Geometrical Optics

Question 1:

Question 2:

No, because this formula is only valid when the angles of incidence and refraction are small. When the angles of incidence and refraction are large, we cannot use relations sini = tani and sinr = tanr to derive the formula for the apparent depth.

Question 3:

There is only one such situation. Light rays go undeviated through a prism only when the prism is kept in a medium that has the same refractive index as that of the prism itself. In all other situations, there will always be some minimum deviation in the path of a light ray passing through the prism.

Question 4:

A diamond shines more because of the phenomenon of total internal reflection (TIR). The refractive index of diamond is ≈2.4 and that of glass is ≈1.5.
By using the relation $\mu =\frac{1}{\sin C}$, where C is the critical angle, we get refractive index (μ).
For a large refractive index, the value of the critical angle is small. Thus, for the diamond, the critical angle for total internal reflection is much smaller than that of the glass. Therefore, a great percentage of incident light gets internally reflected several times before it emerges out.

Question 5:

The beam of light will seem to appear and disappear, just like the twinkling of the stars. This is because the refractive index of the material in the slab fluctuates slowly with time. So, the light rays get refracted differently with time and this causes them to get concentrated during certain times, or get diffused at other times.

Question 6:

No, a plane mirror can never form a real image. This is because in all possible situations of image formation, the light rays never actually meet after getting reflected, but they only appear to meet behind the mirror, forming a virtual image always.

Question 7:

No, the paper will not burn even after sufficient time. This is because the piece of paper is placed at the position of a virtual image of a strong light source, so light rays are not actually converging at that point, but seem to converge. As no light rays are concentrating on the point, the paper will never burn.
Instead, if the paper is placed at the position of a real image, it will burn as the light rays are actually concentrating on the point. Also, when we have the real image of a virtual source, it will also cause the burning of paper due to same reason; the light rays are actually meeting at the point.

Question 8:

Yes, a virtual image can be photographed by a camera. This is because the light rays emitting from a virtual image and reaching the lens of the camera are real. A virtual image is formed from a reflecting surface after reflection of real incident rays, so these rays enter the camera to provide effects on the photographic film. Just like a virtual image can be seen by us in a mirror with our eyes, it can be photographed by a camera.

Question 9:

The special function of a convex mirror is that it creates the image of a distant object that is reduced in size, is upright or erect and always lies within the virtual focal length of the mirror. A plane mirror cannot do this. Also, as the image is formed within the focal length, the image is close to the mirror as well as is small in size, enabling the driver to clearly view the nearer vehicles behind the motor vehicle.

Question 10:

The image of the object moves slower compared to the object. It can be explained using the mirror formula :$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
We know that for a convex mirror, the object distance (u) is positive, image distance (v) is negative and the focal length (f) is also negative. Thus mirror formula of a convex mirror is:$\frac{1}{u}-\frac{1}{v}=-\frac{1}{f}$
As u = +ve
$$\begin{gathered} \frac{1}{v}-\frac{1}{f}>0 \\ \frac{1}{v}>\frac{1}{f} \\ v<f \end{gathered}$$
Therefore, the image is always formed within the focal length of the mirror. Thus, the distance moved by the image is much slower than the distance moved by the object.

Question 11:

When viewed from the water, the friend will seem taller than his usual height.

Let actual height be h and the apparent height be h'.
Here, the refraction is taking place from rarer to denser medium and a virtual image is formed.
Using
$\frac{{\mu }_1}{-u}+\frac{{\mu }_2}{v}=\frac{{\mu }_2-{\mu }_1}{R}$
Where refractive index of water is μ₂ and refractive index of air is μ₁.
u and v are object and image distances, respectively.
R  is the radius of curvature, here we will take it as ∞.
$$\begin{gathered} \frac{{\mu }_1}{-u}+\frac{{\mu }_2}{v}=\frac{{\mu }_2-{\mu }_1}{\infty } \\ \frac{{\mu }_1}{u}=\frac{{\mu }_2}{v} \\ v=\frac{{\mu }_{2 }}{{\mu }_1}\times u \end{gathered}$$
As μ₁= 1
v = u × μ₂
We know magnification is given by:
$m=\frac{v}{u}$
Putting the value of v in the above equation:
$$\begin{gathered} m=\frac{u\times {\mu }_2}{u} \\ m={\mu }_2 \end{gathered}$$
As the magnification is greater than 1, so the apparent height seems to be greater than actual height.

Question 12:

Proof:
$$\begin{gathered} \frac{{\mu }_2}{v}-\frac{{\mu }_1}{u} = \frac{{\mu }_2-{\mu }_1}{R} \\ \mathrm{Now} R=\infty \\ \frac{{\mu }_2}{v}-\frac{{\mu }_1}{u} = \frac{{\mu }_2-{\mu }_1}{\infty } \\ \frac{{\mu }_2}{v}-\frac{{\mu }_1}{u} =0 \\ \frac{{\mu }_2}{v}=\frac{{\mu }_1}{u} \\ \frac{{\mu }_1}{{\mu }_2} = \frac{u}{v} \\ \mathrm{But} \frac{u}{v}=\frac{\mathrm{Real} \mathrm{depth}/\mathrm{height}}{\mathrm{Apparent} \mathrm{depth}/\mathrm{height}} \\ \therefore \frac{\mathrm{Real} \mathrm{depth}/\mathrm{height}}{\mathrm{Apparent} \mathrm{depth}/\mathrm{height}}=\frac{{\mu }_{\mathit{1}}}{{\mu }_{\mathit{2}}} \end{gathered}$$

Question 13:

Yes. Using the lens maker formula we can show it. We know that the formula is:
$\frac{1}{f}=\left(\frac{{\mu }_2}{{\mu }_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Where, f is the focal length of the thin converging lens.
μ₂ and μ₁ are refractive indexes of lens and air respectively.
R₁ and R₂ are the radius of curvature of convex and plane surfaces respectively.
Here, R₂ = ∞ because of the plane surface.

Case 1
When the convex surface is facing the object, we have:
$$\begin{gathered} \frac{1}{f}=\left(\frac{{\mu }_2}{{\mu }_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{\infty }\right) \\ \frac{1}{f}=\left(\frac{{\mu }_2}{{\mu }_1}-1\right)\frac{1}{R_1} ...\left(\mathrm{i}\right) \end{gathered}$$

Case 2

When the plane surface is facing the object, we have:
$$\begin{gathered} \frac{1}{f}=\left(\frac{{\mu }_2}{{\mu }_1}-1\right)\left(\frac{1}{\infty }-\frac{1}{R_1}\right) \\ \frac{1}{f}=-\left(\frac{{\mu }_2}{{\mu }_1}-1\right)\frac{1}{R_1} ...\left(\mathrm{ii}\right) \end{gathered}$$
For Case 1, the focal length is positive and for the Case 2 the focal length is negative. Thus, the image distance is different in both cases.

Question 14:

Yes, we can say so with certainty, as only a diverging lens can cause a parallel beam of light to diverge.

Question 15:

It will act as a diverging lens. This is because the refractive index of air is less than that of water. Thus, light will diverge after passing through the bubble.

Question 16:

Let the two converging lenses be L₁ and L₂, with focal lengths f₁ and f₂ respectively.
To reduce the aperture of a parallel beam of light without losing the energy of the light and also increase the intensity, we have to place the lens L₂ within the focal range of L_1.

Question 17:

No, the focal length of mirror will not change. This is because the focal length of a mirror does not depend on the refractive index of the medium in which the light rays travel.

Question 18:

Yes, the focal length of the lens will change. This is because its focal length is dependent on the medium in which the light rays travel.

Question 19:

No, mirrors cannot give rise to chromatic aberration. This is because chromatic aberration occurs due to the refraction of different colours of light. In case of mirrors, refraction of light does not take place.

Question 1:

No, there will not be a significant chromatic aberration because a laser light is monochromatic in nature. Therefore, all light will get converged at the same point.

Question 2:

(a) All the reflected rays meet at a point when produced backward.

Here, the angle of reflection is equal to the angle of incidence. Therefore, all rays get reflected to converge at a single point to form the point image of the point source.

Question 3:

(b) light goes from optically denser medium to rarer medium

    In this case the incident angle is greater than critical angle, so the light gets reflected back into the same denser medium, instead of being refracted .

Question 4:

(c) form nearly a point image of a point source

     Since, when reflected back, they meet at a single point forming a point image of a point source.

Question 5:

(d) 15 cm behind the mirror

Since u = − 30 cm
v = ?
f = 30 cm

From the mirror formula:
$$\begin{gathered} \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\ \mathrm{or}\frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ \Rightarrow \frac{1}{v}=\frac{1}{30}-\frac{1}{-30} \\ \frac{1}{v}=\frac{1}{30}+\frac{1}{30} \\ \frac{1}{v}=\frac{2}{30} \\ \therefore v=15 \mathrm{cm} \end{gathered}$$

Question 6:

(a) is plane

    This is because the reflected rays are still parallel, which is only possible if the mirror is a plane mirror. A spherical mirror will either converge or diverge the reflected rays.

Question 7:

(c) is certainly real if the object is virtual

As a virtual object is the virtual image of the object, it is always formed beyond the focus of the concave mirror. Thus, the image formed by a concave mirror of the virtual object will always be real.

Question 8:

(d) two images form, one at O' and the other at O"

Light will be refracted differently in both mediums on the right side. Thus, two images will be formed, one at O' due to refraction from medium μ₁ and another at O" due to refraction from medium μ₃.μ1

Question 9:

(c) $\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}$

The thickness of lens will cause a shift in the position of object and image from the ideal condition, which will be:
$$\begin{gathered} u_{\mathrm{new}}=u +t(1 + \frac{1}{\mu } ) \mathrm{and} v_{\mathrm{new}}=v - t(1 + \frac{1}{\mu } ) \\ or u_{\mathrm{new}}=u + t + \frac{t}{\mu } \mathrm{and} v_{\mathrm{new}}=v - t + \frac{t}{\mu } \end{gathered}$$
As $\frac{t}{\mu }$ will be quite small, it can be ignored.
Thus, u_new = u + t and v_new = v − t
We get the new lens formula:
$\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}$

Question 10:

(b) f = R

As from lens maker formula:
$$\begin{gathered} \frac{1}{f}=(m - 1)(\frac{1}{R_1}-\frac{1}{R_2}) \\ \Rightarrow \frac{1}{f}=(1.5 - 1)(\frac{1}{R}-\frac{1}{-R}) (\mathrm{As} \mathrm{given} R_1=R_2=R) \\ \frac{1}{f}=(0.5)(\frac{1}{R}+\frac{1}{R}) \\ \frac{1}{f}=(0.5)\left(\frac{2}{R}\right) \\ f = R \end{gathered}$$=R

Question 11:

(c) 2 f

Since the object is placed at 2 f, the image of the object will be formed at distance of 2 f from a converging lens.
It can also be shown from the lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Here, u = − 2 f and f = f
On putting the respective values we get:
$$\begin{gathered} \frac{1}{v}-\frac{1}{-2f}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{2f} \\ =\frac{1}{2f} \end{gathered}$$
Therefore, image distance v = 2 f

Question 12:

(d) first increases then decreases

Since all beams of light are parallel to the principal axis, they will converge at the focus of a converging lens. Thus, the intensity of light increases till one reaches the focus and then starts decreasing as one moves beyond it.

Question 13:

(a) 2 D

The lens is cut into two equal parts by a plane perpendicular to the principal axis. Thus, the radius of curvature of both the lenses become half of its original value.
As,
 $$\begin{gathered} f=\frac{R}{2} \\ P=\frac{1}{f} \end{gathered}$$
Radius of curvature becomes half. Therefore, the focal also reduces to half its original value.
Thus, the power (P) of the two cut-lenses will be equal.
2 P = 4 D
P = 2 D

Question 14:

(c) 4 D

As the lens is cut along the principal axis, the two new lens will act as two double convex lenses. Even after cutting the lens, the radius of the curvature of the two new lens will remain the same as that of the original lens, with the power equal to the original lens.

Question 15:

(c) increases

The focal length of the combination will increase. For finding out the combination of lens we have the formula:
$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{\mathrm{d}}{f_1{f}_2}$
Where, F is the focal length for the combination
d is the separation between two lenses
Here, d = 0
$$\begin{gathered} \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2} \\ F = \frac{f_1{f}_2}{f_1+f_2} \end{gathered}$$
Hence, the focal length will increase.

Question 16:

(a) a convergent lens

Since the refractive index of lens is more than that of of water and both the sides are convex, it will act like a normal convergent lens (though its focal length will change).

Question 17:

(b) a divergent lens

Since the refractive index of lens is less than that of of water and both the sides are convex, it will act like a normal diverging lens (though its focal length will change to negative).

Question 18:

(b) 2.5 cm

As the focal length of the lens is 20 cm and object distance is 40 cm from the lens, the image is formed at the centre of curvature at the right side of the lens.

From right angled triangle ABC,
$$\begin{gathered} \tan \alpha =\frac{\mathrm{AB}}{\mathrm{BC}} \\ \tan \alpha =\frac{5 }{40} \\ \alpha ={\tan }^{-1}\left(\frac{5 }{40}\right) \end{gathered}$$
and from right angled triangle, we have:
$$\begin{gathered} \tan \alpha =\frac{\mathrm{DE}}{\mathrm{DC}} \\ \tan \alpha =\frac{h}{20} \\ \mathrm{as} \mathrm{DB}-\mathrm{BC}=\mathrm{DC}=20 \mathrm{cm} \end{gathered}$$

Putting the value of angle alpha, we get:
$$\begin{gathered} \tan \left({\tan }^{-1}\left(\frac{5 }{40}\right)\right)=\frac{h}{20} \\ \frac{5 }{40}=\frac{h}{20} \\ \Rightarrow h=2.5 \mathrm{cm} \end{gathered}$$

Question 1:

(d) chromatic aberration

When light rays of different colours do not converge at the same point after passing through a converging lens, it is called chromatic aberration. This happens because a lens has different refractive indices for different colours, i.e, for different wavelengths of light.

Question 2:

Using sign conventions, given,
Distance of object from mirror, u = − 30 cm,
Radius of curvature of concave mirror R = − 40 cm



Using the mirror equation,

$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{2}{R} \\ \Rightarrow \frac{1}{v}=\frac{2}{R}-\frac{1}{u} \\ \Rightarrow \frac{1}{v}=\frac{2}{-40}-\frac{1}{-30}=\frac{1}{-20}+\frac{1}{30} \\ \Rightarrow \frac{1}{v}=\frac{-30+20}{30\times 20}=\frac{-10}{30\times 20} \\ \Rightarrow \frac{1}{v}=-\frac{1}{60} \end{gathered}$$

or, v = − 60 cm
Hence, the required image will be located at a distance of 60 cm in front of the concave mirror.

Question 3:

Given,
Height of the object, h₁ = 20 cm,
Distance of image from screen v = −5.0 m = −500 cm,



Since, we know that
$$\begin{gathered} -\frac{v}{u}=\frac{h_2}{h_1} \\ \mathrm{or} \frac{-(-500)}{u}=\frac{50}{20} \end{gathered}$$
Where 'u' is the distance of object from screen.
(As the image is inverted)
$$\begin{gathered} \mathrm{or} u=-500\times \frac{2}{5} \\ =-200 \mathrm{cm}=-2.0 \mathrm{m} \end{gathered}$$

Using mirror formula,
$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \mathrm{or} \frac{1}{-5}+\frac{1}{-2}=\frac{1}{f} \end{gathered}$$
$$\begin{gathered} \mathrm{or} -\frac{1}{f}=\frac{7}{10} \\ \mathrm{or} f=-\frac{10}{7}=-1.44 \mathrm{m} \end{gathered}$$
Hence, the required focal length of the concave mirror is 1.44 m.

Question 4:

Using sign conventions, given,
Focal length of the concave mirror:
f = −20 cm
As per the question,
Magnification (m) is:
$m=-\frac{v}{u}=2$
⇒ v = −2u



 Case I (Virtual image):

Using mirror formula,
$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{2u}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{2u}=\frac{1}{f} \\ \Rightarrow u=\frac{f}{2}=10 \mathrm{cm} \end{gathered}$$

Case II (Real image)

Using mirror formula,
$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow -\frac{1}{2u}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{3}{2u}=\frac{1}{f} \\ \Rightarrow u=\frac{3f}{2}=30 \mathrm{cm} \end{gathered}$$
Hence, the required positions of objects are 10 cm or 30 cm from the concave mirror.

Question 5:

Given,
Height of the object, h₁ = 1 cm
Focal length of the concave mirror, f = 7.5 cm = $\frac{15}{2} \mathrm{cm}$
Magnification is given as,
$$\begin{gathered} m=-\frac{v}{u}=\frac{h_i}{h_0}=0.6 \\ \mathrm{Using} \mathrm{mirror} \mathrm{equation}, \\ \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{0.6u}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{5}{3u}-\frac{1}{u}=\frac{2}{15} \\ \Rightarrow u=5 \mathrm{cm} \end{gathered}$$

Hence, the distance of the object from the mirror is 5 cm.

Question 6:

Given,
Height (h₁) of the candle flame taken as object AB = 1.6 cm
Diameter of the ball bearing (d) = 0.4 cm
So, radius = 0.2 cm
Distance of object, u = 20 cm



Using mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{2}{R}$

Putting the values according to sign conventions, we get,
$$\begin{gathered} \frac{1}{(-20)}+\frac{1}{v}=\frac{2}{0.2} \\ \Rightarrow \frac{1}{v}=\frac{1}{20}+10 \\ \Rightarrow v=0.1 \mathrm{cm} \mathrm{or} 1.0 \mathrm{mm} \mathrm{insi}\mathrm{de} \mathrm{the} \mathrm{ball} \mathrm{bearing}. \\ \mathrm{Magnification}=m \\ =\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=-\frac{v}{u}=m \\ =\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{200}=\frac{1}{200} \\ \Rightarrow \mathrm{A}\text{'}\mathrm{B}\text{'}=\frac{\mathrm{AB}}{200}=+\frac{1.6}{200}=+0.08 \mathrm{cm} (+0.008 \mathrm{cm}) \end{gathered}$$

Hence, the distance of the image is 1 cm.
Height of the image is 0.008 cm.

Question 7:

Given,
Height of the object AB, h₁ = 3 cm
Distance of the object from the convex mirror, u = −7.5 cm
Focal length of the convex mirror (f) = 6 cm
Using mirror formula,
$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \end{gathered}$$
Putting values according to sign convention,
$$\begin{gathered} \frac{1}{v}=\frac{1}{6}-\frac{1}{(-7.5)} \\ \frac{1}{v}=\frac{1}{6}+\frac{1}{7.5}=\frac{3}{10} \\ \Rightarrow v=\frac{10}{3} \mathrm{cm} \end{gathered}$$
Magnification = m = $-\frac{v}{u}=\frac{10}{7.5\times 3}$
$$\begin{gathered} \Rightarrow \frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=\frac{10}{7.5\times 3} \\ \Rightarrow \mathrm{A}\text{'}\mathrm{B}\text{'}=\frac{100}{75}=\frac{4}{3}=1.33 \mathrm{cm} \end{gathered}$$
Where A'B' is the height of the image.

Hence, the required location of the image is $\frac{10}{3} \mathrm{cm}$ from the pole and image height is 1.33 cm. Nature of the image is virtual and erect.

Question 8:

Given,
Radius of curvature of concave mirror, R = 20 cm
So its focal length will be f = $\frac{R}{2}$ = −10 cm
For part AB of the U shaped wire, PB = 30 + 10 = 40 cm



Therefore, u = −40 cm
Using mirror formula,
$\Rightarrow \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
${\displaystyle \frac{1}{v}}=-\frac{1}{10}-\frac{1}{-40}=-\frac{3}{40}$
$\Rightarrow v=\frac{-40}{3}=13.3 \mathrm{cm}$
$\mathrm{So}, \mathrm{PB}\text{'}=13.3 \mathrm{cm}$
$m=\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=\frac{-\left(v\right)}{u}$
$$\begin{gathered} =\frac{-(-13.3)}{-40}=-\frac{1}{3} \\ \Rightarrow \mathrm{A}\text{'}\mathrm{B}\text{'}=-\frac{10}{3}=-3.33 \mathrm{cm} \end{gathered}$$

For part CD of the U shaped wire, PC = 30 cm
Therefore, u = −30 cm
Again, using mirror equation:
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$$\begin{gathered} {\displaystyle \frac{1}{v}}=-\frac{1}{10}-\frac{1}{-30} \\ {\displaystyle \frac{1}{v}}=-\frac{1}{10}+\frac{1}{30}=-\frac{1}{15} \end{gathered}$$
⇒ v = −15 cm = PC'
$m=\frac{\mathrm{C}\text{'}\mathrm{D}\text{'}}{\mathrm{CD}}=-\frac{v}{u}$
$=-\frac{(-15)}{-30}=-\frac{1}{2}$
⇒ C'D' = 5 cm
⇒ B'C' = PC' − PB'
= 15 − 13.3 = 1.7 cm

Hence, the total length of the U- shaped wire is A'B' + B'C' + C'D'
= (3.3) + (1.7) + 5 = 10 cm

Question 1:

Given,
Distance of the man's face (here, taken as object), u = −25 cm
According to the question, magnification, m = 1.4
   $$\begin{gathered} m=\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=-\frac{v}{u} \\ \Rightarrow 1.4=-\frac{\left(v\right)}{-25} \\ \Rightarrow \frac{14}{10}=\frac{v}{25} \\ \Rightarrow v=\frac{25\times 14}{10}=35 \mathrm{cm} \end{gathered}$$

Using equation of mirror, we get:
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\Rightarrow \frac{1}{f}=\frac{1}{35}-\frac{1}{25}$
$=\frac{5-7}{175}=-\frac{2}{175}$
⇒ f = −87.5

Hence, the required focal length of the concave mirror is 87.5 cm.

Question 2:

(a) reflection
(b) refraction

When the light strikes on a surface nearly parallel to it, it then bends by a small and fixed angle after reflection. Also, when the light travels from one medium to another with slight differences in their refractive indices, it bends by a small angle. Thus, the bending of light by a small but fixed angle can be the case of either reflection or refraction.

Question 3:

(b) If the final rays are converging, we have a real image.

This is because a real image is formed by converging reflected/refracted rays from a mirror/lens.

Question 4:

(a) Pole
(c) Radius of curvature
(d) Principal axis

Paraxial rays are the light rays close to the principal axis. The focus of the spherical mirror for paraxial rays are different from the focus for marginal rays. So, the focus depends on whether the rays are paraxial or marginal. â€‹The pole, radius of curvature and the principal axis of a spherical mirror do not depend on paraxial or marginal rays.

Question 5:

(c) the object is real but the image is virtual
(d) the object is virtual but the image is real

The virtual image of a real object and the real image of a virtual object are always erect.

Question 6:

(c) the image will not be shifted
(d) the intensity of the image will decrease

If the upper half portion of the convex lens is painted, then only the intensity of the image will decrease, as the amount of light passing through the lens will decrease. Also, there will be no shift in the position of the image because all the parameters remain the same. 

Question 7:

(a) there will be an image at O₁
​(b) there will be an image at O₂ 

The lens L₁ converges the light at point O₁ and lens L₂ converges the light at O₂. As the upper half of lens L₃ has a refractive index equal to that of L₁, it will converge the light at O₁ and thus the image will be formed at O₁. Also, the lower half of lens L₃ has a refractive index equal to that of lens L₂, it will converge the light at O₂ and thus the image will be formed at O₂.

Question 9:

(b) must be greater than 20 cm

Let the image be formed at a distance of x cm from the lens.
Therefore, the distance of the object from the lens, u​, will be = (40 − x) cm
From lens formula:
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{f}=\frac{1}{x}-\frac{1}{40-x} \\ \Rightarrow \frac{1}{f}=\frac{40-x-x}{40x-x^2} \\ \Rightarrow f=\frac{40x-x^2}{40-2x} \\ \Rightarrow f(40-2x)=40x-x^2 \\ \Rightarrow x^2-2fx-40x+40f=0 \\ \Rightarrow x^2-(2f+40)x+40f=0 \end{gathered}$$

Therefore, we get x as:

$x=\frac{(2f+40)\pm \sqrt{(2f+40{)}^2-160f}}{2}$

Question 10:

Given,
Focal length of the concave mirror, f = − 7.6 m
Distance between earth and moon taken as object distance, u = −3.8 × 10⁵ km
Diameter of moon = 3450 km
Using mirror equation:
$\Rightarrow \frac{1}{v}=\frac{1}{u}+\frac{1}{f}$
$\therefore \frac{1}{v}+\left(-\frac{1}{3.8\times {10}^8}\right)=\left(-\frac{1}{7.6}\right)$

As we know, the distance of moon from earth is very large as compared to focal length it can be taken as ∞.
Therefore, image of the moon will be formed at focus, which is inverted.
$\Rightarrow \frac{1}{v}=-\frac{1}{7.6}$
⇒ v = −7.6 m

We know that magnification is given by:
$m=-\frac{v}{u}=\frac{d_{\mathrm{image}}}{d_{\mathrm{object}}}$
$\Rightarrow \frac{-(-7.6)}{(-3.8\times {10}^8)}=\frac{d_{image}}{3450\times {10}^3}$
$d_{\mathrm{image}}=-\frac{3450\times 7.6\times {10}^3}{3.8\times {10}^8}$
= $-$0.069 m = $-$6.9 cm
Hence, the required diameter of the image of moon is 6.9 cm.

Question 11:

Given,
Distance of the circle from the mirror taken as object distance, u = −30 cm,
Focal length of the concave mirror, f = −20 cm



Using mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}+\left(-\frac{1}{30}\right)=-\frac{1}{20}$

$\Rightarrow \frac{1}{v}=\frac{1}{30}-\frac{1}{20}=\frac{1}{60}$
⇒ v = $-$60 cm
Therefore, image of the circle is formed at a distance of 60 cm in front of the mirror.
We know magnification (m) is given by:
$m=-\frac{v}{u}=\frac{R_{\mathrm{image}}}{R_{\mathrm{object}}}$
$\Rightarrow -\frac{(-60)}{(-30)}=\frac{R_{\mathrm{image}}}{2}$
Where R_object and R_image are radius of the object and radius of the image, respectively.
⇒ R_image = 4 cm
Hence, the required radius of the circle formed by the image is 4 cm.

Question 12:

Given,
A concave mirror of radius 'R' kept on a horizontal table.
'h' is height up to which the water is poured into the concave mirror.
Let the object be placed at height 'x' above the surface of water.
We know if we place the object at the centre of curvature of the mirror, then the image itself will be formed at the centre of curvature.
Therefore, the apparent position of the object with respect to the mirror should be at the centre of curvature so that the image is formed at the same position.



Since,
$\frac{\mathrm{Real} \mathrm{depth}}{\mathrm{Apparent} \mathrm{depth}}=\frac{1}{\mu }$
(with respect to mirror)
$$\begin{gathered} \mathrm{Now}, \frac{x}{R\mathit{-}h}\mathit{=}\frac{\mathit{1}}{\mu } \\ \mathit{\Rightarrow }\mathit{ }x\mathit{=}\frac{R\mathit{-}h}{\mu } \end{gathered}$$
Hence, the object should be placed at $\frac{R-h}{\mathrm{\mu }}$ above the water surface.

Question 13:

Given,
Two converging mirrors having equal focal length 'f '.
Both the mirrors will produce one image under two conditions:

Case-1 
When the point source is at the centre of curvature of the mirrors, i.e, at a distance of '2f ' from each mirror, the images will be produced at the same point 'S'. Therefore, d = 2f + 2f = 4f


Case-II 
When the point source 'S' is at focus, i.e., at a distance 'f ' from each mirror, the rays from the source after reflecting from one mirror will become parallel and so these parallel rays, after the reflection from the other mirror the object itself. Thus, only one image is formed.
Here d = f + f = 2f

Question 14:

Given,
Converging mirror M₁ with focal length (f₁) = 20 cm
Converging mirror M₂ with focal length (f₂) = 20 cm
f₁ = f₂ = 20 cm = f
Point source is at a distance of 30 cm from M₁.



As the 1^st reflection is through the mirror M₁,
u = −30 cm
f = −20 cm

Using mirror equation:
$\Rightarrow \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}+\frac{1}{-30}=-\frac{1}{20}$
$$\begin{gathered} \Rightarrow \frac{1}{v}=-\frac{1}{20}+\frac{1}{30} \\ \Rightarrow \frac{1}{v}=-\frac{1}{60} \\ \Rightarrow v=-60 \mathrm{cm} \end{gathered}$$

and for the 2^nd reflection at mirror M₂,
u = 60 − (30 + x) = 30 − x
v = − x, f = 20 cm

Again using the mirror equation:
$\Rightarrow \frac{1}{30-x}-\frac{1}{x}=\frac{1}{20}$

$\Rightarrow \frac{x-30+x}{x(30-x)}=\frac{1}{20}$
⇒ 40x − 600 = 30x − x²
⇒ x² + 10x − 600 = 0

$\Rightarrow x=\frac{10+50}{2}=\frac{40}{2}$
= 20 cm or − 30 cm
∴ Total distance between the two lines is 20 + 30 = 50 cm

(b) Location of the image formed by the single reflection from M₂ = 60 $-$ 50 = 10

Thus, the image formed by the single reflection from M₂ is at a distance of 10 cm from mirror M_2.

Question 15:

Given,
Angle of incidence, i = 45°
Angle of refraction, r = 30°
Using Snell's law,
$\frac{\sin i}{\sin r}=\frac{3\times {10}^8}{v}$
$=\frac{\sin 45°}{\sin 30°}=\frac{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}{\left({\displaystyle \frac{1}{2}}\right)}$
$=\frac{2}{\sqrt{2}}=\sqrt{2}$
$$\begin{gathered} \mathrm{So}, \\ v=\frac{3\times {10}^8}{\sqrt{2}} \mathrm{m}/s \end{gathered}$$

Let x be the distance travelled by light in the slab.
Now,
$x=\frac{1 \mathrm{m}}{\cos 30°}=\frac{2}{\sqrt{3}} \mathrm{m}$
We know:
Time taken $=\frac{\mathrm{Distance}}{\mathrm{Speed}}$
$=\frac{2}{\sqrt{3}}\times \frac{\sqrt{2}}{3\times {10}^8}$
= 0.54 × 10⁻⁸
= 5.4 × 10⁻⁹ s

Question 16:

Given,
Length of the pole = 1.00 m
Water level of the swimming pool is 50.0 cm higher than the bed.
Refractive index (μ) of water = 1.33



According to the figure, shadow length = BA' = BD + DA'= 0.5 + 0.5 tan r
Using Snell's law:
$$\begin{gathered} \mathrm{Now} 1.33=\frac{\sin 45°}{\sin r} \\ \Rightarrow \sin r=\frac{1}{1.33\sqrt{2}}=0.53 \\ \Rightarrow \cos r=\sqrt{1-{\sin }^2 r} \\ =\sqrt{1-(0.53{)}^2}=0.85 \\ \mathrm{So}, \tan r=0.6235 \end{gathered}$$

Therefore, shadow length of the pole  = (0.5)$\times$(1 + 0.6235) = 0.81175 m
= 81.2 cm

Question 17:

Given,
Depth of the lake = 2.5 m
Refractive index (μ) of water = $\frac{4}{3}$
When the sun is just setting, θ is approximately = 90Ëš



Therefore, incidence angle is 90Ëš
Using Snell's law:

$\therefore \frac{\sin i}{\sin r}=\frac{{\mathrm{\mu }}_2}{{\mathrm{\mu }}_1}$
$\Rightarrow \frac{1}{\sin r}=\frac{{\displaystyle \frac{4}{3}}}{1}$
$\Rightarrow \sin r=\frac{3}{4}$
$\Rightarrow r=49°$

From the figure, W'O = x is the distance of the shadow
$\mathrm{Thus}, \left({\displaystyle \frac{x}{2.5}}\right)=\tan r=1.15$
$\Rightarrow x=2.5\times 1.15=2.8 \mathrm{m}$

Question 18:

Given,
Thickness of the glass slab, d = 2.1 cm
Refractive index, μ = 1.5
Shift due to the glass slab is given by,
$∆t=\left[1-\left(\frac{1}{\mathrm{\mu }}\right)\right]d$
$=\left[1-\left(\frac{1}{1.5}\right)\right]\left(2.1\right)$
$=\frac{1}{3}\left(2.1\right)=0.7 \mathrm{cm}$
Hence, the microscope should be shifted by 0.70 cm to focus the object 'P' again.