram how unpolarised light gets linearly polarised by scattering. (CBSE Delhi 2015)
Answer:
The two lights will be allowed to pass through a polariser. When the polarizer is rotated in the path of these two light beams, the intensity of light remains the same in all the orientations of the polariser, then the light is unpolarised. But if the intensity of light varies from maximum to minimum then the light beam is a polarised light beam.

It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

Question 46.
(a) Explain how a diffraction pattern is formed due to interference of secondary wavelets of light waves from a slit.
Answer:
(a) Diffraction at single slit: Consider monochromatic plane wavefront WW’ incident on the slit AB as shown in Fig. Imagine the slit to be divided into a large number of very narrow strips of equal width parallel to the slit. When the wavefront reaches the slit, each narrow strip parallel to the slit can be considered to be a source of Huygen’s secondary wavelets.

Central Maxima. Let us first consider the effect of all the wavelets at point O. All wavelets cover the same distance and reach 0 in the same phase. The wavelets superimpose constructively and give maximum intensity at O.

Position of secondary minima. Consider the intensity at a point P at an angle θ with the original direction.
Draw AL ⊥ BP.
In ΔABL


or
sin θ = \(\frac{λ}{a}\)

For second minima, BL = 2A
or a sin θ = 2λ

Similarly, for nth minima a sin θ = nλ
or sin θ = \(\frac{nλ}{a}\) , where n = 1, 2, ……

Position of secondary maxima: If path difference BL = \(\frac{3λ}{2}\), then slit AB can be divided into three equal parts and the path difference between wavelets from corresponding points in the first two parts will be \(\frac{λ}{2}\) and hence cancels each other’s effect and produces destructive Interference and the third path will produce its maxima at P called first secondary maxima.

Similarly, if BL = \(\frac{5λ}{2}\), we get second secondary maxima and so on.

For nth maxima
a sin θ = (2n + 1)\(\frac{λ}{a}\)
or
sin θ = \(\frac{(2 n+1) \lambda}{2 a}\)

Thus we find that the intensity of the central fringe is maximum whereas that of other fringes fall off rapidly in either direction from the centre of the fringe pattern.

(b) Sodium light consists of two wavelengths, 5900 Å and 5960 Å. If a slit of width 2 × 10^-4 m is Illuminated by sodium light, find the separation between the first secondary maxima of the diffraction pattern of the two wavelengths on a screen placed 1.5m away. (CBSE 2019C)
Answer:
The separation between the first secondary maxima of the diffraction pattern of two wavelengths is:

Question 47.
(a) Derive Snell’s law on the basis of Huygen’s wave theory when light Is travelling from a denser to a rarer medium.
Answer:
The ray diagram is as shown.


(b) Draw the sketches to differentiate between plane wavefront and spherical wavefront.
Answer:
Spherical wavefront and plane wavefront

Question 48.
The figure drawn here shows the geometry of path differences for diffraction by a single-slit of width a.

Give appropriate ‘reasoning’ to explain why the intensity of light is
(a) Maximum at the central point C on the screen.
(b) (Nearly) zero for point P on the screen when θ = λ / a.
Hence write an expression for the total linear width of the central maxima on a screen kept at a distance D from the plane of the slit. (CBSE Delhi 2016C)
Answer:
(a) At central point C, the angle is zero, all path differences are zero. Hence all the parts of the slit contribute to the same phase. This gives the maximum intensity at point C.
(b)

From figure
NP – LP = NQ= a sin θ = aθ When θ = λ/a

Then path difference NP – LP = aθ = λ.
Hence MP – LP = NP – MP = λ/2

It implies that the contribution from corresponding points in two halves of the slit has a phase difference of π. Therefore, contributions from two halves cancel each other in pairs, resulting in a zero net intensity at point P on the screen. Half angular width of central maxima = λ/a Half linear width = λD/a

Linear width of central maxima = 2λD/a

Question 49.
Two polaroids, P₁ and P₂, are ‘set-up’ so that their ‘pass axis is ‘crossed’ with respect to each other. AQ third Polaroid, P₃ is now introduced between these two so that its ‘pass axis makes an angle with the ‘pass axis of P_1. A beam of unpolarized light, of Intensity I, Is Incident on P₁ If the Intensity of light that gets transmitted through this combination of three polaroids, Is I’, find the ratio \(\left(\frac{I^{\prime}}{I}\right)\) when θ equals: (i) 30°, (ii) 45° (CBSE Delhi 2016C)
Answer:
The diagram is as shown.

The intensity of unpolarized light is given as l.
It becomes half after passing through Polaroid P₁ (l/2)
Using Malus law for the intensity of light passing through P3 we have
l₁ = \(\left(\frac{l}{2}\right)\) cos²θ

This intensity l₁ is Incident on P₂, hence the intensity of light coming out of P₂ will be

Question 50.
What is the effect on the Interference pattern observed In Young’s double-slit experiment In the following cases:
(a) Screen is moved away from the plane of the slits,
(b) Separation between the slits is Increased and
(c) Widths of the slits are doubled. Give the reason for your answer.
Answer:
The fringe width is given by the expression
β = \(\frac{Dλ}{d}\)

(a) When D Is Increased, the fringe width Increases.
(b) When d 1s Increased the fringe width decreases.
(c) If the width w of the slits Is changed then Interference occurs only If \(\frac{1}{w}\) > \(\frac{1}{d}\) remains satisfied, where d is the distance between the slits.

Question 51.
Two slits In Young’s double-slit experiment are illuminated by two different lamps emitting light? Will you observe the Interference pattern? Justify your answer. Find the ratio of Intensities at two points on a screen In Young’s double-slit experiment, when waves from two slits have a path difference of (i) 0 and (ii) λ/4.
Answer:
The two sources act as Independent sources of light and hence can never be coherent. In a light, source light is produced by billions of atoms under proper excitation condition and each atom acts independently of the other atoms. Thus there Is no coherence ‘ between these two Independent sources hence no interference.

The phase difference corresponding to the two paths are Φ = 0 and Φ = π/2

Now intensity at the screen when the phase difference is Φ = 0 is
l_A = l₁ +l₂ + 2\(\sqrt{l_{1} l_{2}}\)cos Φ
or
l_A = l + l + 2\(\sqrt{l l}\)cos Φ = 4l

Now intensity at the screen when the phase difference is Φ = 90° Is
l_B = l₁ +l₂ + 2\(\sqrt{l_{1} l_{2}}\)cos Φ
or
l_B = l + l + 2\(\sqrt{l l}\)cos 90° = 2l
Therefore, ratio of intensities is
\(\frac{l_{A}}{l_{B}}=\frac{4l}{2l}\) = 2

Question 52.
(a) In a single-slit diffraction pattern, how does the angular width of the central maximum vary, when
(i) the aperture of the slit Is Increased?
(ii) distance between the slit and the screen is decreased?
Justify your answer In each case.
Answer:
The angular width of the central maxima in a single-slit diffraction pattern is given by 2θ = \(\frac{2λ}{a}\) where λ is the wavelength of light and ‘a’ the slit width.
(i) When the aperture of the slit is increased the angular width decreases.
(ii) When the distance between the slit and the screen is decreased, the angular width will remain the same but the linear width will increase.

(b) How Is the diffraction pattern different from the interference pattern obtained In Young’s double¬slit experiment? (CBSE Delhi 2011C)
Answer:
The difference is shown in the table:

Diffraction	Interference
1. It is due to interference between the wavelets starting from two parts of the same wavefront.	1. It is a superposition of the two waves starting from two coherent sources.
2. The intensity of the consecutive bands goes on decreasing.	2. All bright fringes have the same intensity.
3. Fringes have poor contrast.	3. Fringes have good contrast.
4. Diffraction fringes are not of the same width.	4. Fringes may or may not be of the same width.

Question 53.
(a) Can two independent monochromatic light sources be used to obtain a steady interference pattern? Justify your answer. (CBSE 2019C)
Answer:
No, because the phase difference between the light waves from two independent sources keeps on changing continuously and for a steady interference pattern, the phase difference between the waves should remain constant with time. Hence two independent monochromatic light sources cannot produce a steady interference pattern.

(b) In Young’s double-slit experiment, explain the formation of interference fringes and obtain an expression for the fringe width.
Answer:
Interference of light

Let S₁, S₂ be the two fine slits illuminated by a monochromatic source S of wavelength λ.

The intensity of light at any point P on the screen at a distance D from the slit depends upon the path difference between S₂P and S₁P.
∴ Path difference = S₂P – S₁P
= (S₂A + AP) – S₁P

Path difference = S₂A = d sin θ
Since θ is small, sin θ can be replaced by tan θ.
∴ Path difference = d tan θ = d\(\frac{y}{D}\)

Constructive interference [Bright Fringes]. For bright fringes, the path difference should be equal to integral multiple of A.
∴ \(\frac{dy}{D}\) = nλ
or
y = n\(\frac{λD}{d}\)

For nth fringe, let us write y as y_n
so y_n = n\(\frac{λD}{d}\)

The spacing between two consecutive bright fringes is equal to the width of a dark fringe.
∴ Width of the dark fringe.
β = y_n – y_n-1 = \(\frac{nλD}{d}\) – (n – 1)\(\frac{λD}{d}\) = \(\frac{λD}{d}\) …(i)

Destructive interference. [Dark Fringes].
For dark fringes, the path difference should be an odd multiple of λ/2.
∴ \(\frac{d}{D}\)y = (2n + 1)λ/2
or
y’ = \(\frac{D}{d}\)(2n + 1)\(\frac{λ}{2}\) = \(\frac{(2 n+1) \lambda D}{2 d}\)

For nth fringe, writing y as yn, we get
y’_n = \(\frac{(2 n+1) \lambda D}{2 d}\).

The spacing between two consecutive dark fringes is equal to the width of a bright fringe.
∴ Width of the bright fringe.

From Eqs. (i) and (ii), we find that dark and bright fringes are of same width given by
β = \(\frac{λD}{d}\).

(c) In an interference experiment using monochromatic light of wavelength A, the intensity of light of point, where the path difference is X, on the screen is K units. Find out the Intensity of light at a point when path difference is λ/4. (CBSE 2019C)
Answer:
The intensity of light on the screen where the waves meet having phase difference Φ is:


Question 54.
(a) Two monochromatic waves emanating from two coherent sources have the displacements represented by y₁ = a cos ωt and y₂ = a cos (ωt + Φ), where Φ is the phase difference between the two displacements. Show that the resultant intensity at a point due to their superposition is given by l = 4l_o cos² Φ/2, where l_o = a².
Answer:
Let the displacements of the waves from the sources S₁ and S₂ at a point on the screen at any time t be given by y₁ = a cos ωt and y₂ = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves. By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is A = 2a cos (Φ/2)

Therefore the intensity at that point is
l = A² = 4a² cos² \(\frac{Φ}{2}\)

(b) Hence obtain the conditions for constructive and destructive interference. (CBSE AI 2014C)
Answer:
Bright fringes: For bright fringes I = max, therefore Φ = 0° or cos Φ = +1
or
Φ = 2 n π
Dark fringes: For dark fringes l = 0,
therefore Φ = π or cos Φ = -1
or
Φ = (2n + 1) π

Question 55.
A parallel beam of monochromatic light falls normally on a narrow slit of width ‘a’ to produce a diffraction pattern on the screen placed parallel to the plane of the slit. Use Huygens’ principle to explain that
(a) the central bright maxima is twice as wide as the other maxima.
Answer:
The width of the central maxima is the distance between the first secondary minima on either side of the centre of the screen. The width of the central maxima is twice the angle 6 subtended by the first minima on either side of the central maxima.

Now sin θ = \(\frac{λ}{a}\). Since θ is small there, a fore it can be replaced by tan 0, hence
tan θ = \(\frac{λ}{a}\) = \(\frac{y}{L}\)
or
y = \(\frac{Lλ}{a}\). This gives the distance of the first secondary min¬ima on both sides of the centre of the screen. Therefore, the width of the central maxima is 2y, hence
2y = \(\frac{2Lλ}{a}\)

(b) the Intensity falls as we move to successive maxima away from the centre on either side. (CBSE Delhi 2014C)
Answer:
The maxima in diffraction pattern are formed at (n +1 /2) λ/a, with n = 2, 3, etc. These become weaker with increasing n, since only one third, one- fifth, one-seventh, etc., of the slit contributes in these cases.

Question 56.
Answer the following questions:
(a) In a double-slit experiment using the light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.
Answer:

(b) Light of wavelength 5000 A propagating 1n air gets partly reflected from the surface of the water. How will the wavelengths and frequencies of the reflected and refracted light be affected? (CBSE Delhi 2015)
Answer:
No change in the wavelength and frequency of reflected light. In the case of refracted light, there is no change in frequency but the wavelength becomes 1/1.33 times the original wavelength.

Question 57.
(a) Good quality sun-glasses made of polaroids are preferred over ordinary coloured glasses. Justify your answer,
Answer:
A Polaroid sun-glass limits the light entering the eye, thus providing a soothing effect.

(b) Two polaroids and P₂ are placed In crossed positions. A third Polaroid P₃ is kept between P₁ and P₂ such that the pass axis of P₃ is parallel to that of P₁. How would the Intensity of light l₂ transmitted through P₂ vary as P₃ Is rotated? Draw a plot of Intensity l₂ Vs the angle ‘θ’, between pass axes of P₁ and P₃. (CBSE AI 2015C)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁ Then the intensity of light after passing through the second polarizer P₃ will be
l = l_o cos² θ
where θ is the angle between pass axes of P₁ and P₃. Since P₁ and P₂ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ).

Hence the intensity of light emerging from P2 will be

Therefore, the transmitted intensity will be maximum when θ = π/4

For graph

Question 58.
In a single-slit diffraction pattern, how does the angular width of central maximum change, when
(a) slit width is decreased,
(b) distance between the slit and screen is increased, and
(c) light of smaller visible wavelength is used? Justify your answer in each case.
Answer:
We know that the angular width of the central maximum of the diffraction pattern of a single-slit is given by
w = \(\frac{2Dλ}{a}\).

(a) If slit width ‘a’ is decreased, the angular width will increase because
x ∝ \(\frac{1}{a}\)

(b) Increase in distance between the slit and the screen does not affect the angular width of diffraction maxima. However, linear width of the maxima
w = \(\frac{2Dλ}{a}\) will increase.

(c) If the light of a smaller visible wavelength is used, the angular width is decreased because x ∝ λ.

Question 59.
Light, from a sodium lamp, is passed through two polaroid sheets P₁ and P₂ kept one after the other. Keeping P₁ fixed, P₂ is rotated so that its ‘pass axis can be at different angles, θ, with respect to the pass-axis of P₁
An experimentalist records the following data for the Intensity of light coming out of P₂ as a function of the angle θ.

(a) l_o = Intensity of beam falling on P₁ One of these observations is not in agreement with the expected theoretical variation of l, Identify this observation and write the correct expression.
Answer:
The observation \(\frac{l_{0}}{2 \sqrt{2}}\) is not correct. It should be l_o/4.

(b) Define the Brewster angle and write the expression for it in terms of the refractive index of the medium.
Answer:
It is the angle of incidence at which the refracted and the reflected rays are perpendicular to each other. It is related to the refractive index as tan i_p = n

Question 60.
(a) Light, from a monochromatic source, is made to fall on a single-slit of variable width. An experimentalist records the following data for the linear width of the principal maxima ‘ on a screen kept at a distance of 1 m from the plane of the slit.

Use any two observations from this data to estimate the value of the wavelength of light used.
Answer:
The width of the central maxima is given by the expression β = \(\frac{2Lλ}{a}\)
or
λ₁ = \(\frac{βa}{2L}\)

Using the values of the first observation we have
λ₁ = \(\frac{6 \times 10^{-3} \times 0.1 \times 10^{-3}}{2 \times 1}\) = 0.3 × 10^-6

Using the values of the second observation we have
λ₁ = \frac{3 \times 10^{-3} \times 0.2 \times 10^{-3}}{2 \times 1} = 0.3 × 10^-6

Thus the wavelength of light used is λ = 0.3 × 10^-6 m

(b) Show that the Brewster angle iB for a given pair of transparent media is related to their critical angle ic through the relation i_c = sin^-1 (ωt i_B)
Answer:
We know that

Question 61.
For a single-slit of width, “a” the first minimum of the interference pattern of monochromatic light of wavelength λ occurs at an angle of λ/a. At the same angle λ/a, we get a maximum for two narrow slits separated by a distance ‘a’. Explain. (CBSE Delhi 2014)
Answer:
The path difference between two secondary wavelets is given by nλ = a sin θ. Since θ is a very small sin θ = 0. So, for the first-order diffraction n = 1, the angle is λ/a. Now we know that θ must be very small θ = 0 (nearly) because of which the diffraction pattern is minimum.

Now for interference case, for two interfering waves of intensity l1 and l2 we must have two slits separated by a distance. We have the resultant intensity,
l = l₁ + l₂ + 2\(\sqrt{l_{1} l_{2}}\) cos θ

Since θ = 0° (nearly) corresponding to angle λ/a so cos θ = 1 (nearly)
So,
l = l₁ + l₂ + 2\(\sqrt{l_{1} l_{2}}\) cos θ
l = l₁ + l₂ + 2\(\sqrt{l_{1} l_{2}}\)

We see the resultant intensity is the sum of the two intensities, so there is a maxima corresponding to the angle λ/a.

This is why, at the same angle of λ/a, we get a maximum for two narrow slits separated by a distance “a”.

Question 62.
Show using a proper diagram of how unpolarised light can be linearly polarized by reflection from a transparent glass surface. (CBSE AI 2018, Delhi 2018)
Answer:
An ordinary beam of light, on reflection from a transparent medium, becomes partially polarised. The degree of polarisation increases as the angle of incidence is increased. At a particular value of the angle of incidence, the reflected beam becomes completely polarised. This angle of incidence is called the polarising angle (i_p).

Question 63.
Answer the following questions:
(a) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
Answer:
The size reduces by half according to the relation: size = λ/d. Intensity increases fourfold.

(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
Answer:
The intensity of interference fringes in a double-slit arrangement is modulated by the diffraction pattern of each slit.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
Answer:
Waves diffracted from the edge of the circular obstacle interfere constructively at the centre of the shadow producing a bright spot.

(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily? (NCERT)
Answer:
For diffraction or bending of waves by obstacles/apertures by a large angle, the size of the latter should be comparable to the wavelength. If the size of the obstacle/aperture is much too large compared to the wavelength, diffraction is by a small angle. Here the size is of the order of a few metres. The wavelength of light is about 5 × 10^-7 m, while sound waves of, say, 1 kHz frequency have a wavelength of about 0.3 m. Thus, sound waves can bend around the partition while light waves cannot.

Question 64.
(a) When an unpolarized light of intensity l0 is passed through a polaroid, what is the intensity of the linearly polarised light? Does it depend on the orientation of the polaroid? Explain your answer.
Answer:
(a) The intensity of the linearly polarised light would be l_o/2.
No, it does not depend on the orientation.
Explanation: The polaroid will let the component of the unpolarized light, parallel to its pass axis, to pass through it irrespective of its orientation.

(b) A plane polarised beam of light is passed through a polaroid. Show graphically the variation of the intensity of the transmitted light with an angle of rotation of the polaroid incomplete one rotation. (CBSE Delhi 2018C)
Answer:
We have l = l_o cos² θ
∴ The graph is as shown below

Question 65.
(a) If one of two identical slits producing interference in Young’s experiment is covered with glass so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the Interference pattern.
Answer:
As intensity is directly proportional to the square of the amplitude.


(b) What kind of fringes do you expect to observe if white light is used Instead of monochromatic light?
Answer:
The central fringe will be white and the remaining will be coloured fringes of different width in the VIBGYOR sequence.

Question 66.
Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum? (CBSE Delhi 2015)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁ Then the intensity of light after passing through the second polarizer P₂ will be
l = l_o cos² θ
where 0 is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P3 will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be

The transmitted intensity will be maximum when 2θ = π/2 or θ = π/4

Question 67.
Derive the expression for the fringe width In Young’s double-slit experiment.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r₁ and r₂ are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o0 = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,

y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe

y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,

y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, a width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

Question 68.
(a) Using Huygens’ principle, draw a diagram to show the propagation of a wave-front originating from a monochromatic point source.
Answer:

(b) Describe diffraction of light due to a single-slit. Explain the formation of a pattern of fringes obtained on the screen and plot showing a variation of intensity with path difference in single-slit diffraction.
Answer:
Let us discuss the nature of the Fraunhofer diffraction pattern produced by a single-slit. Let us examine the waves coming from the various portions of the slit, as shown in the figure. According to Huygens principle, each portion of the slit acts as a source of waves. Hence, light from one portion of the slit can interfere with the light of another portion, and the resultant intensity on the screen will depend on the direction of θ. The secondary waves coming from the different parts of the slit interfere to produce either maxima or minima, thereby giving rise to a diffraction pattern.

For the diagram

2y = \(\frac{2Lλ}{a}\)

Question 69.
What is the effect on the interference fringes in Young’s double-slit experiment due to each of the following operations:
(a) the screen is moved away from the plane of the slits;
Answer:
The angular separation of the fringes remains constant (= λ/d). The actual separation of the fringes increases in proportion to the distance of the screen from the plane of the slits

(b) the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength;
Answer:
The separation of the fringes (and also angular separation) decreases.

(c) the separation between the two slits is Increased
Answer:
The separation of the fringes (and also angular separation) decreases.

(d) the source slit is moved closer to the double-slit plane;
Answer:
Let s be the size of the source and S its distance from the plane of the two slits. For interference, fringes to be seen the
condition \(\frac{s}{S}<\frac{\lambda}{d}\) should be satisfied otherwise interference patterns produced by different parts of the source overlap and no fringes are seen. Thus as S decreases (i.e. the source slit is brought closer), the interference pattern gets less and less sharp and when the source is brought too close for this condition to be valid the fringes – disappear. Till this happens, the fringe separation remains fixed.

(e) the width of the source slit is Increased;
Answer:
Same as in (d). As the source slit width increases fringe pattern gets less and less sharp. When the source slit is so wide that the condition \(\frac{s}{S}<\frac{\lambda}{d}\) is not satisfied the interference pattern disappears.

(f) the monochromatic source is replaced by a source of white light? (In each operation take all parameters other than the one specified to remain unchanged.)
Answer:
The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are in the same position. Therefore, the central fringe is white. For a point P for which S₂P – S₁P= λ_b /2, where (λ_b = 400 nm) represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear red in colour. Slightly farther away from where S₂Q – S₁Q = λ_b = λ_r/2 where λ_r (800 nm) is the wavelength for the red colour, the fringe will be predominantly blue. Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.

Question 70.
What is the diffraction of light? Draw a graph showing the variation of intensity with the angle in a single-slit diffraction experiment. Write one feature which distinguishes the observed pattern from the double-slit interference pattern. How would the diffraction pattern of a single-slit be affected when:
(i) the width of the slit is decreased?
(ii) the monochromatic source of light is replaced by a source of white light?
Answer:
Diffraction is the bending of light around obstacles or openings. It is a consequence of the wave nature of light. For diffraction to take place the obstacle should be of the order of the wavelength of light.

The intensity distribution for the single-slit diffraction pattern is as shown.

The intensity of all bright fringes is the same in Young’s interference pattern, but in diffraction, the intensity of bright fringes falls off on both sides of the central fringe.
We know that the angular width of the central maximum of the diffraction pattern of a single-slit is given by = \(\frac{2Dλ}{a}\)
(i) If slit width ‘a’ is decreased, the angular width will increase because
x ∝ \(\frac{1}{a}\)

(ii) When monochromatic light is replaced by white light, all the seven wavelengths form their own diffraction pattern, so coloured fringes are formed. The first few fringes are visible but due to overlapping the clarity of fringes decreases as the order increases.

Question 71.
(a) Define a wavefront. Using Huygens’ geometrical construction, explain with the help of a diagram how the plane wavefront travels from the instant t₁ to t₂ in the air.
Answer:
(a) Wavefront: It is the locus of the medium or points of a medium that is in the same phase of disturbance
First, consider a plane wave moving through free space as shown in the figure. At t = 0, the wavefront is indicated by the plane labelled AA’. In Huygens’s construction, each point on this wavefront is considered a point source. For clarity, only a few points on AA’ are shown. With these points as sources for the wavelets, we draw circles each of radius cΔt, where c is the speed of light in free space and Δt is the time of propagation from one wavefront to the next. The surface drawn tangent to these wavelets is the plane BB’, (Wavefront at a later time t). Here A₁ A₂ = B₁ B₂ = C₁C₂ = C_τ

(b) A plane wavefront is incident on a convex lens. Explain, with the help of the diagram, the shape of the refracted wavefront formed. (CBSE AI 2019)
Answer:
Spherical wavefront

Question 72.
(a) In Young’s double-slit experiment, derive the condition for
(i) constructive Interference and
(ii) destructive Interference at a point on the screen.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a wavelength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r₁ and r₂ are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,

y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe

y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,

y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

(b) A beam of light consisting of two wavelengths, 800 nm and 600 nm, is used to obtain the interference fringes in Young’s double-slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide. (CBSE Al 2012)
Answer:
Let at a distance y from central maxima the bright fringes due to wavelengths l1 and l2 coincide first time. For this to happen, nl₁ = (n + 1)l₂, where n is an integer.
or
\(\frac{n+1}{n}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{800 \times 10^{-9}}{600 \times 10^{-9}}=\frac{4}{3}\) solving for n we have n = 3

It means that at distance nth (3^th) maxima for wavelength λ₁ is just coinciding with (n + 1)^th (4^th) maxima for wavelength l₂

Question 73.
(a) How does an unpolarized light incident on a Polaroid get polarised? Describe briefly, with the help of a necessary diagram, the polarisation of light by reflection from a transparent medium.
(b) Two polaroids ‘A’ and ‘B’ are kept in a crossed position. How should a third Polaroid ‘C’ be placed between them so that the intensity of polarised light transmitted by Polaroid B reduces to 1/8th of the intensity of unpolarized light incident on A? (CBSE AI 2012)
Answer:
(a) The polariser allows only those vibrations of light to pass which are parallel to the pass axis of the polariser and blocks the remaining vibrations. Thus the light vibrations are restricted to only one plane.

Suppose an unpolarised light beam is an incident on a surface as shown in (figure a). The beam can be described by two electric field components, one parallel to the surface (the dots) and the other perpendicular to the first and to the direction of propagation (the arrows). It is found that the parallel components reflect more strongly than the other component, this results in a partially polarised beam. Furthermore, the refracted ray is also partially. polarised. Now suppose the angle of incidence 6, is varied until the angle between the reflected and the refracted beam is 90° (fig b). At this particular angle of incidence, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised.

(b) Let l0 be the intensity of unpolarized light incident on Polaroid A. Then l_o/2 will be the intensity of polarised light after passing through the first polariser A. Then the intensity of light after passing through second polariser B will be l = (l_o/2)cos²θ

where θ is the angle between pass axes of A and B. Since A and B are crossed the angle between the pass axes of B and C will be (π/2 – θ). Hence the intensity of light emerging from C will be

Therefore, we have
\(\frac{l_{0}}{8}=\frac{l_{0}}{8}\) sin² 2θ or sin 2θ = 1 or 2θ = 90°
i.e. θ = π/4

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 74.
(a) Describe any two characteristic features which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young’s double-slit experiment.
(b) In the diffraction due to a single slit experiment, the aperture of the slit is 3 mm. If monochromatic light of wavelength 620 nm is incident normally on the slit, calculate the separation between the first order minima and the 3^rd order maxima on one side of the screen. The distance between the slit and the screen is 1.5 m. (CBSE Delhi 2019)
Answer:
(a) (i) Interference pattern has a number of equally spaced bright and dark bands, while the diffraction pattern has a central bright maximum which is twice as wide as the other maxima.

(ii) In interference pattern, the intensity of all bright fringes is the same, while in diffraction pattern intensity of bright fringes goes on decreasing with the increasing order of the maxima.

Let the displacements of the waves from the sources S₁ and S₂ at a point P on the screen at any time t be given by y₁ = a cos cot and y₂ = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves.

By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is
A = 2a cos (Φ/2) …(4)

Therefore the intensity at that point is
l = A² = 4a² cos²\(\frac{Φ}{2}\) …(5)

(b) Given a = 3mm = 3 × 10^-3m, λ = 620nm = 620 × 10^-9 m, D = 1.5 m Distance of first-order minima from centre of the central maxima = X_D1 = λ_D/a

Distance of third order maxima from centre of the central maxima X_D3 = 7λ_D/2a

Distance between first order minima and third order maxima = X_D3 – X_D1

Question 75.
(a) In Young’s double-slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence deduce the expression for the fringe width.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S_D1 and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S2 and S1 at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r1 and r2 are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. Therefore, the condition for dark fringes, or destructive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,

y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe

y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,

y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

(b) Show that the fringe pattern on the screen is actually a superposition of single-slit diffraction from each slit.
Answer:
It is a broader diffraction peak in which there appear several fringes of smaller width due to the double-slit interference pattern. This is shown below:

(c) What should be the width of each slit to obtain 10 maxima of the double-slit pattern within the central maximum of the single-slit pattern, for the green light of wavelength 500 nm, if the separation between two slits is 1 mm? (CBSE AI 2015)
Answer:
Given 10 b = width of central maxima in diffraction pattern λ = 500 nm, d= 1 mm,
Now 10\(\frac{Dλ}{d}\) = \(\frac{Dλ}{a}\)
or
a = \(\frac{d}{5}=\frac{1}{5}\) = 0.2 mm

Question 76.
(a) Define a wavefront. How is it different from a ray?
Answer:
A wavefront is defined as the locus of all adjacent points at which the phase of vibration of a physical quantity associated with the wave is the same.

It is in two dimensions while a ray is in one dimension.

(b) Depict the shape of a wavefront in each of the following cases.
(i) Light diverging from a point source,
Answer:

(ii) Light emerging out of a convex lens when a point source is placed at its focus.
Answer:

(c) Using Huygen’s construction of secondary wavelets, draw a diagram showing the passage of a plane wavefront from a denser into a rarer medium. (CBSE AI 2015C)
Answer:
The diagram is as shown

Question 77.
(a) Why does unpolarised light from a source show a variation in intensity when viewed through a Polaroid which is rotated? Show with the help of a diagram how unpolarised light from sun got linearly polarised by scattering.
(b) Three identical Polaroid sheets P1, P2 and P3 are oriented so that the pass axis of P2 and P3 are inclined at angles of 60° and 90° respectively with the pass axis of P1. A monochromatic source S of unpolarised light of intensity l0 is kept in front of the polaroid sheet P! as shown in the figure. Determine the intensities of light as observed by the observer at O when Polaroid P3 is rotated with respect to P2 at angles θ = 30° and 60°. (CBSE AI 2016)

Answer:
(a) When light passes through a Polaroid it absorbs all vibrations which are not parallel to its pass axis.
It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

(b) Let l0 be the intensity of light before passing P₁ Intensity of light coming out of P₁ will be l_o/2.

Then the intensity of light after passing through second poloroid (analyser) is
l₂ = \(\frac{l_{0}}{2}\)cos²60° = \(\frac{l_{0}}{8}\)

When P₃ is rotated with respect to P₂ at an angle of 30°, then angle be¬tween the pass axis of P₂ and P₃ will be θ = 30°+ 30° = 60°.
Hence l3 = l2 cos²60° = \(\frac{l_{0}}{8} \times \frac{1}{4}=\frac{l_{0}}{32}\)
or
θ = 30° – 30° = 0°

Therefore, l₃ = l₂ cos² 0° = l₂ = l_o/8
When P₃ is rotated with respect to P₂ at an angle of 60°, then the angle between the pass axis of P₂ and P₃ will be
θ = 30° + 60° = 90°

Question 78.
(a) Derive an expression for path difference in Young’s double-slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r₁ and r₂ are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,
y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe
y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,
y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

(b) The Intensity at the central maxima in Young’s double-slit experiment is l0. Find out the intensity at a point where the path difference is λ/6, λ/4 and λ/3. (CBSE AI 2016)
Answer:
Let l be the intensity of light coming out of each slit. Since waves at the central maxima are in phase, therefore, we have

Question 79.
(a) Define a wavefront. Using Huygens’ Principle, verify the laws of reflection at a plane surface.
Answer:
The wavefront is a locus of points that oscillate in the same phase.

Consider a plane wavefront AB incident obliquely on a plane reflecting surface MM^–. Let us consider the situation when one end A of wavefront strikes the mirror at an angle i but the other end B has still to cover distance BC. The time required for this will be t = BC/c.

According to Huygen’s principle, point A starts emitting secondary wavelets and in time t, these will cover a distance c t = BC and spread. Hence, with point A as centre and BC as radius, draw a circular arc. Draw tangent CD on this arc from point C. Obviously, the CD is the reflected wavefront inclined at an angle ‘r’. As incident wavefront and reflected wavefront, both are in the plane of the paper, the 1^st law of reflection is proved.

To prove the second law of reflection, consider ΔABC and ΔADC. BC = AD (by construction),
∠ABC = ∠ADC = 90° and AC is common.

Therefore, the two triangles are congruent and, hence, ∠BAC = ∠DCA or ∠i = ∠r, i.e.the angle of reflection is equal to the angle of incidence, which is the second law of reflection.
Or
The refractive index of medium 2, w.r.t. medium 1 equals the ratio of the sine of the angle of incidence (in medium 1) to the sine of the angle of refraction (in medium 2), The diagram is as shown.

From the diagram

(b) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? Explain.
Answer:
Size of central maxima reduces to half, (Size of central maxima = 2λD/d) Intensity increases.
This is because the amount of light, entering the slit, has increased and the area, over which it falls, decreases.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the obstacle. Explain why.
Answer:
This is because of the diffraction of light. Light gets diffracted by the tiny circular obstacle and reaches the centre of the shadow of the obstacle.
β = \(\frac{Dλ}{d}\).

(i) When D is decreased, the fringe width decreases.
(ii) When d is Increased, the fringe width decreases.

Question 80.
The following figure shows an experiment set up similar to Young’s double-slit experiment to observe interference of light.

Here SS₂ – SS₁ = λ/4
Write the condition of (i) constructive,
(ii) destructive Interference at any point P in terms of path difference Δ = S₂P – S₁P.

Does the central fringe observed in the above set up tie above or below O? Give reason in support of your answer.
Yellow light of wavelength 6000°A produces fringes of width 08 mm in Young’s double-slit experiment. What will be the fringe width if the light source is replaced by another monochromatic source of wavelength 7500° A and separation between the slits is doubled?
Answer:
Given SS₂ – SS₁ = λ/4
Now path difference between the two waves from slits S₁ and S₂ on reaching point P on screen is

(a) For constructive interference at point P both difference Δx = nλ.
Therefore

where n = 0, 1, 2, 3, …

(b) For destructive interference at point P path difference

For central bright fringe, putting n = 0 in equation (1), we get
\(\frac{y d}{D}=-\frac{\lambda}{4}\)
or
y = – \(\frac{\lambda D}{4 d}\)

The -ve sign indicates that the central bright fringe will be observed below centre O of the screen, at distance below it.
Given λ₁ = 6000°A, β₁ = 0.8 mm,
λ₂ = 7500°A, β₂ = ?, d₂ = 2d₁

Using the expression
β = \(\frac{Dλ}{d}\) we have
\(\frac{\beta_{2}}{\beta_{1}}=\frac{\lambda_{2} d_{1}}{\lambda_{1} d_{2}}=\frac{7500}{6000} \times \frac{d_{1}}{2 d_{1}}\)

Solving we have β₂ = 0.5 mm

Question 81.
(a) There are two sets of apparatus of Young’s double-slit experiment. Inset A, the phase difference between the two waves emanating from the slits does not change with time, whereas in set B, the phase difference between the two waves from the slits changes rapidly with time. What difference will be observed in the pattern obtained on the screen in the two setups?
(b) Deduce the expression for the resultant intensity in both the above- mentioned setups (A and B), assuming that the waves emanating from the two slits have the same amplitude A and same wavelength λ.
Or
(a) The two polaroids, in a given setup, are kept ‘crossed’ with respect to each other. A third polaroid, now put in between these two polaroids, can be rotated. Find an expression for the dependence of the intensity of light I, transmitted by the system, on the angle between the pass axis of the first and the third polaroid. Draw a graph showing the dependence of l on θ.
(b) When an unpolarized light is an incident on a plane glass surface, find the expression for the angle of incidence so that the reflected and refracted light rays are perpendicular to each other. What is the state of polarization, of reflected and refracted light, under this condition?
Answer:
(a) Set A: Stable interference pattern, the positions of maxima and minima, does not change with time.
Set B: Positions of maxima and minima will change rapidly with time and an average uniform intensity distribution will be observed on the screen.

(b) Let the displacements of the waves from the sources S1 and S2 at a point on the screen at any time t be given by y₁ = a cos ωt and y₂ = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves. By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is A = 2a cos (Φ/2)

Therefore, the intensity at that point is l = A² = 4a² cos²\(\frac{Φ}{2}\) – 4l_o cos²\(\frac{Φ}{2}\)
Since Φ = 0, l.e. there is no phase difference, hence l = 4 l_o

Inset B, the intensity will be given by the average intensity
l = 4l_o\(\left(\cos ^{2} \frac{\phi}{2}\right)\) = 2l_o
Or
(a) Let P₁ and P₂ be the two crossed Polaroids and P₃ be the polaroid kept between the two. Let l_o, be the intensity of polarised light after passing through the first Polaroid P₁. Then the Intensity of light after passing through the second Polaroid P₂ will be,
l = l_o cos² θ
where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₂ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – 0). Hence the intensity of light emerging from P2₂ will be

The transmitted intensity will be maximum when π = π/4. The graph is as shown

(b) The ray diagram is as shown.

When light Is incident at a certain angle called Brewster angle, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised. The angle of incidence at which this occurs is called the polarising angle i_B.

From figure we see that at the polaris-ing angle
i_B + 90° + r = 180°
or
r = 90° – i_B.

Using Snell’s law we have
μ = \(\frac{\sin i_{\mathrm{B}}}{\sin r}\)

Now sin r = (90° – i_B), therefore the above expression becomes

Nature of polarisation: Reflected light and linearly polarised

Numerical Problems:

Formulae for solving numerical problems

• Fringe width is given by β = \(\frac{Dλ}{d}\)
• Brewster’s Law: μ = \(\frac{\sin \theta_{p}}{\cos \theta_{p}}\) = tan θ_p
• Intensity of light coming out of a polariser l = l_o cos² θ
• If l₁ and l₂ are the intensities of light emitted by the two sources, w₁ and w₂ be the widths of the two slits, a₁ and a₂ be the amplitudes of the waves from the two slits then,
  \(\frac{l_{1}}{l_{2}}=\frac{w_{1}}{w_{2}}=\frac{a_{1}^{2}}{a_{2}^{2}}\)
• lf l_max and lmin be the intensities of light at the maxima and the minima the
  \(\frac{l_{\max }}{l_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}\)

Question 1.
Laser light of wavelength 630 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of laser light that produces interference fringes separated by 8.1 mm using the same pair of slits. (CBSE Al 2011C)
Answer:
Given λ₁ = 630 nm, β₁ = 7.2 mm, β₂ = 8.1 mm, λ₂ =?

We know that β = \(\frac{Dλ}{d}\), for the same value of D and d we have β₂ ∝ λ,
Therefore, we have

Question 2.
What is the speed of light in a denser medium of polarising angle 30°? (CBSE Delhi 2019)
Answer:

Question 3.
Laser light of wavelength 640 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of light that produces interference fringes separated by 8.1 mm using the same arrangement. Also, find the minimum value of the order (n) of the bright fringe of the shorter wavelength which coincides with that of the longer wavelength. (CBSE AI 2012C)
Answer:
Given λ₁ = 640 nm, λ₂ = ?, b₁ = 7.2 mm and b₂ = 8.1 mm

Using the expression β = \(\frac{Dλ}{a}\) we have

Now n fringes of shorter wavelength will coincide with (n – 1) fringes of Longer wavelength
n₁λ₁ = n₂λ₂
or
n × 640 = (n – 1) × 720
Solving for n we have n =9

Question 4.
Yellow light (λ = 6000 Å) illuminates a single-slit of width 1 × 10 m. Calculate
(a) the distance between the two dark tines on either side of the central maximum, when the diffraction pattern is viewed on a screen kept 1.5 m away from the slit;
(b) the angular spread of the first diffraction minimum. (CBSE AI 2012C)
Answer:

Question 5.
Find the ratio of Intensities of two points P and Q on the screen in Young’s double-slit experiment when the waves from sources S₁ and S₂ have a phase difference of (i) π/3 and (11) π/2.
Answer:
Intensity at the screen when the phase difference is Φ = π/3 is

Now intensity at the screen when the phase difference is Φ = 90° is

Therefore, ratio of intensities is
\(\frac{l_{p}}{l_{Q}}=\frac{3l}{2l}=\frac{3}{2}\)

Question 6.
In young’s double-slit experiment, two sifts are separated by 3 mm distance and illuminated by the light of wavelength 480 nm. The screen is at 2 m from the plane of the slits. Calculate the separation between the 8th bright fringe and the 3N dark fringes observed with respect to the central bright fringe.
Answer:
Using the formuLa

Question 7.
Two coherent sources have Intensities In the ratio 25: 16. Find the ratio of the Intensities of maxima to minima, after the interference of light occurs.
Answer:
In the given problem

Question 8.
The ratio of intensities of maxima and minima in an interference pattern is found to be 25: 9. Calculate the ratio of light intensities of the sources producing this pattern.
Answer:
Given

Question 9.
In Young’s double-slit experiment using the light of wavelength 400 nm, Interference fringes of width X are obtained. Th. the wavelength of light is Increased to 600 nm and the separation between the slits Is halved. If on. wants the observed fringe width on the screen to b. the same in the two cases, find the ratio of the distance between the screen and the plan. of the Interfering sources In the two arrangements.
Answer:
Let D₁ be the distance between the screen and the sources when a tight of wavelength 400 nm Is used.

β = \(\frac{Dλ}{a}\)
or
X = \(\frac{D_{1} \times 400 \times 10^{-9}}{d}\) …(i)

Let D₂ be the distance between the screen and the sources to obtain the same fringe width, when light of wavelength 600 nm is used. Then,
X = \(\frac{D_{2} \times 600 \times 10^{-9}}{d}\) ….(ii)

From the equations (i) and (ii), we have
\(\frac{D_{1}}{D_{2}}=\frac{600 \times 10^{-9}}{400 \times 10^{-9}}\) = 1.5

Question 10.
In Young’s slit experiment, Interference fringes are observed on a screen, kept at D from the slits. If the screen Is moved towards the slits by 5 × 10^-2 m, the change in fringe width Is found to be 3 × 10^-5 m. If the separation between the slits is 10-3 m, calculate the wavelength of the light used.
Answer:
Given ΔD = 5 × 10^-2-2 m, Δβ = 3 × 10^-5-5 m, d = 10^-3 m,

Using the relation β = \(\frac{Dλ}{a}\) we have for the two cases
β₁ = \(\frac{D_{1} \lambda}{d}\) and
β₂ = \(\frac{D_{2} \lambda}{d}\) subtracting we have
β₁ – β₂ = \(\frac{λ}{d}\)(D1 – D2)
or

Question 11.
A slit of width ‘a’ is illuminated by monochromatic light of wavelength 700 nm at Normal Incidence. Calculate the value of ‘a’ for the position of
(a) the first minimum at an angle of diffraction of 30°.
(b) first maximum at an angle of diffraction of 30°.
Answer:
Given X = 700 nm = 7 × 10^-7 m

Question 12.
Estimate the angular separation between . first order maximum and third order minimum of the diffraction pattern due to a single-slit of width 1 mm, when light of wavelength 600 nm is incident normal on it. (CBSE AI 2015C)
Answer:
Given d = 1 mm = 10^-3 m,
λ = 600 nm = 6 × 10^-7 m,
For first order maxima
d sin θ = n λ
or
sin θ = \(\frac{n \lambda}{d}=\frac{6 \times 10^{-7}}{10^{-3}}\) = 6 × 10^-4
or
θ_max = 1.047°
Now for minima we have
d sin θ = (2n+1 )\(\frac{λ}{2}\)

For third-order minima we have n = 3
Therefore we have

sin θ = (2 × 3 + 1)\(\frac{λ}{2d}\) = \(\frac{7λ}{2d}\)
= \(\frac{7 \times 6 \times 10^{-7}}{2 \times 10^{-3}}\)
= 21 × 10^-4
or
θ_min = 6.397°

Therefore, angular separation is
θ_min – θ_max = 6.397 – 1.047 = 5.35°