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

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

Q: How important is chapter 9 in physics of Class 12?

This chapter has the highest weightage in the board exam, thus scoring well in this chapter will help in scoring in the final paper.

Q: How are these questions solved?

Each and every question is solved in complete detail covering the basics, the numericals and the diagrams in NCERT exemplar Class 12 Physics solutions chapter 9.

Q: What all things one will learn in Class 12 Physics chapter 9 Ray Optics and Optical Instruments?

One will cover almost every topic related to ray optics; like reflection, refraction, dispersion, TIR, optical instruments etc.

Edited By Safeer PP | Updated on Sep 14, 2022 01:03 PM IST | #CBSE Class 12th

NCERT Exemplar Class 12 Physics solutions chapter 9 is one of the most scoring chapters from the view of exams and gives an insight on various different topics related to visible light. NCERT Exemplar Class 12 Physics chapter 9 solutions deals with various phenomena relating to light and explains the concept of reflection of light by spherical mirrors. Class 12 Physics NCERT Exemplar solutions chapter 9 also contains various mathematical derivations for different terms relating to such concepts, such as for obtaining the focal length of spherical mirrors and mirror equations which help in obtaining different values out of some given information. NCERT chapter ray optics also explains the phenomena of refraction and total internal reflection in detail along with its practical application in the surrounding nature. Students can make use of NCERT Exemplar Class 12 Physics solutions chapter 9 PDF download for preparing for exams.

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This Story also Contains

NCERT Exemplar Class 12 Physics Solutions Chapter 9 MCQI
NCERT Exemplar Class 12 Physics Solutions Chapter 9 MCQII
NCERT Exemplar Class 12 Physics Solutions Chapter 9 Very Short Answer
NCERT Exemplar Class 12 Physics Solutions Chapter 9 Short Answer
NCERT Exemplar Class 12 Physics Solutions Chapter 9 Long Answer
NCERT Exemplar Class 12 Physics Solutions Chapter 9 Ray Optics and Optical Instruments - Main Subtopic
NCERT Exemplar Class 12 Physics Chapter Wise Links
Important Topics To Cover For Exams From NCERT Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Also see - NCERT Solutions for Class 12 Physics

NCERT Exemplar Class 12 Physics Solutions Chapter 9 MCQI

Question: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 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 option is (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 because of the relation $A=r_{1}+r_{2}$ . This also applies to both $i_{1}$ and $i_{2}$ .

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 $\delta =A\left ( \mu -1 \right )A$
$=\left ( 1.5-1 \right )\times 5^{o}=2.5^{o}$
The angle of the prism is $5^{o}$ . The ray emr=erges 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: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 emerging colour would be (d) 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: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 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: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 (b) may 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: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 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 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 least which facilitates total internal reflection for the beam of blue light.

Question: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 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 act as a convex lens irrespective of the side on which the object lies.

Question:7

The phenomena 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 answer is the option (b)
The phenomena are very similar to total internal reflection of light in the air during a mirage, option (b). 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: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?

Answer:

(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

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?

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

Answer:

The correct path followed by the ray of light is an option (b) 2.
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:10

A car is moving with 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 answer is the option (d).
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 principle axis. The negative sign indicates that the image always moves in the direction opposite to that of the object

Question: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
A.

Answer:

The 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 is bending just the way shown in option (a) that is on the same side of the normal.

NCERT Exemplar Class 12 Physics Solutions Chapter 9 MCQII

Question: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:

All three (a, b, c) are the correct options.
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 water surface from the submerged object before reaching to the observer, it bends away from the normal and the angle formed by the from 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 incident increase while moving right, and it becomes equal to the critical angle. Thus, due to total internal reflection, some point of the object present far from the edge may not be visible.

Question: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:

(a, d) option is the right answer because till the angle of incidence is less than the critical angle, i.e. incident angle on AD of the ray from the pin, pin will always seem nearer to the point A.

But when the angle of incident is larger than the critical angle, the light undergoes total internal reflection and is no visible through AD.

Question: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:

Formed because of light scattered into the region interfere destructively the dark Alexander's band lie between the primary and secondary rainbow; thus, (a, d) are the correct options. The angle of the primary rainbow in the observer's eyes is from 41° to 42°, and in the secondary rainbow, the angle in the observer's eyes lies between 51° to 54° with respect to the incident light ray.
Therefore, the overall angle of scattered rays with reference to incident light of the sun ranges from 42° and 50°.

Question: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:

Right options are - (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: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 right answers to this question are 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: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: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: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: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:

Key concept : Coordinate convention: At the first surface (+upward and -ve downward)
$\frac{\mu _{2}}{h'}-\frac{\mu _{1}}{(-h)}=\frac{\mu _{2}-\mu _{1}}{\infty }$ (infinity because the surface is plane)
or $h'=\frac{\mu _{2}}{\mu _{1}}h.$ The negative sign shows that it is on the side of the object

$h'$ is the apparent depth of O after refraction from interface.

The position of image O after the refraction from surface-1. If seen from $u_{2}$ , the apparent depth is $h_{1}$

$h_{1}=\frac{\mu _{2}}{\mu _{1}}\frac{h}{3}$
The negative sign indicates that it is on the side of the object.
Since the image formed by surface-1. It will act as an object for surface-2. If seen from $u_{3}$ , the apparent depth is $h_{2}$ .
Similarly, the image formed by medium 2, $O_{2}$ acts as an object for medium 3
$h_{2}=-\frac{\mu _{3}}{\mu _{2}}\left (\frac{\mu _{2}}{\mu _{1}}\frac{h}{3}+\frac{h}{3} \right )=-\frac{h}{3}\left (\frac{\mu _{3}}{\mu _{2}}+\frac{\mu _{2}}{\mu _{1}} \right )$
Finally the image formed by surface -2 will act as an object for surface-2. If seen from outside, the apparent depth is $h_{3}$
$h_{3}=-\frac{1}{\mu _{3}}\left [\frac{h}{3}+ \frac{h}{3}\left (\frac{\mu _{3}}{\mu _{2}}+\frac{\mu _{2}}{\mu _{1}} \right ) \right ]=-\frac{h}{3}\left ( \frac{1}{\mu _{1}}+\frac{1}{\mu _{2}}+\frac{1}{\mu _{3}} \right )$
Hence apparent depth of odt is $\frac{h}{3}\left ( \frac{1}{\mu _{1}}+\frac{1}{\mu _{2}}+\frac{1}{\mu _{3}} \right )$
Important point:
Apparent depth (distance of final image from final surface)
$=\frac{t_{1}}{n_{1\; real}}+\frac{t_{2}}{n_{2\; real}}+\frac{t_{3}}{n_{3\; real}}+...+\frac{t_{n}}{n_{n\; real}}$

Question: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 prism angle.

NCERT Exemplar Class 12 Physics Solutions Chapter 9 Short Answer

Question: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: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
$\mu \times \sin \; i=1 \times \; sin \; r$
$\mu \times \frac{a-R}{\sqrt{d^{2}+\left ( a-R \right )^{2}}}=1\times \frac{a+R}{\sqrt{d^{2}+\left ( a+R \right )^{2}}}$
$\Rightarrow d=\frac{\mu \left ( a^{2}-b^{2} \right )}{\sqrt{\left ( a+r \right )^{2}-\mu \left ( a-r \right )^{2}}}$

Question: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 in two parts parallel to the principle axis. In such a case the intensity of image that has been formed by every part will be lower in comparison to that of the complete lens.
If there was no cut, then the object would have been at the 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 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 ate 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: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 confocal length of the length in laboratory
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
If we take a convex lens L place between an object O and a screen S. The distance between the object and the screen would be D and the position of the object and screen are held fix. 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: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.

Answer:

Key concept :

In the above figure, ray 1 strikes the surface at an angle less than critical angle c and gets refracted in the rarer medium. Ray 2 strikes the surface, making an angle greater than the critical angle and gets internally reflected. All the rays strike at the circle of illuminance gets refracted in a rarer medium. The observer in this field will only see the light coming out of from the C.O.I.
In the above figure, O is the small dot at the bottom of the jar. The ray from the dot emerges out of the circular patch of water surface of diameter AB till the angle of incidence for the rays OA and AON exceeds the critical angle.

Rays of light incident at an angle greater than $i_{C}$ are totally reflected with in water and consequently cannot emerge out of the water surface.
As $\ sin \; i_{C}=\frac{1}{\mu }\Rightarrow \tan\; i_{C}=\frac{1}{\sqrt{\mu ^{2}-1}}$
Now, $\frac{d}{2h}=\tan\; i_{e}$
$\Rightarrow d=\frac{2h}{\sqrt{\mu ^{2}-1}}$
This is the required expression of d.

Question: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:

Key concepts :

$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$
in terms of power $P=P_{1}+P_{2}$
(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
$\frac{1}{x_{n}}+\frac{1}{0.02}=64\; or\; \frac{1}{x_{n}}+50=64$
$\frac{1}{x_{n}}=14\Rightarrow x_{n}=\frac{1}{14}m=0.07 m$
(iii) With glasses $P_{n}^{'}=P{_{f}}^{'}+4=54$
$54=\frac{1}{x{_{n}}^{'}}+\frac{1}{0.02}=\frac{1}{x{_{n}}^{'}}+50$
$\frac{1}{x{_{n}}^{'}}=4\Rightarrow x{_{n}}^{'}=\frac{1}{4}m=0.25\; m$

NCERT Exemplar Class 12 Physics Solutions Chapter 9 Long Answer

Question: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 more 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:29

The mixture 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 enters 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: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: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.

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: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
$w(b)=k_{1}ln\left ( \frac{k_{2}}{b} \right ) b_{min}<b<b_{max}$
$=k_{1}ln\left ( \frac{k_{2}}{b_{min}} \right )b<b_{min}$
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 taken by ray to travel from S to $P_{1}$ is
$t_{1}=\frac{SP_{1}}{c}=\frac{\sqrt{u^{2}+b^{2}}}{c}$
Or $t_{1}=\frac{u}{c}\left ( 1+\frac{1}{2}\frac{b^{2}}{u^{2}} \right )$ assuming $b<<1$
The time required to travel from $P_{1}$ to O is
$t_{2}=\frac{P_{1}O}{c}=\frac{\sqrt{v^{2}+b^{2}}}{c}=\frac{v}{c}\left ( 1+\frac{1}{2}\frac{b^{2}}{v^{2}} \right )$
The time required to travel through the lens is
$t_{l}=\frac{(n-1)w(b)}{c}$
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,
$t=\frac{1}{c}\left ( u+v+\frac{1}{2}\frac{b^{2}}{D}+\left ( n-1 \right )\left ( w_{0}+\frac{b^{2}}{\alpha } \right ) \right )$
Fermat's principle gives the time taken should be minimum. For that first derivative should be zero.
$\frac{dt}{db}=0=\frac{b}{CD}-\frac{2(n-1)b}{c\alpha }$
$\alpha =2(n-1)D$
Thus, a convergent lens is formed if $\alpha =2(n-1)D.$ This is independent of and hence all paraxial rays from s will converge at O i.e., for rays $b<<n\; and\; b<<v$
Since, $\frac{1}{D}=\frac{1}{u}+\frac{1}{v}$ , the focal length is D.
(ii) In this case, differentiating expression of time taken t e.r.t. b.
$t=\frac{1}{c}\left ( u+v+\frac{1}{2}\frac{b^{2}}{D}+(n-1)k_{1}ln\left ( \frac{k_{2}}{b} \right ) \right )$
$\frac{dt}{db}=0=\frac{b}{D}-(n-1)\frac{k_{1}}{b}$
$\Rightarrow b^{2}=(n-1)k_{1}D$
$\therefore b=\sqrt{(n-1)k_{1}D}$
Thus, all rays passing at a height b shall contribute to the image. The ray paths make an angle.
$\beta =\frac{b}{v}=\frac{\sqrt{(n-1)k_{1}D}}{v}=\sqrt{\frac{(n-1)k_{1}uv}{v^{2}(u+v)}}=\sqrt{\frac{(n-1)k_{1}u}{(u+v)v}}$ This is the required expression.

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

Introduction
Reflection of light by spherical surfaces
Sign convention
Focal length of spherical mirrors
The Mirror equation
Refraction
Total internal reflection
Total internal reflection in nature and its technological applications
Refraction at spherical surface and by lenses
Refraction at a spherical surface
Refraction by a lens
Power of a lens
Combination of thin lenses in contact
Refraction through a prism
Dispersion by a prism
Some natural phenomena due to sunlight
The rainbow
Scattering of light
Optical instruments
The eye
The microscope
Telescope

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NCERT Exemplar Class 12 Physics Chapter Wise Links

Chapter 1 Electric Charges and Fields

Chapter 2 Electrostatic Potential and Capacitance

Chapter 3 Current Electricity

Chapter 4 Moving Charges and Magnetism

Chapter 5 Magnetism and Matter

Chapter 6 Electromagnetic Induction

Chapter 7 Alternating Current

Chapter 8 Electromagnetic Waves

Chapter 10 Wave Optics

Chapter 11 Dual Nature of Radiation and Matter

Chapter 12 Atoms

Chapter 13 Nuclei

Chapter 14 Semiconductor Electronics

Chapter 15 Communication Systems

Important Topics To Cover For Exams From NCERT Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

· NCERT Exemplar Solutions For Class 12 Physics chapter 9 covers questions related to the phenomena of refraction by the lenses and how formulas are derived out of it to calculate the Power and combination of lenses required.

· Class 12 Physics NCERT Exemplar solutions chapter 9 also explores the diagrammatic representation of total internal reflection in different objects of different shapes such as a prism, diamond, etc.

· Most importantly, NCERT Exemplar Class 12 Physics solutions chapter 9 gives full detailed insight into different optical instruments such as the eye, the microscope, and the telescope, the phenomena associated, the principles used, and the detailed process of such happenings.

NCERT Exemplar Class 12 Solutions

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Frequently Asked Question (FAQs)

1. How important is chapter 9 in physics of Class 12?

This chapter has the highest weightage in the board exam, thus scoring well in this chapter will help in scoring in the final paper.

2. How are these questions solved?

Each and every question is solved in complete detail covering the basics, the numericals and the diagrams in NCERT exemplar Class 12 Physics solutions chapter 9.

3. What all things one will learn in Class 12 Physics chapter 9 Ray Optics and Optical Instruments?

One will cover almost every topic related to ray optics; like reflection, refraction, dispersion, TIR, optical instruments etc.

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NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

NCERT Class 12 Physics Chapter 8 Notes Electromagnetic Waves - Download PDF

NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

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Questions related to CBSE Class 12th

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I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

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Read Complete Answer

Sajal Trivedi 29 January,2024

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

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Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

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Nuthalapati Vyshnavi 21 August,2022

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

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Muskan Dhalwal 17 August,2022

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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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