Light Waves

Question 1:

Question 2:

White light is a composition of seven colours. These are violet, indigo, blue, green, yellow, orange and red, collectively known as VIBGYOR. Spectrum of white light consists of  sever colour bands. Each band consists of some range of wavelengths or frequencies. 
For orange colour : (590 nm to 620 nm)
For red colour: (620 nm to 780 nm)

So, the colour of 620 nm and 780 nm lights may be different. But the colour of 620 nm light and 621 nm light is same.

Question 3:

Colour of light will depend only on the frequency of light and not on the wavelength of the light. So, light will appear red to an observer under water. 

Question 4:

The width of the central band is inversely proportional to the slit width. So, as the width of the slit is increased, the central band will become less wider and further bands will start merging in them. Hence, diffraction effects will be visible less clearly.

Question 5:

Light waves have the property of travelling in a straight line, unlike sound waves. When we put a cardboard between the light source and our eyes, the light waves are obstructed by the cardboard and cannot reach our eyes, which doesn't happen when the cardboard is inserted between sound source and our ear.

Question 6:

To receive TV signals transmitted from Delhi in Patna directly, one has to use antennas of great height, which will cost much. On the other hand, transmission of signals with the help of satellites requires only high frequency waves and can be done easily.

Question 7:

Young's double slit experiment can be performed with sound waves, as the sound waves also show interference pattern. To get a reasonable "fringe pattern", the separation of the slits should be of the order of the wavelength of the sound waves used.
In this experiment, the bright and dark fringes can be detected by measuring the intensity of sound using a microphone or any other detector.

Question 8:

Interference pattern can be studied with waves of unequal intensity.
 ${\rm I}={{\rm I}}_{\mathit{1}}+{{\rm I}}_{\mathit{2}}+\mathit{2}\sqrt{\left[{{\rm I}}_{\mathit{1}}\times {{\rm I}}_{\mathit{2}}\right]}\mathit{cos}\left(\varphi \right)$ ,
where $\varphi =\mathrm{phase} \mathrm{difference}$.
In case of waves of unequal intensities, the contrast will not be clear, as the minima will not be completely dark.

Question 9:

The interference pattern can be produced by any two coherent waves moving in the same direction. It cannot be concluded from the interference phenomenon that light is a transverse wave, as sound waves that are longitudinal in nature also interfere.

Question 10:

In order to get interference, the sources should be coherent, i.e. they should emit wave of the same frequency and a stable phase difference. Two candles that are placed close to each other are distinct and cannot be considered as coherent sources. Two independent sources cannot be coherent. So, two different laser sources will also not serve the purpose. 

Question 11:

The fringe width in Young's double slit experiment depends on the separation of the slits.
$\chi =\frac{\lambda D}{d}$,
where
           $$\begin{gathered} \lambda =\mathrm{wavelength} \\ \chi =\mathrm{fringe} \mathrm{width} \\ D=\mathrm{distance} \mathrm{between} \mathrm{slits} \mathrm{and} \mathrm{screen} \\ d=\mathrm{separation} \mathrm{between} \mathrm{slits} \end{gathered}$$
On increasing d, fringe width decreases. If the separation is increased too much, the fringes will merge with each other and the fringe pattern won't be detectable.

Question 1:

The violet filter will allow only violet light to pass through it. Now, if the double slit experiment is performed with the white light and violet light, the fringe pattern will not be the same as obtained by just using white light as the source. To have interference pattern, the light waves entering from the slits should be monochromatic. So, in this case, the violet light will superimpose with only violet light (of wavelength 400 nm) in such a way that the bright bands will be of violet colour and the minima will be completely dark.

Question 2:

(c) both a particle and a wave phenomenon
       
Light shows photoelectric effect and Compton effect, which depicts its particle nature. It also shows interference and diffraction, which depicts the wave nature of light. 

Question 3:

(d) neither on elasticity nor on inertia

The speed of light in any medium depends on the refractive index of that medium, which is an intensive property. Hence, speed of light is not affected by the elasticity and inertia of the medium.

Question 4:

(d) electric field

Light consists of mutually perpendicular electric and magnetic fields. So, the equation of a light wave is represented by its field vector.

Question 5:

(d) Polarization 

Reflection, interference and diffraction are the phenomena shown by both transverse waves and longitudinal waves. Polarization is the phenomenon shown only by transverse waves.

Question 6:

(c) its wavelength decreases but frequency remains unchanged

Frequency of a light wave, as it travels from one medium to another, always remains unchanged, while wavelength decreases.

Decrease in the wavelength of light entering a medium of refractive index $\mu$ is given by
 $$\begin{gathered} {\lambda }_{{\rm M}}=\frac{\lambda }{\mu }, \\ where {\lambda }_{{\rm M}}=wavelength in medium \\ \lambda =wavelength in vacuum \\ \mu =refractive index \end{gathered}$$

Question 7:

(b) Frequency

Frequency of a light wave doesn't change on changing the medium of propagation of light. 

Question 8:

(a) ${\lambda }_a>{\lambda }_f$

An electromagnetic wave bends round the corners of an obstacle if the size of the obstacle is comparable to the wavelength of the wave. An AM wave has less frequency than an FM wave. So, an AM wave has a higher wavelength than an FM wave and it bends round the corners of a 1 m $\times$ 1m board.
λ

Question 9:

(d) A laser

Among the given sources, laser is the best coherent source providing monochromatic light with constant phase difference.

Question 10:

(d) ${\cos }^{-1}\left(1/\sqrt{3}\right)$
On writing the given equation in the plane equation form lx + my + nz = p,
where l² + m² + n² = 1 and p>0, we get:

$\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=\frac{c}{\sqrt{3}}$
If $\theta$ is the angle between the normal and +X axis, then 
$$\begin{gathered} \cos \theta =\frac{1}{\sqrt{3}} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{gathered}$$

Question 11:

(a) plane

Wave travelling from a distant source always has plane wavefront.

Question 12:

(a) point source

Intensity of a point source obeys the inverse square law.
Intensity of light at distance r from the point source is given by
$I=S/\left(4{\mathrm{\pi r}}^2\right)$ ,
where S is the source strength.

Question 13:

(d) having a constant phase difference
For light waves emitted by two sources of light to remain coherent, the initial phase difference between waves should remain constant in time. If the phase difference changes continuously or randomly with time, then the sources are incoherent.  

Question 1:

(d) interference of light 

Interference effect is produced by a thin film ( coating of a thin layer of a translucent material on a medium of different refractive index which allows light to pass through it))In the present case, oil floating on water forms a thin film on the surface of water, leading to the display of beautiful colours in daylight because of the interference of sunlight.

Question 2:

Given:
Range of wave length is
$400 \mathrm{nm}<\lambda <700 \mathrm{nm}$
We know that frequency is given by $f=\frac{c}{\lambda }$,
$\mathrm{where} \mathrm{c}=\mathrm{speed} \mathrm{of} \mathrm{light}=3\times {10}^8 \mathrm{m}/\mathrm{s}$
          f is the frequency
          λ is the wavelength
We can write wavelength as:
$$\begin{gathered} \frac{1}{700 \mathrm{nm}}<\frac{1}{\lambda }<\frac{1}{400 \mathrm{nm}} \\ \Rightarrow \frac{1}{7\times {10}^{-7} \mathrm{m}}<\frac{1}{\lambda }<\frac{1}{4\times {10}^{-7} \mathrm{m}} \\ \frac{3\times {10}^8}{7\times {10}^{-7}} \mathrm{Hz}<\frac{c}{\lambda }<\frac{3\times {10}^8}{4\times {10}^{-7}} \mathrm{Hz} \\ \Rightarrow 4.3\times {10}^{14 }\mathrm{Hz}<\frac{c}{\lambda }<7.5\times {10}^{14} \mathrm{Hz} \\ \Rightarrow 4.3\times {10}^{14} \mathrm{Hz}<f<7.5\times {10}^{14} \mathrm{Hz} \end{gathered}$$

Hence, frequency of the range of light that is visible to an average human being is $4.3\times {10}^{14} \mathrm{Hz} \mathrm{to}7.5\times {10}^{14} \mathrm{Hz}$.

Question 3:

Given:
Wavelength of sodium light in air, ${\lambda }_{\mathrm{a}}=589 \mathrm{nm}=589\times {10}^{-9} \mathrm{m}$
Refractive index of water, μ_w= 1⋅33
We know that $f=\frac{c}{\lambda }$,
$\mathrm{where} c=\mathrm{speed} \mathrm{of} \mathrm{light}=3\times {10}^8 \mathrm{m}/\mathrm{s}$
          f  = frequency
          λ = wavelength
(a) Frequency in air, $f_{\mathrm{air}}=\frac{c}{{\lambda }_{\mathrm{a}}}$
$$\begin{gathered} f_{\mathrm{air}}=\frac{3\times {10}^8}{589\times {10}^{-9}} \\ =5.09\times {10}^{14} \mathrm{Hz} \end{gathered}$$

(b)
Let wavelength of sodium light in water be ${\lambda }_{\mathrm{w}}$.
We know that
$\frac{{\mathrm{\mu }}_{\mathrm{a}}}{{\mathrm{\mu }}_{\mathrm{w}}}=\frac{{\lambda }_{\mathrm{\omega }}}{{\lambda }_{\mathrm{a}}}$,
where μ_a is the refractive index of air which is equal to 1 and
           λ_w is the wavelength of sodium light in water.
$$\begin{gathered} \Rightarrow \frac{1}{1.33}=\frac{{\lambda }_{\mathrm{\omega }}}{589\times {10}^{-9}} \\ \Rightarrow {\lambda }_{\mathrm{\omega }}=443 \mathrm{nm} \end{gathered}$$

(c) Frequency of light does not change when light travels from one medium to another.
$$\begin{gathered} \therefore f_{\mathrm{\omega }}=f_{\mathrm{a}} \\ =5.09\times {10}^{14} \mathrm{Hz} \end{gathered}$$

(d) Let the speed of sodium light in water be ${\nu }_{\mathrm{\omega }}$
     and speed in air, v_a = c.
     Using $\frac{{\mu }_{\mathrm{a}}}{{\mu }_{\mathrm{\omega }}}=\frac{{\nu }_{\mathrm{\omega }}}{{\nu }_{\mathrm{a}}}$, we get:
$$\begin{gathered} {\nu }_{\mathrm{\omega }}=\frac{{\mu }_{\mathrm{a}}c}{{\mu }_{\mathrm{\omega }}} \\ =\frac{3\times {10}^8}{\left(1.33\right)}=2.25\times {10}^8 \mathrm{m}/\mathrm{s} \end{gathered}$$

Question 4:

Given:
Refractive index of fused quartz for light of wavelength 400 nm is 1.472.
And refractive index of fused quartz for light of wavelength 760 nm is 1.452.
We known that refractive index of a material is given by
μ = $\frac{\mathrm{Speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{vacuum}}{\mathrm{Speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{the} \mathrm{material}}=\frac{c}{v}$
Let speed of light for wavelength 400 nm in quartz be v₄₀₀.
So,
$$\begin{gathered} 1.472=\frac{3\times {10}^8}{v_{400}} \\ \Rightarrow v_{400}=2.04\times {10}^8 \mathrm{m}/\mathrm{s} \end{gathered}$$
Let speed of light of wavelength 760 nm in quartz be v₇₆₀.
Again, $\frac{1.452}{1}=\frac{3\times {10}^8}{{\nu }_{760}}$
$\Rightarrow v_{760}=2.07\times {10}^8 \mathrm{m}/\mathrm{s}$

Question 5:

Given:
Speed of yellow light in liquid (v_L)= 2⋅4 × 10⁸ m s⁻¹
And  speed of yellow light in air speed = v_a 
Let μ_L be the refractive index of the liquid
Using, ${\mathrm{\mu }}_{\mathrm{L}}=\frac{\mathrm{speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{vaccum}}{\mathrm{velocity} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{medium}}=\frac{c}{v_{\mathrm{L}}}$
${\mathrm{\mu }}_{\mathrm{L}}=\frac{3\times {10}^8}{\left(2.4\right)\times {10}^8}=1.25$

Hence, the required refractive index is 1.25.

Question 14:

Given:
Separation between two narrow slits, d = 1 cm = 10⁻² m
Wavelength of the light, $\lambda =5\times {10}^{-7} \mathrm{m}$
Distance of the screen, $D=1 \mathrm{m}$
(a)
    We know that separation between two consecutive maxima =  fringe width (β).
That is, $\beta =\frac{\lambda D}{d}$    ...(i)
       $$\begin{gathered} =\frac{5\times {10}^{-7}\times 1}{{10}^{-2}} \mathrm{m} \\ =5\times {10}^{-5} \mathrm{m}=0.05 \mathrm{mm} \end{gathered}$$

(b)
Separation between two consecutive maxima = fringe width
∴ $\beta =1 \mathrm{mm}={10}^{-3} \mathrm{m}$
Let the separation between the sources be 'd'
Using equation (i), we get:
$$\begin{gathered} d\text{'}=\frac{5\times {10}^{-7}\times 1}{{10}^{-3} } \\ \Rightarrow d\text{'}=5\times {10}^{-4} \mathrm{m}=0.50 \mathrm{mm}. \end{gathered}$$

Question 15:

(c) 9 : 4

Ratio of maximum intensity and minimum intensity is given by
$$\begin{gathered} \frac{I_{max}}{I_{min}}=\frac{{\left(\sqrt{I_1} +\sqrt{I_2}\right)}^2}{{\left(\sqrt{I_1}-\sqrt{I_2}\right)}^2}=\frac{25}{1} \\ \Rightarrow \sqrt{I_1}=3 and \sqrt{I_2}=2 \\ \Rightarrow I_1=9 and I_2=4 \end{gathered}$$
Then,
$\frac{I_1}{I_2}=\frac{9}{4}$

Question 16:

(b) I₀/​4

Total intensity coming from the source is I₀  which is present at the central maxima. In case of two slits, the intensity is getting distributed between the two slits and for a single slit, the amplitude of light coming from the slit is reduced to half  which leads to 1/4th of intensity. 

Question 17:

(c) remain same

On the introduction of a transparent sheet in front of one of the slits, the fringe pattern will shift slightly but the width will remain the same.

Question 1:

(a) the fringe width will decrease

As fringe width is proportional to the wavelength and wavelength of light is inversely proportional to the refractive index of the medium,
Here,   $$\begin{gathered} {\lambda }_{{\rm M}}=\lambda /\eta \\ {\lambda }_{{\rm M}}=\mathrm{wavelength} \mathrm{in} \mathrm{medium} \\ \lambda =\mathrm{wavelength} \mathrm{in} \mathrm{vacuum} \\ \eta =\mathrm{refractive} \mathrm{index} \mathrm{of} \mathrm{medium} \end{gathered}$$
Hence, fringe width decreases when Young's double slit experiment is performed under water.

Question 2:

(a) in vacuum
(c) in a material medium

Light is an electromagnetic wave that can travel through vacuum or any optical medium.

Question 3:

(b) Speed of light in water is smaller than its speed in vacuum.
(c) Light shows interference. 

Snell's Law, which states that the speed of light reduces on moving from a rarer to a denser medium, can be concluded from Huygens' wave theory and interference of light waves is based on the wave properties of light.

Question 4:

(b) have zero average value
(c) are perpendicular to the direction of propagation of light
(d) are mutually perpendicular

Light is an electromagnetic wave that propagates through its electric and magnetic field vectors, which are mutually perpendicular to each other, as well as to the direction of propagation of light. The average value of both the fields is zero.

Question 5:

(c) find the new position of a wavefront
(d) explain Snell's Law

Huygen's wave theory explains the origin of points for the new wavefront proceeding successively. It also explains the variation in speed of light on moving from one medium to another, i.e. it proves Snell's Law.

Question 6:

(c) v_{A =} v_{B =} v_C  
(d) ${\nu }_B=\frac{1}{2}\left(v_A+v_C\right)$

Question 7:

 (a) v_A > v_B > v_C
(d) v_B= (v_A + v_C )/2 

In any other medium, the speed of light is given by $v=c/\eta , \mathrm{where} \mathrm{\eta } \mathrm{is} \mathrm{the} \mathrm{refractive} \mathrm{index} \mathrm{of} \mathrm{the} \mathrm{medium}$ and according to Doppler effect,  for an observer moving towards the source ,speed of light appears to be  more than the other two cases. On the other hand, it will be least when the observer is moving away from the source. 

Question 8:

(a) x = c

​The wave is travelling along the X-axis. So, it'll have planar wavefront perpendicular to the X-axis.

Question 9:

(b) consecutive fringes will come closer

Fringe width,  $\beta =\lambda D/d$.
Wavelength of red light is greater than wavelength of violet light; so, the fringe width will reduce.

Question 10:

(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(d) The fringe next to the central will be violet.

The superposition of all the colours at the central maxima gives the central band a white colour. As we go from the centre to corner, the fringe colour goes from violet to red. There will not be a completely dark fringe, as complete destructive interference does not take place.

Question 6:

(a) (i) and (ii)
(d) (iii) and (iv).

The waves are travelling with the same frequencies and varying by constant phase difference. 

Question 7:

Given:
Separation between consecutive dark fringes = fringe width (β) = 1 mm = 10⁻³ m
Distance between screen and slit (D) = 2.5 m
The separation between slits (d) = 1 mm = 10⁻³ m
Let the wavelength of the light used in experiment be λ.
We know that
$\beta =\frac{\lambda D}{d}$
$$\begin{gathered} {10}^{-3} \mathrm{m}=\frac{2.5\times \lambda }{{10}^{-3}} \\ \Rightarrow \lambda =\frac{1}{2.5 }{10}^{-6} \mathrm{m} \\ =4\times {10}^{-7} \mathrm{m}=400 \mathrm{nm} \end{gathered}$$

Hence, the wavelength of light used for the experiment is 400 nm.

Question 8:

Given:
Separation between the two slits, $d=1 \mathrm{mm}={10}^{-3} \mathrm{m}$
Wavelength of the light used, $\lambda =5.0\times {10}^{-7} \mathrm{m}$
Distance between screen and slit, $D=1 \mathrm{m}$
(a) The distance of the centre of the first minimum from the centre of the central maximum, x = $\frac{\mathrm{width} \mathrm{of} \mathrm{central} \mathrm{maxima}}{2}$
That is, $x=\frac{\beta }{2}=\frac{\lambda D}{2d}$    ...(i)
       $$\begin{gathered} =\frac{5\times {10}^{-7}\times 1}{2\times {10}^{-3}} \\ =2.5\times {10}^{-4} \mathrm{m}=0.25 \mathrm{mm} \end{gathered}$$
(b) From equation (i),
fringe width, $\beta =2\times x=0.50 \mathrm{mm}$
So, number of bright fringes formed in one centimetre (10 mm) = $\frac{10}{0.50}=20$.

Question 9:

Given:
Separation between two narrow slits, $d=0.8 \mathrm{mm}=0.8\times {10}^{-3} \mathrm{m}$
Wavelength of the yellow light, $\lambda =589 \mathrm{nm}=589\times {10}^{-9} \mathrm{m}$
Distance between screen and slit, $D=2.0 \mathrm{m}$
Separation between the adjacent bright bands = width of one dark fringe
That is, $\beta =\frac{\lambda D}{d}$    ...(i)
$$\begin{gathered} \Rightarrow \beta =\frac{589\times {10}^{-9}\times 2}{0.8\times {10}^{-3}} \\ =1.47\times {10}^{-3} \mathrm{m} \\ =1.47 \mathrm{mm}. \end{gathered}$$

Hence, the adjacent bright bands in the interference pattern are 1.47 mm apart.  

Question 10:

Given:
Wavelength of the blue-green light, $\lambda =500\times {10}^{-9} \mathrm{m}$
Separation between two slits, $d=2\times {10}^{-3} \mathrm{m},$
Let angular separation between the consecutive bright fringes be θ.
$$\begin{gathered} \mathrm{Using} \theta =\frac{\mathit{ }\beta }{D}=\frac{\lambda D}{dD}=\frac{\lambda }{d},\text{ we get:} \\ \theta =\frac{500\times {10}^{-9}}{2\times {10}^{-3}} \\ =250\times {10}^{-6} \\ =25\times {10}^{-5} \mathrm{radian} \mathrm{or} 0.014° \end{gathered}$$
Hence, the angular separation between the consecutive bright fringes is 0.014 degree. 

Question 11:

Given:
Wavelengths of the source of light,
${\lambda }_1=480\times {10}^{-9} \mathrm{m} \mathrm{and} {\lambda }_2=600\times {10}^{-9} \mathrm{m}$
Separation between the slits, $d=0.25 \mathrm{mm}=0.25\times {10}^{-3} \mathrm{m}$
Distance between screen and slit, $D=150 \mathrm{cm}=1.5 \mathrm{m}$
We know that the position of the first maximum is given by
 $y=\frac{\lambda D}{d}$
So, the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths = y₂ − y₁
$y_2-y_1=\frac{D\left(y_2-y_1\right)}{d}$

$$\begin{gathered} \Rightarrow y_2-y_1=\frac{1.5}{0.25\times {10}^{-3}}\left(600\times {10}^{-9}-480\times {10}^{-9}\right) \\ y_2-y_1=72\times {10}^{-5} \mathrm{m} =0.72 \mathrm{mm} \end{gathered}$$

Question 12:

Let the separation between the slits be d and distance between screen from the slits be D.
Suppose, the m^th bright fringe of violet light overlaps with the n^th bright fringe of red light.
Now, the position of the m^th bright fringe of violet light, y_v = $\frac{m{\lambda }_{v}D}{d}$
Position of the n^th bright fringe of red light, y_r = $\frac{n{\lambda }_{r}D}{d}$
For overlapping, y_{v =}y_r .
So, as per the question,
$$\begin{gathered} \frac{m\times 400\times \mathrm{D}}{d}=\frac{n\times 700\times \mathrm{D}}{d} \\ \Rightarrow \frac{m}{n}=\frac{7}{4} \end{gathered}$$
Therefore, the $7^{\mathrm{th}}$ bright fringe of violet light overlaps with the 4^th bright fringe of red light.
It can also be seen that the 14^th violet fringe will overlap with the 8^th red fringe.
Because, $\frac{m}{n}=\frac{7}{4}=\frac{14}{8}$

Question 13:

Given:
The refractive index of the plate is $\mu$.
Let the thickness of the plate be 't' to produce a change in the optical path difference of $\frac{\lambda }{2}$.
We know that optical path difference is given by $\left(\mathrm{\mu }-1\right)t$.
$$\begin{gathered} \therefore \left(\mathrm{\mu }-1\right)t=\frac{\lambda }{2} \\ \Rightarrow t=\frac{\lambda }{2\left(\mathrm{\mu }-1\right)} \end{gathered}$$
Hence, the thickness of a plate is $\frac{\lambda }{2\left(\mathrm{\mu }-1\right)}$.

Question 14:

Given:
Refractive index of the plate is μ.
The thickness of the plate is t.
Wavelength of the light is λ.
(a)
When the plate is placed in front of the slit, then the optical path difference is given by $\left(\mathrm{\mu }-1\right)t$.

(b) For zero intensity at the centre of the fringe pattern, there should be distractive interference at the centre.
So, the optical path difference should be = $\frac{\lambda }{2}$
$$\begin{gathered} \mathrm{i}.\mathrm{e}. \left(\mu -1\right) t=\frac{\lambda }{2} \\ \Rightarrow t=\frac{\lambda }{2 \left(\mu -1\right)} \end{gathered}$$

Question 15:

Given:
Refractive index of the paper, μ = 1.45
The thickness of the plate, $t=0.02 \mathrm{mm}=0.02\times {10}^{-3} \mathrm{m}$
Wavelength of the light, $\lambda =620 \mathrm{nm}=620\times {10}^{-9} \mathrm{m}$
We know that when we paste a transparent paper in front of one of the slits, then the optical path changes by $\left(\mathrm{\mu }-1\right)t$.
And optical path should be changed by λ for the shift of one fringe.
∴ Number of fringes crossing through the centre is
$$\begin{gathered} n=\frac{\left(\mathrm{\mu }-1\right)t}{\lambda } \\ =\frac{\left(1.45-1\right)\times 0.02\times {10}^{-3}}{620\times {10}^{-9}} \\ =14.5 \end{gathered}$$

Hence, 14.5 fringes will cross through the centre if the paper is removed.

Question 16:

Given:
Refractive index of the mica sheet,μ = 1.6
Thickness of the plate, $t=1.964 \mathrm{micron}=1.964\times {10}^{-6} \mathrm{m}$
Let the wavelength of the light used = λ.
Number of fringes shifted is given by
$n=\frac{\left(\mathrm{\mu }-1\right)t}{\lambda }$
So, the corresponding shift in the fringe width equals the number of fringes multiplied by the width of one fringe.
                                         $$\begin{gathered} \mathrm{Shift} =n \times \beta \\ =\frac{\left(\mu -1\right)t}{\lambda }\times \frac{\lambda D}{d} \\ =\frac{\left(\mu -1\right)t\times D}{d} ...\left(\mathrm{i}\right) \end{gathered}$$
As per the question, when the distance between the screen and the slits is doubled,
i.e. $D\text{'}=2D$,
fringe width, $\beta =\frac{\lambda D\text{'}}{d}=\frac{\lambda 2D}{d}$
According to the question, fringe shift in first case = fringe width in second case.
$$\begin{gathered} \mathrm{So}, \frac{\left(\mathrm{\mu }-1\right)t\times \mathrm{D}}{d}=\frac{\lambda 2\mathrm{D}}{d} \\ \Rightarrow \lambda =\frac{\left(\mu -1\right) t}{2} \\ =\frac{\left(1.6-1\right)\times \left(1.964\right)\times {10}^{-6}}{2} \\ =589.2\times {10}^{-9}=589.2 \mathrm{nm} \end{gathered}$$

Hence, the required wavelength of the monochromatic light is 589.2 nm.

Question 17:

Given:
The thickness of the strips = $t_1=t_2=t=0.5 \mathrm{mm}=0.5\times {10}^{-3} \mathrm{m}$
Separation between the two slits,$\mathit{ }d=0.12 \mathrm{cm}=12\times {10}^{-4} \mathrm{m}$
The refractive index of mica, μ_m = 1.58 and of polystyrene, μ_p = 1.58
Wavelength of the light, $\mathrm{\lambda }=590 \mathrm{nm}=590\times {10}^{-9} \mathrm{m},$
Distance between screen and slit, D = 1 m

(a)
We know that fringe width is given by
     $\beta =\frac{\lambda D}{d}$ 
$$\begin{gathered} \Rightarrow \beta =\frac{590\times {10}^{-9}\times 1}{12\times {10}^{-4}} \\ =4.9\times {10}^{-4} \mathrm{m} \end{gathered}$$

(b) When both the mica and polystyrene strips are fitted before the slits, the optical path changes by
$$\begin{gathered} ∆x=\left({\mathrm{\mu }}_{\mathrm{m}}-1\right) t- \left({\mathrm{\mu }}_{\mathrm{p}}-1\right) t \\ =\left({\mathrm{\mu }}_{\mathrm{m}}-{\mathrm{\mu }}_{\mathrm{p}}\right) t \\ =\left(1.58-1.55\right)\times \left(0.5\right) \left({10}^{-3}\right) \\ =\left(0.015\right)\times {10}^{-3} \mathrm{m} \end{gathered}$$
∴ Number of fringes shifted, n = $\frac{∆x}{\lambda }$.
$\Rightarrow n=\frac{0.015\times {10}^{-3}}{590\times {10}^{-9}}=25.43$
∴ 25 fringes and 0.43^th of a fringe.
⇒ In which 13 bright fringes and 12 dark fringes and 0.43^th of a dark fringe.
So, position of first maximum on both sides is given by
On one side,
$$\begin{gathered} x=\left(0.43\right)\times 4.91\times {10}^{-4} \left(\because \beta =4.91\times {10}^{-4} \mathrm{m}\right) \\ =0.021 \mathrm{cm} \end{gathered}$$
On the other side,
$$\begin{gathered} x\text{'}=\left(1-0.43\right)\times 4.91\times {10}^{-4} \\ =0.028 \mathrm{cm} \end{gathered}$$

Question 18:

Given:
Refractive index of the two slabs are µ₁ and µ₂.
Thickness of both the plates is t.
When both the strips are fitted, the optical path changes by
$$\begin{gathered} ∆x=\left({\mathrm{\mu }}_1-1\right) t- \left({\mathrm{\mu }}_2-1\right) t \\ =\left({\mathrm{\mu }}_1-{\mathrm{\mu }}_2\right) t \end{gathered}$$
For minimum at P₀, the path difference should be $\frac{\lambda }{2}$.
$\mathrm{i}.\mathrm{e}. ∆x=\frac{\lambda }{2}$
$$\begin{gathered} \mathrm{So},\frac{\lambda }{2}=\left({\mathrm{\mu }}_1-{\mathrm{\mu }}_2\right)t \\ \Rightarrow t=\frac{\lambda }{2\left({\mathrm{\mu }}_1-{\mathrm{\mu }}_2\right)} \end{gathered}$$
Therefore, minimum at point P₀ is $\frac{\lambda }{2\left({\mathrm{\mu }}_1-{\mathrm{\mu }}_2\right)}$.

Question 19:

Given:
The thickness of the thin paper, $t=0.02 \mathrm{mm}=0.02\times {10}^{-3} \mathrm{m}$
Refractive index of the paper, $\mathrm{\mu }=1.45$.
Wavelength of the light,  $\lambda =600 \mathrm{nm}=600\times {10}^{-9} \mathrm{m}$
(a)
Let the intensity of the source without paper = I₁
and intensity of source with paper =I₂
Let a₁ and a₂ be corresponding amplitudes.
As per the question,
$I_2=\frac{4}{9}{I}_1$
We know that
$$\begin{gathered} \frac{I_1}{I_2}=\frac{{a_1}^2}{{a_2}^2} \left(\because I\propto a^2\right) \\ \Rightarrow \frac{a_1}{a_2}=\frac{3}{2} \end{gathered}$$
Here, a is the amplitude.
$$\begin{gathered} \mathrm{We} \mathrm{know} \text{that } \frac{I_{\max }}{I_{\min }}=\frac{{\left(a_1+a_2\right)}^2}{{\left(a_1-a_2\right)}^2}. \\ \Rightarrow \frac{I_{\max }}{I_{\min }}=\frac{{\left(3+2\right)}^2}{{\left(3-2\right)}^2} \\ =\frac{25}{1} \\ \Rightarrow I_{\max }:I_{\min }=25 : 1 \end{gathered}$$

(b)
Number of fringes that will cross through the centre is given by $n=\frac{\left(\mu -1\right)t}{\lambda }$.
$$\begin{gathered} \Rightarrow n=\frac{\left(1.45-1\right)\times 0.02\times {10}^{-3}}{600\times {10}^{-9}} \\ =\frac{0.45\times 0.02\times {10}^4}{6}=15 \end{gathered}$$

Question 20:

Given:
Separation between two slits, $d=0.28 \mathrm{mm}=0.28\times {10}^{-3} \mathrm{m}$
Distance between screen and slit (D) = 48 cm = 0.48 m
Wavelength of the red light, ${\lambda }_a=700 \mathrm{nm} \mathrm{in} \mathrm{vaccum} =700 \times {10}^{-9} \mathrm{m}$
Let the wavelength of red light in water = ${\lambda }_{\omega }$
We known that refractive index of water (μ_w =4/3), 
μ_w = $\frac{\mathrm{Speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{vacuum}}{\mathrm{Speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{the} \mathrm{water}}$
$$\begin{gathered} \mathrm{So}, {\mathrm{\mu }}_{\mathrm{w}}=\frac{v_a}{v_{\omega }}=\frac{{\lambda }_a}{{\lambda }_{\omega }} \\ \Rightarrow \frac{4}{3}=\frac{{\lambda }_a}{{\lambda }_{\omega }} \\ \Rightarrow {\lambda }_{\omega }= \frac{3{\lambda }_a}{4}=\frac{3\times 700}{4}=525 \mathrm{nm} \end{gathered}$$
So, the fringe width of the pattern is given by
$$\begin{gathered} \mathrm{\beta }=\frac{{\lambda }_{\omega }D}{d} \\ =\frac{525\times {10}^{-9}\times \left(0.48\right)}{\left(0.28\right)\times {10}^{-3}} \\ =9\times {10}^{-4}=0.90 \mathrm{mm} \end{gathered}$$

Hence, fringe-width of the pattern formed on the screen is 0.90 mm.