NCERT Exemplar Class 12 Physics Solutions Chapter 9 Ray Optics and Optical Instruments

NCERT Exemplar Class 12 Physics Solutions Chapter 9 MCQ I

Question:9.1

A ray of light incident at an angle on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is made of a material of a refractive index 1.5, the angle of incidence is
A. $7.5^{o}$
B. $5^{o}$
C.$15^{o}$
D. $2.5^{o}$

Answer:

The correct answer is the option (A) $7.5^{o}$
As it is a thin prism, the distance between the refracting surfaces is non-accountable and (A), the prism angle is also very small. Since angle (A) is small, it implies that both $r_{1}$ and $r_{2}$ are also small.

According to snell's law,

1 $\sin i_{1}=\mu \; \sin r_{1}\Rightarrow i_{1}=\mu\; r_{1}$
Also, 1 $\sin i_{2}=\mu \; \sin r_{2}\Rightarrow i_{2}=\mu\; r_{2}$
Therefore, deviation,

$\delta =\left ( i_{1}-r_{1} \right )+\left ( i_{2}-r_{2} \right )$
$\Rightarrow \; \; \; \; \delta =\left ( i_{1}+i_{2} \right )-\left ( r_{1}+r_{2} \right )=\left ( r_{1}+r_{2} \right )\left ( \mu -1 \right )$
$\Rightarrow \delta =A\left ( \mu -1 \right )$
Since, deviation $\delta =A\left ( \mu -1 \right )A$
$\delta=\left ( 1.5-1 \right )\times 5^{o}=2.5^{o}$
The angle of the prism is $5^{o}$. The ray emrerges from refracting face of a prism normally.
Then, $i_{2}=r_{2}=0$
As $A= r_{1}+r_{2}\Rightarrow r_{i}=A\; or\; r_{1}=5^{o}$
But $i_{1}=\mu .r_{1}=\frac{3}{2}\times 5=7.5^{o}$

Question:9.2

A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is
A. blue
B. green
C. violet
D. red

Answer:

The correct answer is the option (D)

The emerging colour would be red, because of the relation $v=f\lambda ,$ which provides the velocity of a wave. During the changing of the mediums, the frequency of the wave remains constant. So, $v\; \alpha \lambda$ or wavelength is directly proportional to speed. As red colour has the highest wavelength in the spectra, it also has the highest speed according to tot eh relation. Hence, when the light travels in the slab, the first colour to emerge is red.

Question:9.3

An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image
A. moves away from the lens with a uniform speed 5 m/s.
B. moves away from the lens with uniform acceleration.
C. moves away from the lens with a non-uniform acceleration.
D. moves towards the lens with a non-uniform acceleration.

Answer:

The correct answer is the option (C).
The image will moves away from the lens with a non-uniform acceleration), in this case, the objects are moving with a uniform speed of 5 m/s towards a convergent lens from the left, thus with a non-uniform acceleration, the image moves away from the lens. The image starts at a uniform speed but gets accelerated going from 2F to F; the image goes from 2F to infinity. When the image is at 2F, both the object and image will have the same speed.

Question:9.4

A passenger in an aeroplane shall
A. never see a rainbow.
B. may see a primary and a secondary rainbow as concentric circles.
C. may see a primary and a secondary rainbow as concentric arcs.
D. shall never see a secondary rainbow.

Answer:

The answer is the option (B)
A passenger has an aeroplane has the possibility to see a primary and a secondary rainbow as concentric circles because the plane is at an extremely high altitude.
In case of an object moving towards the convex lens with a constant speed from infinity to focus, the image will move slower in the beginning and then faster.
$v_{i}=\left ( \frac{f}{f+u} \right )^{2}v_{o}$
And if an object approaches the lens, the image moves away from the lens with a non-uniform acceleration.

Question:9.5

You are given four sources of light each one providing a light of a single colour – red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90°. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?
A.The beam of red light would undergo total internal reflection.
B. The beam of red light would bend towards normal while it gets refracted through the second medium.
C. The beam of blue light would undergo total internal reflection.
D. The beam of green light would bend away from the normal as it gets refracted through the second medium.

Answer:

The correct answer is the option (C).
The beam of blue light would undergo total internal reflection, so option (c) is the correct answer.
According to the Cauchy relationship, smaller the wavelength higher the refractive index and consequently smaller the critical angle

We know $v=f\lambda,$ the frequency of wave remains unchanged with medium hence $v\alpha \lambda$
The critical angle,
$\sin \; C=\frac{1}{\mu }$
Also, velocity of light,
$v\alpha \frac{1}{\mu }$
According to VIBGYOR, among all given sources of light, the blue light have the smallest wavelength.

As $\lambda _{blue}<\lambda _{yellow}$, hence $v _{blue}<v _{yellow}$, it means $\mu _{blue}<\mu _{yellow}$
It means the critical angle for blue is less than yellow colour, the critical angle is the least, which facilitates total internal reflection for the beam of blue light.

Question:9.6

The radius of curvature of the curved surface of a Plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will
A. act as a convex lens only for the objects that lie on its curved side.
B. act as a concave lens for the objects that lie on its curved side.
C. act as a convex lens irrespective of the side on which the object lies.
D. act as a concave lens irrespective of side on which the object lies.

Answer:

The correct answer is the option (C) act as a convex lens irrespective of the side on which the object lies
Key concept : The relation between $f,\mu ,R_{1}$ and $R_{2}$ is known as lens maker's formula and it is
$\frac{1}{f}=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )$
$R_{1}=\infty ,R_{2}=-R$
$f=\frac{R}{\left ( \mu -1 \right )}$

Here, R=20 cm, $\mu =1.5$.

On substituting the values, we get
$f=\frac{R}{\mu -1}=\frac{20}{15-1}=40\; cm$
As $f>0$ means converging nature.

Therefore, the lens acts as a convex lens irrespective of the side on which the object lies.

Question:9.7

The phenomenon involved in the reflection of radio waves by ionosphere is similar to
A. reflection of light by a plane mirror.
B. total internal reflection of light in air during a mirage.
C. dispersion of light by water molecules during the formation of a rainbow.
D. scattering of light by the particles of air.

Answer:

The correct answer is the option (B)
The phenomena are very similar to total internal reflection of light in the air during a mirage. The Ionosphere layer of atmosphere around earth is responsible for reflecting the radio waves. And as the angle of incidence is greater than critical, it is very similar to total internal reflection of light in the air during a mirage.

Question:9.8

The direction of ray of light incident on a concave mirror is shown by PQ while directions in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4 (Fig 9.1). Which of the four rays correctly shows the direction of reflected ray?

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

The correct answer is the option (B)

The ray PQ incident to the concave mirror passes through the focus F, after reflection, as shown in the diagram is supposed to be parallel to the principal axis.
Important Note:
To find out the exact extended graphical location of the image of any object, one can draw any of the given two rays:
1. Draw a ray reflected through the focus of the mirror, which is initially parallel to the principal axis (1).
2. When a ray is drawn through the centre of curvature, it reflects back along with itself (3).
3. Draw a ray that is initiated through the focus; it will be reflected parallel to the principal axis (2).
4. A ray drawn incident to the pole gets reflected back in a symmetrical manner.

Question:9.9

The optical density of turpentine is higher than that of water while its mass density is lower. Fig 9.2. shows a layer of turpentine floating over water in a container. For which one of the four ray’s incident on turpentine in Fig 9.2, the path shown is correct?

Fig 9.2
A. 1
B. 2
C. 3
D. 4

Answer:

The correct answer is the option (B)

The Snell's law describes the relationship between the angle of incidence and the angle of refraction.
$\mu _{1}\sin \theta_{1}=\mu _{2}\sin \theta_{2}=\text {Constant}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(1)$
Where $\mu _{1}\; \text {and}\; \mu _{2}$ are refractive indices of the two media.

In this diagram, the ray of light is going from an optically rarer medium to turpentine which an optically denser medium, then the ray bends towards the normal, which means $\theta _{1}> \theta _{2}$, but when it goes in the opposite direction, i.e. optically denser medium turpentine to rarer medium water, the ray bends away from the normal.

Question:9.10

A car is moving at a constant speed of $60\; km\; h^{-1}$ on a straight road. Looking at the rearview mirror, the driver finds that the car following him is at a distance of 100 m and is approaching with a speed of $5\; km\; h^{-1}$. In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every 2 s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct?
A. The speed of the car in the rear is $65\; km\; h^{-1}$.
B. In the side mirror the car in the rear would appear to approach with a speed of $5\; km\; h^{-1}$ to the driver of the leading car.
C. In the rearview mirror the speed of the approaching car would appear to decrease as the distance between the cars decreases.
D. In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases.

Answer:

The correct answer is the option (A).
When an object is placed in front of a mirror, all the positions of objects in the front of a mirror, image is virtual, erect and smaller in size. And as the object moves towards the pole, magnification increases and tends to unity at the pole.

An object moving along the principal axis – on differentiating the mirror formula with respect to time we get,
$\frac{dv}{dt}=-\left ( \frac{v}{u} \right )^{2}\frac{du}{dt}=-\left ( \frac{f}{u-f} \right )^{2}\frac{du}{dt}$
where $\frac{dv}{dt}$ is the velocity of the image moving along the principal axis, and $\frac{du}{dt}$ is the velocity of the object along the principal axis. The negative sign indicates that the image always moves in the direction opposite to that of the object

Question:9.11

There are certain material developed in laboratories which have a negative refractive index (Fig. 9.3). A ray incident from air (medium 1) into such a medium (medium 2) shall follow a path given by

Answer:
The correct answer is the option (A)
When a material has a negative refractive index, it follows the Snell's law in an opposite manner. And when an incident ray from one medium (air) falls on them, it bends just the way shown in option (a) that is on the same side of the normal.

NCERT Exemplar Class 12 Physics Solutions Chapter 9 MCQ II

Question:9.12

Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough, the object looks distorted because
A. the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.
B. the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air.
C. some of the points of the object far away from the edge may not be visible because of total internal reflection.
D. water in a trough acts as a lens and magnifies the object.

Answer:

The correct answers are the options (A,B,C)
When the ray of light passes from water into the air from the pencil it is refracted, the beam bends away from the normal when going from high to low refractive index.
In case of light getting refracted from the water surface from the submerged object before reaching the observer, it bends away from the normal and the angle formed by the image of the object is smaller than the real angle formed by the object in the air. This also affects the apparent depth as it becomes nearer to the surface of the water when points are close to the edge as compared to the points far from the edge.
The angle of incidence increases while moving right, and it becomes equal to the critical angle. Thus, due to total internal reflection, some points of the object present far from the edge may not be visible.

Question:9.13

A rectangular block of glass ABCD has a refractive index 1.6. A pin is placed midway on the face AB (Fig. 9.4). When observed from the face AD, the pin shall


A. appear to be near A.
B. appear to be near D.
C. appear to be at the centre of AD.
D. not be seen at all.

Answer:

The correct answers are the options (A, D)

As long as angle of incidence on AD of the ray emanating from the pin is less than the critical angle, the pin shall appear to be near A.

When angle of incidence on AD of the ray emanating from pins is greater than the critical angle, the light suffers from total internal reflection and cannot be seen through AD .

Question:9.14

Between the primary and secondary rainbows, there is a dark band known as Alexandra’s dark band. This is because
A. light scattered into this region interfere destructively.
B. there is no light scattered into this region.
C. light is absorbed in this region.
D. angle made at the eye by the scattered rays with respect to the incident light of the sun lies between approximately 42° and 50°.

Answer:

The correct answers are the options (A,D)

Alexandar's dark band lies between the primary and secondary rainbows, formed due to light scattered into this region interfere destructively.

Since, primary rainbows subtends an angle nearly $41^{\circ}$ to $42^{\circ}$ at observer's eye, whereas, secondary rainbows subtends an angle nearly $51^{\circ}$ to $54^{\circ}$ at observer's eye w.r.t. incident light ray.
So, the scattered rays with respect to the incident light of the sun lie between approximately $42^{\circ}$ and $50^{\circ}$.

Question:9.15

A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in
A. a larger angle to be subtended by the object at the eye and hence viewed in greater detail.
B. the formation of a virtual erect image.
C. increase in the field of view.
D. infinite magnification at the near point.

Answer:

The correct answers are the options (A, B)
A magnifying glass simply contains a convex lens of very small focal length.

For magnification when the final image is formed at D and $\infty$ $\left ( \text {i.e}, m_{D}\; \text {and}\; m_{\infty } \right ).$
$m_{D}=\left ( 1+\frac{D}{f} \right )_{max}$ and $m_{\infty }=\left (\frac{D}{f} \right )_{min}$
By using a magnifying glass, any objects can be brought closer to the eye as compared to the normal near point. The angle formed by the object in the eye is larger and thus, is more detailed. The image formed is virtual erect and is also enlarged.

Question:9.16

An astronomical refractive telescope has an objective of focal length 20m and an eyepiece of focal length 2cm.
A. The length of the telescope tube is 20.02m.
B. The magnification is 1000.
C. The image formed is inverted.
D. An objective of a larger aperture will increase the brightness and reduce chromatic aberration of the image.

Answer:

The correct answers are the options (A,B and C)
Key concept :

• Use to see heavenly bodies.
• $f_{objective}>f_{eye\; lens}$ and $d_{objective}>d_{eye\; lens}$
• The intermediate image is real, inverted and small.
• The final image is virtual, inverted and small.
• Magnification:
• $m_{D}=-\frac{f_{0}}{f_{e}}\left ( 1+\frac{f_{e}}{D} \right )$and $m_{\infty }=-\frac{f_{0}}{f_{e}}$
• Length :
• $L_{D}=f_{0}+u_{e}=f_{0}+\frac{f_{e}D}{f_{e}+D}$and $L_{\infty }=f_{0}+f_{e}$

The length of the telescope tube is
$f_{0}+f_{e}=20+(0.02)=20.02\; m$
Also,
$m=\frac{20}{0.02}=1000$
Also, the image formed is inverted.

NCERT Exemplar Class 12 Physics Solutions Chapter 9 Very Short Answer

Question:9.17

Will the focal length of a lens for red light be more, same or less than that for blue light?

Answer:

The refractive index depends on the colour of light or wavelength of light.
Cauchy's equation : $\mu =A+\frac{B}{\lambda ^{2}}+\frac{C}{\lambda ^{4}}+......$
As $\lambda _{red}>\lambda _{blue}$, hence $\mu _{red}<\mu _{blue}$
Hence parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.
By lens maker's formula,
$\frac{1}{f}=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )$
The refractive index for red light is less than that for blue light, $\mu _{red}<\mu _{blue}$
Hence $\frac{1}{f_{red}}<\frac{1}{f_{blue}}\Rightarrow f_{red}>f_{blue}$
Thus, the focal length for red light will be greater than that for blue light.

Question:9.18

The near vision of an average person is 25cm. To view an object with an angular magnification of 10, what should be the power of the microscope?

Answer:

In the given question, the least distance of distinct vision of an average person is 25 cm, in order to view an object with magnification 10,
Here,
$v=D=25\; cm\; and\; u=f$
But the magnification
$m=\frac{v}{u}=\frac{D}{f}$
$\Rightarrow f=\frac{D}{m}=\frac{25}{10}=2.5 \; cm=0.025\; m$
But power
$P=\frac{1}{f(in\; m)}=\frac{1}{0.025}=40\; D$
This is the required power of the lens.

Question:9.19

An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?

Answer:

For a given object position if the focal length of the lens does not change, the image position remains unchanged.
By lens maker's formula,

$\frac{1}{f}=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )$
For this position $R_{1}$ is positive
and $R_{2}$ is negative. Hence focal length at this position
$\frac{1}{f_{1}}=\left ( \mu -1 \right )\left ( \frac{1}{\left ( +R_{1} \right )}-\frac{1}{\left ( -R_{2} \right )} \right )=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right )$
Now the lens is reversed,
At this position, $R_{2}$ is positive and $R_{1}$ is negative. Hence focal length at this position is
$\frac{1}{f_{2}}=\left ( \mu -1 \right )\left ( \frac{1}{\left ( +R_{2} \right )}-\frac{1}{\left ( -R_{1} \right )} \right )=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right )$
We can observe the focal length of the lens does not change in both positions, hence the image position remains unchanged.

Question:9.20

Three immiscible liquids of densities $d_{1}>d_{2}>d_{3}$ and refractive indices $\mu _{1}>\mu_{2}>\mu_{3}$ are put in a beaker. The height of each liquid column is $\frac{h}{3}$. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

Answer:

\begin{aligned}
&\text { Let the apparent depth be } \mathrm{O}_1 \text { for the object seen from } m_2 \text {, then }\\
&\mathrm{O}_1=\frac{\mu_2}{\mu_1} \frac{h}{3}
\end{aligned}

Since, apparent depth $=$ real depth /refractive index $\mu$.
Since, the image formed by Medium $1, \mathrm{O}_2$ act as an object for Medium 2.
If seen from $\mu_3$, the apparent depth is $\mathrm{O}_2$.
Similarly, the image formed by Medium $2, \mathrm{O}_2$ act as an object for Medium 3

$
\begin{aligned}
O_2 & =\frac{\mu_3}{\mu_2}\left(\frac{h}{3}+O_1\right) \\
& =\frac{\mu_3}{\mu_2}\left(\frac{h}{3}+\frac{\mu_2 h}{\mu_1 3}\right)=\frac{h}{3}\left(\frac{\mu_3}{\mu_2}+\frac{\mu_2}{\mu_1}\right)
\end{aligned}
$
Seen from outside, the apparent height is

$
\begin{aligned}
O_3 & =\frac{1}{\mu_3}\left(\frac{h}{3}+O_2\right)=\frac{1}{\mu_3}\left[\frac{h}{3}+\frac{h}{3}\left(\frac{\mu_3}{\mu_2}+\frac{\mu_3}{\mu_1}\right)\right] \\
& =\frac{h}{3}\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3}\right)
\end{aligned}
$
This is the required expression of apparent depth.

Question:9.21

For a glass prism $\mu=\sqrt{3}$ the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.

Answer:

The refractive index of prism angle A and angle of minimum deviation $\delta _{m}$ is given by
$\mu=\frac{\sin\left [ \frac{\left ( A+\delta _{m} \right )}{2} \right ]}{\sin\left ( \frac{A}{2} \right )}$
Here we are given, $\delta _{m}=A$
Substituting the value, we have
$\mu=\frac{\sin A}{\sin \frac{A}{2}}$
$\Rightarrow \mu=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}}=2 \cos\frac{A}{2}$
$\Rightarrow \mu=\frac{\sin A}{\sin \frac{A}{2}}=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}}=2 \cos\frac{A}{2}$
For the given value of refractive index,
We have, $\cos\frac{A}{2}=\frac{\sqrt{3}}{2}\Rightarrow \frac{A}{2}=30^{o}$
Or $A=60^{o}$
This is the required value of the prism angle.

NCERT Exemplar Class 12 Physics Solutions Chapter 9 Short Answer

Question:9.22

A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f, what will be the length of the image? You may take $L<<\left | v-f \right |$

Answer:

Thin mirror formula :
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$u=$ object distance and $v=$ image distance
Since, the object distance is ,$u$. Let us consider the two ends of the object be at distance $u_{1}$ and $u_{2}$ respectively, so that $du=\left | u_{1}-u_{2} \right |=L.$ Hence size of the image can be written as $dv$.
By differentiating both sides
$\left ( -\frac{du}{u^{2}} \right )+\left ( -\frac{dv}{v^{2}} \right )=0\Rightarrow \frac{dv}{v^{2}}=-\frac{du}{u^{2}}$
or
$dv=-\left ( \frac{v^{2}}{u^{2}} \right )du \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(i)$
As $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\Rightarrow v=\frac{fu}{u-f}$
Or $\frac{v}{u}=\frac{f}{u-f} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(ii)$
From (i) and (ii)

$dv=-\left ( \frac{f}{u-f} \right )^{2}du$
But $du=L,$

Hence the length of the image is
$\frac{f^{2}}{(u-f)^{2}}L$
This is the required expression for the length of the image.

Question:9.23

A circular disc of radius ‘R’ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius ‘a’ (Fig. 9.5). The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index $\mu$ and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

Answer:

As shown in the figure, AM is in the direction of the incidence ray before filling the liquid. After filling liquid in the bowl, BM is in the direction of the incident ray. Refracted in both cases is same as that along MN.

Let the disc is separated by O at a distance d as shown in the figure. Also, considering the angle
$P=90^{o}, OM=a,CB=R,BP=a{-}R,AP=a+R$
Here, in $\Delta BMP,$
$\sin\; i=\frac{BP}{BM}=\frac{a-R}{\sqrt{d^{2}+\left ( a-R \right )^{2}}}\; \; \; \; \; \; \; \; \; \; \; ....(i)$
and
$in \Delta AMP,\; \cos \left ( 90^{o}-\alpha \right )=\sin \; \alpha =\frac{a+R}{\sqrt{d^{2}+\left ( a+R \right )^{2}}}\; \; \; \; \; \; \; \; \; \; \; \; .....(ii)$
But on applying Snell's law at point M
$\mu \times \sin \; i=1 \times \; sin \; \alpha$
$\mu \times \frac{a-R}{\sqrt{d^{2}+\left ( a-R \right )^{2}}}=1\times \frac{a+R}{\sqrt{d^{2}+\left ( a+R \right )^{2}}}$
$\Rightarrow d=\frac{\mu \left ( a^{2}-b^{2} \right )}{\sqrt{\left ( a+r \right )^{2}-\mu \left ( a-r \right )^{2}}}$

Question:9.24

A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at $\left ( 0,0 \right )$ and an object placed at $\left ( -50\; cm,0 \right )$. Find the coordinates of the image.

Answer:

The focal length will remain the same for each of the parts in case an asymmetric lens is cut into two parts parallel to the principal axis. In such a case, the intensity of the image that has been formed by every part will be lower in comparison to that of the complete lens.
If there were no cut, then the object would have been at a height of 0.5 cm from the principal axis OO'

The top part is placed at $\left ( 0,0 \right )$ and an object placed at $\left ( -50\; cm,0 \right )$. There is no effect on the focal length of the lens.

Applying the lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50}\Rightarrow v=50\; cm$
Magnification is
$m=\frac{v}{u}=-\frac{50}{50}=-1$
Therefore, the image would have been formed at 50 cm from the pole and 0.5 cm below the principal axis. Thus, with respect to the X-axis passing through the edge of the cut lens, the coordinates of the image are $(50\; cm, -1 \; cm)$.

Question:9.25

In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

Answer:

This method is known as the "Displacement method", is also used to find the focal length of the lens in a laboratory
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
If we take a convex lens L and place it between an object O and a screen S. The distance between the object and the screen would be D and the positions of the object and the screen are held fixed. Let the distance of position 1 from the object be $x_{1}$. Then the distance of the screen from the lens is $D-x_{1}$.

Therefore, $u=-x_{1}$ and $v=+\left ( D-x_{1} \right ).$
Placing it in the lens formula,
$\frac{1}{D-x_{1}}-\frac{1}{\left ( -x_{1} \right )}=\frac{1}{f}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(i)$
At position II, let the distance of the lens from the screen be $x_{2}$. Then the distance of the screen from the lens is $D-x_{2}$
Therefore, $u=-x_{2}$ and $v=+\left ( D-x_{2} \right )$.
Placing it in the lens formula
$\frac{1}{D-x_{2}}-\frac{1}{\left ( -x_{2} \right )}=\frac{1}{f}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(ii)$
Comparing Eqs. (i) and (ii), we realize that there are only two solutions:
$1.x_{1}=x_{2}; \text {or}$
$2.D-x_{1}=x_{2}; \text {and}\; D-x_{2}=x_{1}$
The first solution is trivial. Therefore, if the first position of the lens, for a sharp image, is $x_{1}$ fro the object, the second position is at $D-x_{1}$ from the object.
Let the distance between the two positions I and II be d.
From the diagram, it is clear that
$D=x_{1}+x_{2}\; \text {and}\; d=x_{2}-x_{1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(iii)$
On solving, two equations in (iii) we have
$x_{1}=\frac{D-d}{2}$ and $D-x_{1}=\frac{D+d}{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(iv)$
Substituting Eq. (iv) in Eq. (i), we get
$\frac{1}{\left ( \frac{D+d}{2} \right )}-\frac{1}{\left ( -\frac{D-d}{2} \right )}=\frac{1}{f}\Rightarrow f=\frac{D^{2}-d^{2}}{4D}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(v)$
or $d=\sqrt{D^{2}-4Df}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(vi)$
If $u=\frac{D}{2}+\frac{d}{2}$, then the image is at $v=\frac{D}{2}-\frac{d}{2}$.
$\therefore$ The magnification $m_{1}=\frac{D-d}{D+d}$
If $u=\frac{D-d}{2},$ then $v=\frac{D+d}{2},$
$\therefore$ The magnification $m_{2}=\frac{D+d}{D-d}$
Thus, $\frac{m_{2}}{m_{1}}=\left ( \frac{D+d}{D-d} \right )^{2}$
This is the required expression of magnification.

Question:9.26

A jar of height h is filled with a transparent liquid of refractive index $\mu$ (Fig. 9.6). At the center of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the center, the dot is invisible.

Fig 9.6

Answer:

Let $d$ be the diameter of the disc. The spot shall be invisible if the incident rays from the dot at $O$ to the surface at $d / 2$ at the critical angle.
Let $i$ be the angle of incidence.
Using relationship between refractive index and critical angle, then,

$
\sin i=\frac{1}{\mu}
$

Using geometry and trigonometry.

$
\begin{array}{ll}
\text { Now, } & \frac{d / 2}{h}=\tan i \\
\Rightarrow & \frac{d}{2}=h \tan i=h\left[\sqrt{\mu^2-1}\right]^{-1} \\
\therefore & d=\frac{2 h}{\sqrt{\mu^2-1}}
\end{array}
$

This is the required expression of $d$.

Question:9.27

A myopic adult has a far point at 0.1 m. His power of accommodation is 4 diopters. (i) What power lenses are required to see distant objects? (ii) What is his near point without glasses? (iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2 cm.)

Answer:

(i) If for the normal relaxed eye of an average person, the power at the far point be $P_{f}$. The required power
$P_{f}=\frac{1}{f}=\frac{1}{0.1}+\frac{1}{0.02}=60\: D$
By the corrective lens, the object distance at the far point is $\infty$.
The power required is ,
$P'_{f}=\frac{1}{f'}=\frac{1}{\infty }+\frac{1}{0.02}=50\; D$
Now for eye+lens system, we have the sum of the eye and that of the glasses $P_{g}$.
$P'_{f}=P_{f}+P_{g}\Rightarrow 50\; D=60D+P_{g}$
Which gives, $P_{g}=-10D$
(ii) For the normal eye his power of accommodation is 4D. Let the power of the normal eye for near vision be $P_{n}$
Then, $4=P_{n}-P_{f}\; or \; P_{n}=64\; D$
Let his near point be $x_{n'}$ then
$\frac{1}{x_{n}}+\frac{1}{0.02}=64\;$

$or\; \frac{1}{x_{n}}+50=64$
$or, \frac{1}{x_{n}}=14$

$\Rightarrow x_{n}=\frac{1}{14}m=0.07 m$
(iii) With glasses $P_{n}^{'}=P{_{f}}^{'}+4=54$
$\Rightarrow 54=\frac{1}{x{_{n}}^{'}}+\frac{1}{0.02}=\frac{1}{x{_{n}}^{'}}+50$
$\Rightarrow \frac{1}{x{_{n}}^{'}}=4$

$\Rightarrow x{_{n}}^{'}=\frac{1}{4}m=0.25\; m$

NCERT Exemplar Class 12 Physics Solutions Chapter 9 Long Answer

Question:9.28

Show that for a material with refractive index $\mu \geq \sqrt{2}$ light incident at any angle shall be guided along a length perpendicular to the incident face.

Answer:

Let the ray incident on face AB at angle i, after refraction, it travels along PQ and then interact with face AC which is perpendicular to the incident face.

Any ray is guided along AC if the angle ray makes with the face AC $\left ( \phi \right )$ is greater than the critical angle. From figure
$\phi +r=90^{o}$, therefore $\sin\; \phi =\cos\; r\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ......(i)$
If $\phi$ is the critical angle it means, $\sin\; \phi \geq \frac{1}{\mu }\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ......(ii)$
From (i) and (ii),

$\cos \; r \geq \frac{1}{\mu ^{2}}\;$

$or\; 1-\cos ^{2}r\leq 1-\frac{1}{\mu ^{2}}$
i.e. , $\Rightarrow \sin ^{2}r\leq 1-\frac{1}{\mu ^{2}}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(iii)$
Applying Snell's law on face AB,
$1 \sin \; i=\mu \sin\; r$
Or $1 \sin^{2} \; i=\mu^{2} \sin^{2}\; r$

$\Rightarrow \sin^{2}\; r=\frac{1}{\mu ^{2}} \sin^{2}\; i\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(iv)$
From (i) and (ii),

$\frac{1}{\mu ^{2}}\sin^{2}i\leq 1-\frac{1}{u^{2}}$
Or $\sin^{2}i\leq \mu ^{2}-1$
When $i=\frac{\pi }{2}$, then we have smallest angle $\phi$.
If it is greater than the critical angle, then all other angles of incidence shall be greater than the critical angle
Thus, $1\leq \mu ^{2}-1\; or \; \mu ^{2}\geq 2$
$\Rightarrow \mu \geq \sqrt{2}.$ This is the required result.

Question:9.29

The mixture of a pure liquid and a solution in a long vertical column (i.e., horizontal dimensions << vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance d << h, the height of the column.

Answer:

Let us consider a portion of a ray between x and $x+dx$ inside the liquid solution.
Let us assume that the angle of incidence ray be $\theta$ and
Let the ray enter the thin column at the height y.
As a result of the refraction, the ray deviates from its original path and subsequently emerges at $x+dx$ with an angle
$\theta +d\theta$ and at a height $y+dy.$

From Snell's law,
$\mu \left ( y \right )\sin \theta =\mu \left ( y+dy \right )\sin\left ( \theta+d\theta \right )......(i)$

Let refractive index of the liquid at position y be $\mu (y)=\mu,$ then
$\mu\left ( y+dy \right )=\mu+\left ( \frac{d\mu}{dy} \right )dy=\mu+kdy$
Where $k=\left ( \frac{d\mu }{dy} \right )=$ refractive index gradient along the vertical dimension.
Hence from (i), $\mu \sin\theta=\left ( \mu+kdy \right )\sin \left ( \theta+d \theta \right )$
$\mu \sin\theta=\left ( \mu+kdy \right ).\left ( \sin \theta.\cos d \theta+\cos \theta. \sin d\theta \right )$
$\mu \sin\theta=\left ( \mu+kdy \right ).\left ( \sin \theta.1+\cos \theta.d \theta \right )\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(ii)$
For small angle $\sin d \theta \approx d \theta \; and \; \cos d \theta\approx 1$
$\mu \sin \theta=\mu \sin \theta+kdy \; \sin \theta+\mu \cos \theta .d\theta+k\; \cos\; \theta dy.d \theta$
$kdy \sin \theta +\mu \cos \theta.d\theta=0\Rightarrow d \theta=-\frac{k}{\mu}\tan \theta dy$
But $\tan \theta=\frac{dx}{dy}\; and \: k=\left ( \frac{d\mu}{dy} \right )$
$d\theta=-\frac{k}{\mu}\left ( \frac{dx}{dy} \right )dy\Rightarrow d\theta=-\frac{k}{\mu}dx$
Integrating both sides, $\int_{0}^{\delta }d\theta =-\frac{k}{\mu}dx$
$\Rightarrow \delta =-\frac{kd}{\mu }=-\frac{d}{\mu }\left ( \frac{d\mu }{dy} \right )$

Question:9.30

If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refractive index of the medium given by
$n(r)=1+2\frac{GM}{rc^{2}}$
where r is the distance of the point of consideration from the center of the mass of the massive body, G is the universal gravitational constant, M the mass of the body and c the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object.

Answer:

Let us consider two spherical surfaces of radius r and $r + dr$. Let the light be incident at an angle $\theta$ at surface r and $r + dr$ at an angle $\theta+d\theta$.
From Snell's Law,
$n(r)\sin \theta=n(r+dr)\sin(\theta+d\theta)=\left ( n(r)+\left ( \frac{dn}{dr} \right ) dr\right )\left ( \sin \theta.\cos \; d\theta+\cos \theta.\sin \; d\theta \right )$
$\Rightarrow n(r)\sin \theta=\left ( n(r)+\left ( \frac{dn}{dr} \right )dr \right )\left ( \sin \theta+\cos \theta.d\theta \right )$
For small angle, $\sin d\theta\approx d\theta\; and \; \cos d\theta\approx 1$
Ignoring the product of differentials
$\Rightarrow n(r)\sin \theta=n(r).\sin \theta+\left ( \frac{dn}{dr} \right )dr.\sin \theta+n(r).\cos \theta.d\theta$

or we have, $-\frac{dn}{dr}\tan \theta=n(r)\frac{d\theta}{dr}$

$\frac{2GM}{r^{2}c^{2}}\tan \theta=\left ( 1+\frac{2GM}{rc^{2}} \right )\frac{d\theta}{dr}\approx \frac{d\theta}{dr}$

$\int_{0}^{\theta_{0}}d\theta=\frac{2GM}{c^{2}}\int_{-\infty }^{\infty }\frac{\tan \theta dr}{r^{2}}$

Now, $r^{2}=x^{2}+R^{2}\; and \; \tan \theta=\frac{R}{x}$
$2rdr=2xdx$
Now substitution for integrals, we have

$\int_{0}^{\theta_{0}}d\theta=\frac{2GM}{c^{2}}\int_{-\infty }^{\infty }\frac{R}{x}\frac{xdx}{\left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}$
Put $x=R\; \tan \phi$
$dx=R\; sec^{2} \phi \; d\phi$
$\therefore \theta_{0}=\frac{2GMR}{c^{2}}\int_{-\pi /2}^{\pi /2}\frac{R\; \sec ^{2}\phi d\; \phi }{R^{3}\sec ^{3}\phi }$
$\theta_{0}=\frac{2GM}{Rc^{2}}\int_{-\pi /2}^{\pi /2}\cos \phi d\; \phi =\frac{4GM}{Rc^{2}}$
$\Rightarrow \theta_{0}=\frac{4GM}{Rc^{2}}.$ This is the required proof.

Question:9.31

An infinitely long cylinder of radius R is made of an unusual exotic material with refractive index –1 (Fig. 9.7). The cylinder is placed between two planes whose normal are along the y direction. The centre of the cylinder O lies along the y-axis. A narrow laser beam is directed along the y direction from the lower plate. The laser source is at a horizontal distance x from the diameter in the y direction. Find the range of x such that light emitted from the lower plane does not reach the upper plane.

Fig 9.7

Answer:

Here the material has refractive index $-1,\theta _{r}$ is negative and $\theta _{r}^{'}$ positive.
Now, $\left | \theta _{i} \right |=\left | \theta _{r} \right |=\left | \theta _{r}^{'} \right |$ The total deviation of the outcoming ray from the incoming ray is $4\theta _{r}$.

Rays shall not reach the receiving plate if
$\frac{\pi }{2}\leq 4\theta _{i}\leq \frac{3\pi }{2}$
[angles measured clockwise from the y-axis]

On solving,
$\frac{\pi }{8}\leq \theta _{i}\leq \frac{3\pi }{8}$
Now,
$\sin \theta _{i}=\frac{x}{R}$

$\frac{\pi}{8}\leq \sin^{-1}\frac{x}{R}\leq \frac{3\pi}{8}$
or $\frac{\pi}{8}\leq \frac{x}{R}\leq \frac{3\pi}{8}$
Thus, for light emitted from the source shall not reach the receiving plate,
If $\frac{R\pi}{8}\leq x\leq \frac{3R\pi }{8}$

Question:9.32

(i) Consider a thin lens placed between a source (S) and an observer (O) (Fig. 9.8). Let the thickness of the lens vary as $w(b)=w_{o}-\frac{b_{2}}{\alpha }$ where b is the vertical distance from the pole. $w_{o}$ is a constant. Using Fermat’s principle i.e. the time of transit for a ray between the source and observer is an extremum; find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length.

(ii) A gravitational lens may be assumed to have a varying width of the form
$\begin{aligned} w(b) & =k_1 \ln \left(\frac{k_2}{b}\right) & & b_{\min }<b<b_{\max } \\ & =k_1 \ln \left(\frac{k_2}{b_{\min }}\right) & & b<b_{\min }\end{aligned}$
Show that an observer will see an image of a point object as a ring about the center of the lens with an angular radius
$\beta =\sqrt{\frac{\left ( n-1 \right )k_{1}\frac{u}{v}}{u+v}}$

Answer:

(i) The time required to travel from S to $\mathrm{P}_1$ is $t_1=\frac{S P_1}{c}=\frac{\sqrt{u^2+b^2}}{c} \simeq \frac{u}{c}\left(1+\frac{1}{2} \frac{b^2}{u^2}\right)$ assuming $\mathrm{b} \ll \mathrm{u}_0$
The time required to travel from $\mathrm{P}_1$ to O is $t_2=\frac{P_1 O}{c}=\frac{\sqrt{v^2+b^2}}{c} \simeq \frac{v}{c}\left(1+\frac{1}{2} \frac{b^2}{v^2}\right)$
The time required to travel through the lens is $t_1=\frac{(n-1) w(b)}{c}$ where n is the refractive index.
Thus the total time is

$t=\frac{1}{c}\left[u+v+\frac{1}{2} b^2\left(\frac{1}{u}+\frac{1}{v}\right)+(n-1) w(b)\right]$

Put $\frac{1}{D}=\frac{1}{u}+\frac{1}{v}$

Then $t=\frac{1}{c}\left(u+v+\frac{1}{2} \frac{b^2}{D}+(n-1)\left(w_0+\frac{b^2}{\alpha}\right)\right)$

Fermet's principle gives

$
\begin{aligned}
& \frac{d t}{d b}=0=\frac{b}{C D}-\frac{2(n-1) b}{c \alpha} \\
& \alpha=2(n-1) D
\end{aligned}
$

Thus a convergent lens is formed if $\alpha=2(n-1) D$. This is independent of $b$ and hence all paraxial rays from $S$ will converge at O (i.e. for rays $\mathrm{b} \ll \mathrm{n}$ and $\mathrm{b} \ll \mathrm{v}$ ).

Since $\frac{1}{D}=\frac{1}{u}+\frac{1}{v}$, the focal length is D .

(ii) In this case

$
\begin{aligned}
& t=\frac{1}{c}\left(u+v+\frac{1}{2} \frac{b^2}{D}+(n-1) k_1 \ln \left(\frac{k_2}{b}\right)\right) \\
& \frac{d t}{d b}=0=\frac{b}{D}-(n-1) \frac{k_1}{b} \\
& \Rightarrow \mathrm{~b}^2=(n-1) k_1 D \\
& \therefore \mathrm{~b}=\sqrt{(n-1) k_1 D}
\end{aligned}
$

\begin{aligned}
&\text { Thus all rays passing at a height } b \text { shall contribute to the image. The ray paths make an angle }\\
&\beta \simeq \frac{b}{v}=\frac{\sqrt{(n-1) k_1 D}}{v^2}=\sqrt{\frac{(n-1) k_1 u v}{v^2(u+v)}}=\sqrt{\frac{(n-1) k_1 u}{(u+v) v}} .
\end{aligned}