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

## NCERT Solutions For Class 12 Physics Chapter 9 PDF Free Download

The NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics Optical Instruments given here, is one of the best study material that you will come across to score good marks in CBSE class 12 examination. Ray Optics Class 12 Physics NCERT solutions pdf provided here entails important questions,  worksheets and exercises that will help students understand the topic thoroughly.

The NCERT solutions class 12 physics Chapter 9 Ray Optics and Optical Instruments is a prime topic in 12th standard. The NCERT solutions provided here, help students understand the chapter in an easy and interesting way.

The NCERT class 12 Solutions for physics chapter 9 is created by subject experts according to the latest CBSE syllabus (2019-20). Students must practice the solutions regularly to prepare effectively for their board exam.

## NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics Optical Instruments

Ray optics class 12 Physics NCERT pdf is an important topic, for it has a good weightage of marks allotted in the examination. In this chapter, the phenomena of reflection, refraction and dispersion of light, using the ray picture of light is considered. Using the basic laws of reflection and refraction, students shall study the image formation by plane and spherical reflecting and refracting surfaces. They will learn to describe the construction and working of some important optical instruments, including the human eye.

### Topics covered in NCERT Solutions Class 12 Physics Chapter 9

 Section Number Topic 9.1 Introduction 9.2 Reflection Of Light By Spherical Mirrors 9.2.1 Sign convention 9.2.2 Focal Length Of Spherical Mirrors 9.2.3 The Mirror Equation 9.3 Refraction 9.4 Total Internal Reflection 9.4.1 Total Internal Reflection In Nature And Its Technological Applications 9.5 Refraction At Spherical Surfaces And By Lenses 9.5.1 Refraction At A Spherical Surface 9.5.2 Refraction By A Lens 9.5.3 Power Of A Lens 9.5.4 Combination Of Thin Lenses In Contact 9.6 Refraction Through A Prism 9.7 Dispersion By A Prism 9.8 Some Natural Phenomena Due To Sunlight 9.8.1 The Rainbow 9.8.2 Scattering Of Light 9.9 Optical Instruments 9.9.1 The Eye 9.9.2 The Microscope 9.9.3 Telescope

### Class 12 Physics NCERT Solutions Ray Optics Optical Instruments Important Questions

Question 1:

A candle which is 2.5 cm in height is placed 27 cm in front of a concave mirror having radius of curvature 36 cm. Find the distance from the mirror at which the screen should be placed to obtain a sharp image? Detail the size and nature of the image.

By how much the screen have to be moved if the candle is moved towards the mirror?

Height of the candle, h =2.5 cm

Image size=h’

Object distance, u= -27 cm

Radius of the concave mirror, R= -36 cm

Focal length of the concave mirror, f=$\frac{R}{2}$ =-18 cm

Image distance=v

$\frac{1}{u}$+ $\frac{1}{v}$= $\frac{1}{f}$ $\frac{1}{v}$= $\frac{1}{f}$- $\frac{1}{u}$

=$\frac{1}{-18}$- $\frac{1}{-27}$=$\frac{-3+2}{54}$=$-\frac{1}{54}$ $∴$v=-54cm

Therefore, the screen should be placed at a distance of 54 cm away from mirror to obtain a sharp image

Magnification of the Image is:

$m=\frac{h’}{h}=-\frac{v}{u}$ $∴ h’=\frac{-v}{u}\times h$ $=-(\frac{-54}{-27})\times2.5$=-5cm

The height of the candle’s Image Is 5 cm. The image Is inverted and virtual since there is a negative sign.

If we move the candle closer to the mirror the screen will also have to be moved away from the mirror to obtain the image.

Question 2:

A needle which is 4.5cm in size is placed 12cm away from a convex mirror having a focal length of 15cm.  Find the location and the magnification of the image. Also find what happens as the needle is moved away from the mirror

Size of the needle, h1=4.5 cm

Object distance, u=-12 cm

Focal length of the convex mirror, f= 15 cm

Image distance= v

$\frac{1}{u}$+ $\frac{1}{v}$= $\frac{1}{f}$ $\frac{1}{v}$= $\frac{1}{f}$- $\frac{1}{u}$ $\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}$

Therefore

$\frac{60}{9}$=6.7cm

We found that the image of the needle is 6.7 cm away from the mirror and is on the other side of the mirror.

By using the magnification formula:

$m=\frac{h_{2}}{h_{1}}=-\frac{v}{u}$ $h_{2}=\frac{-v}{u}\times h_{1}$ $=\frac{-6.7}{-12}\times 4.5$=+2.5cm

$m=\frac{h_{2}}{h_{1}}=\frac{2.5}{4.5}$=0.56

Height of the image is found to be 2.5 cm. The image is erect, virtual and diminished since it shows a positive sign

If we move the needle farther from the mirror the image will also move away from the mirror, and the size of image will also reduce gradually.

Question 3:

Height of a tank containing water is 12.5cm and by using a microscope the apparent depth of needle lying at the bottom is measured to be 9.4cm. Find the refractive index of water? By what distance would the microscope have to be moved to focus again if water is replaced by a liquid of refractive index 1.63 filled up to the same height?

Actual depth of the needle in water, h1=12.5cm

Apparent depth of the needle in water, h2 =9.4 cm

Refractive Index of water =$\mu$

The value of $\mu$ can be obtained as follows:

$\mu=\frac{h_{1}}{h_{2}}$ $\frac{12.5}{9.4}$=1.33

Hence, 1.33 is the refractive index of water.

Now water is replaced by a liquid of refractive index 1.63

The actual depth of the needle remains the same, but its apparent depth changes.

Let x be the new apparent depth of the needle.

$\mu ‘=\frac{h_{1}}{x}$

Therefore y=$\frac{h_{1}}{\mu ‘}$ $\frac{12.5}{1.63}$=7.67cm

Hence 7.67cm is the new apparent depth of the needle. We found that the value is less than $h_{2}$

Therefore the needle should be moved up to focus again

Distance to be moved to focus=9.4-7.67=1.73cm

Question 4: Refraction of a ray which is incident at $60°$with the normal to a glass-air and water-air interface is shown in the above figures (a) and (b). What will be the angle of refraction in glass when the angle of incidence in $45°$ with the normal to a water-glass interface(c)

For the glass-air interface:

Angle of incidence, i =60°

Angle of refraction, r=35°

According to Snell’s law the refractive index of glass with respect to air is given by:

$\mu_{g}^{a}=\frac{sin\;i}{sin\;r}\\ =\frac{sin\;60°}{sin\;35°}=\frac{0.8660}{0.5736}=1.51$———-(i)

For the air-water interface:

Angle of incidence, i=$60°$

Angle of refraction, r=$47°$

According to Snell’s law the refractive index of water with respect to air is given by:

$\mu_{w}^{a}=\frac{sin\;i}{sin\;r}\\ =\frac{sin\;60°}{sin\;47°}=\frac{0.8660}{0.7314}=1.184$—————–(ii)

With the help of equations (i) and (ii) the relative refractive index of glass with respect to water can be found:

$\mu_{g}^{w}=\frac{\mu_{g}^{a}}{\mu_{w}^{a}}\\ =\frac{1.51}{1.184}=1.275$ Angle of incidence, i= $45°$

Angle of refraction= r

r can be calculated from Snells’s law=

$\frac{sin\;i}{sin\;r}=\mu_{g}^{w}\\ \frac{sin45°}{sin\;r}=1.275\\ sin\;r=\frac{\frac{1}{\sqrt{2}}}{1.275}=0.5546\\ Therefore\;r=sin^{-1}(0.5546)=38.68°$

Therefore, the angle of refraction at the water -glass Interface is $38.68°$.

Question 5:

At the bottom of a tank having a depth of 80cm there is a small bulb placed.  Find the area of the surface of water through which light can come out of the bulb.  Lets consider the bulb to be a point source and refractive index of water is 1.33

Actual depth of the bulb in water, d1=80 cm =0.8 m

Refractive Index of water, $\mu$= 1.33 I = Angle of incidence

r =Angle of refraction – 90°

Emergent light is considered as a circle of radius since the bulb is a point source

R=$\frac{AC}{2}$=AO=OB

Using Snell’s law, we can write the relation for the refractive index of water as:

$\mu=\frac{sin\;r}{sin\;i}\\ 1.33=\frac{sin\;90°}{sin\;i}\\ Therefore\;i=sin^{-1}(\frac{1}{1.33})=48.75°$

The relation:

tan i=$\frac{OC}{OB}=\frac{R}{d_{1}}$

Therefore R=$tan\;48.75° \times 0.8=0.91m$

Area of the surface of water=$nR^{2}=n(0.91)^{2}=2.61m^{2}$ $2.61m^{2}$ is found to be the area of water through which the light from the water can emerge out.

Question 6:

There is a prism which is made of glass of unknown refractive index and a parallel bean of light is incident on the face of the prism. $40°$ is found to be the angle of minimum deviation. Find the refractive index of the material with which the prism is made. The angle of refraction of prism is $60°$. Find the angle of minimum deviation when a parallel beam of light is passed through it when the prism is placed in water. (Refractive index 1.33)

Angle of minimum deviation, $\delta_{m}=40°$

Angle of the prism, A = $60°$

Refractive Index of water, $\mu=1.33$

Refractive Index of the material= $\mu ‘$

The angle of deviation is related to refractive index ($\mu ‘$)

$\mu ‘$=$\frac{sin\frac{A+\delta_{m}}{2}}{sin\frac{A}{2}}$

=$\frac{sin\frac{60°+40°}{2}}{sin\frac{60°}{2}}=\frac{si n 50°}{sin 30°}$=1.532

1.532 is the refractive index of the material of the prism

$\delta_{m}$ is the new angle of minimum deviation since the prism is immersed in water.

The relation shows the refractive index of glass with respect to the water:

$\delta_{g}^{w}=\frac{\mu ‘}{\mu}=\frac{sin\frac{(A+\delta_{m}’)}{2}}{sin\frac{A}{2}} \\ sin\frac{(A+\delta_{m}’)}{2}=\frac{\mu ‘}{\mu}sin\frac{A}{2}\\ sin\frac{(A+\delta_{m}’)}{2}=\frac{1.532}{1.33}\times sin\frac{60°}{2}=0.5759\\ \frac{(A+\delta_{m}’)}{2}=sin^{-1}0.5759=35.16°\\ 60°+\delta_{m}’=70.32°\\ Therefore\;\delta_{m}’=70.32°-60°=10.32°$ $10.32°$ is the new minimum angle of deviation.

Question 7:

A glass with refractive index 1.55 is used to make double convex lenses with both sides having the same radius of curvature. The focal length is 20cm and hence find the radius of curvature required

Refractive Index of glass, $\mu=1.55$

Focal length of the double-convex lens, f=20 cm

Radius of curvature of one face of the lens=R1

Radius of curvature of the other face of the lens=R2

Radius of curvature of the double-convex lens=R

Therefore $R_{1}= R\;and \;R_{2}= – R$

Calculate the value of R

$\frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}} \right ]\\ \frac{1}{20}=(1.55-1)\left[\frac{1}{R}+\frac{1}{R} \right ]\\ \frac{1}{20}=0.55\times\frac{2}{R}\\ Therefore\;R=0.55\times2\times20=22cm$

Hence 22cm is the radius of curvature of the double-convex lens.

Question 8:

At a point P a beam of light converges. A lens can be placed on the way of convergent beam 12 cm from P. Find the point at which the beam converge if the lens is (i)20 cm focal length convex lens and (ii) 16cm focal length concave lens

The object is virtual and the image formed is real.

Object distance, u= +12 cm

(i) Focal length of the convex lens, f =20 cm

Image distance=v

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \frac{1}{v}-\frac{1}{12}=\frac{1}{20}\\ \frac{1}{v}=\frac{1}{20}+\frac{1}{12}=\frac{3+5}{60}=\frac{8}{60}\\ Therefore\;v=\frac{60}{8}=7.5cm$

The image will be formed 7.5cm away from the lens, to the right.

(ii)

Focal length of the concave lens, f =-16 cm

Image distance =v

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \frac{1}{v}=-\frac{1}{16}+\frac{1}{12}\\ \frac{-3+4}{48}=\frac{1}{48}\\ Therefore\;v=48cm$

Hence the image is formed to the right 48cm away from the lens

Question 9:

There is a concave lens of focal length 21cm and an object of size 3cm is placed 14cm in front of it. Analyze the image produced by the lens. What will happen if the object is moved further away from the lens?

Size of the object, $h_{1} = 3 cm$

Object distance, u = -14 cm

Focal length of the concave lens, f = -21 cm

Image distance = v

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \frac{1}{v}=-\frac{1}{12}-\frac{1}{14}\\ \frac{-2-3}{42}=\frac{-5}{42}\\ Therefore\;v=-\frac{42}{5}=-8.4cm$

Hence, the image is formed 8.4cm away and on the other side of the lens. We can understand that the image is virtual and erect from the negative sign.

$m=\frac{Image\;height(h_{2})}{Object\;height(h_{1})}=\frac{v}{u}\\ Therefore\;h_{2}=\frac{-8.4}{-14}\times3=0.6\times3=1.8cm$

1.8cm is the height of the image.

The virtual image will move toward the mirror if the object is moved further away from the lens. With increase in object distance the size of the image decreases.

Question 10:

What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens?

Ignore thickness of the lenses.

Focal length of the convex lens, f1=30cm

Focal length of the concave lens, f2= -20cm

Focal length of the system of lenses= f

The equivalent focal length of a system of two lenses in contact is given as:

$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$ $\frac{1}{f}=\frac{1}{30}+\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}$

Therefore f=-60cm

Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the system of lenses acts as a diverging lens.

Question 11:

A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Focal length of the objective lens, f1=2.0 cm

Focal length of the eyepiece, f2=6.25 cm

Distance between the objective lens and the eyepiece, d =15 cm

(a) Least distance of distinct vision, d’ = 25 cm

Image distance for the eyepiece, $v_{2}$= -25cm

Object distance for the eyepiece= u2

According to the lens formula, we have the relation:

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$ $\frac{1}{u_{2}}=\frac{1}{v_{2}}-\frac{1}{f_{2}}$ $\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$ $u_{2}=-5cm$

Image distance for the objective lens, $v_{1}=d+u_{2}=I5 – 5=10 cm$

Object distance for the objective lens = $u_{1}$

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$ $\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$ $\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$ $u_{1}=-2.5cm$

Hence, the magnifying power of the microscope is 20.

(b) The final Image is formed at infinity.

Therefore image distance for the eyepiece, $v_{2}=\infty$

Object distance for the eyepiece=$u_{2}$

According to the lens formula, we have the relation:

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$ $\frac{1}{\infty}-\frac{1}{u_{2}}=\frac{1}{6.25}$ $u_{2}=-6.25cm$

Image distance for the objective lens, $v_{1} = d + u_{2}=15 -6.25 = 8.75 cm$

Object distance for the objective lens=u1

According to the lens formula, we have the relation:

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$ $\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$ $\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$ $u_{1}=-\frac{17.5}{6.75}=-2.59cm$

Magnitude of the object distance, $|u_{1}|$=2.59 cm

The magnifying power of a compound microscope is given by the relation:

m=$\frac{v_{1}}{|u_{1}|}(\frac{d’}{|u_{2}|}$

=$\frac{8.75}{2.59}\times\frac{25}{6.25}$=13.51

Hence, the magnifying power of the microscope is 13.51.

Question 12:

A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,

Focal length of the objective lens,$f_{o}$= 8 mm = 0.8 cm

Focal length of the eyepiece, $f_{e}$= 2.5 cm

Object distance for the objective lens, $u_{o}$ = -9.0 mm = -0.9 cm

Least distance of distant vision, d=25 cm

Image distance for the eyepiece, $v_{e}$ = -d=-25 cm

Object distance for the eyepiece =$u_{e}$.

Using the lens formula, we can obtain the value of $u_{e}$ as:

$\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}$ $\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}$ $\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}$ $u_{e}=-\frac{25}{11}=-2.27cm$

We can also obtain the value of the image distance for the objective lens (v ) using the lens formula.

$\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}$ $\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}$ $\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}$ $u_{e}=-\frac{25}{11}=-2.27cm$

The distance between the objective lens and the eyepiece=Ind yo =2.27 – 7.2 = 9.47 cmttUl / id) The magnifying power of the microscope is calculated as:

= —7.2(1 +25 —25)=80 + 10)=88 0.9

Hence, the magnifying power of the microscope is 88.

Question 13:

A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal
length 6.0 cm. What is the magnifying power of the telescope? What is the separation
between the objective and the eyepiece?

Focal length of the objective lens, $f_{o}$= 144 cm
Focal length of the eyepiece, $f_{e}$= 6.0 cm
The magnifying power of the telescope is given as:
$\frac{f_{o}}{f_{e}}$

=$\frac{144}{6}$=24

The separation between the objective lens and the eyepiece is calculated as: $f_{o}+f_{e}$=144+6=150cm

Therefore 24 is the magnifying power of the telescope and the separation between the objective lens and the eyepiece is 150cm

Question 14:

(i) A giant refracting telescope at an observatory has an objective lens of focal
length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular
magnification of the telescope?

(ii)If this telescope is used to view the moon, what is the diameter of the image
of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 106 m$,and the radius of lunar orbit is $3.8 \times 108m$.

Focal length of the objective lens, $f_{o}$=15m =$15 \times 102cm$
Focal length of the eyepiece, $f_{e}$= 1.0 cm
(i) The angular magnification of a telescope is given as:$\alpha \frac{f_{o}}{f_{e}}$

=$\frac{15\times10^{2}}{1}$=1500
Hence, the angular magnification of the given refracting telescope is 1500.
(ii) Diameter of the moon, d = $3.48 \times 106m$
Radius of the lunar orbit, $r_{o}$= $3.8 \times 108m$
Let d’ be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by
the image.

$\frac{d}{r_{o}}=\frac{d’}{f_{o}}$ $\frac{3.48\times10^{6}}{3.8\times10^{8}}=\frac{d’}{15}$

Therefore d’=$\frac{3.48}{3.8}\times10^{-2}\times15$ $13.74\times10^{-2}m=13.74cm$

Therefore 13.74cm is the diameter of the moons image formed by the objective lens.

Question 15:
Use the mirror equation to deduce that:
(i) an object placed between f and 2f of a concave mirror produces a real image
beyond 2f.
(ii) a convex mirror always produces a virtual image independent of the location
of the object.
(iii) the virtual image produced by a convex mirror is always diminished in size
and is located between the focus and the pole.
(iv) an object placed between the pole and focus of a concave mirror produces a
virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of images
that one obtains from explicit ray diagrams.]

(i) For a concave mirror, the focal length (f) is negative.
Therefore f < 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
For image distance v, we can write the lens formula as:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$——————(i)

The object lies between f and 2f.

Therefore                    2f<u<f
$\frac{1}{2f}<\frac{1}{u}<\frac{1}{f}$ $-\frac{1}{2f}<-\frac{1}{u}<-\frac{1}{f}$ $\frac{1}{f}-\frac{1}{2f}<-\frac{1}{f}-\frac{1}{u}<0$——————-(ii)

Using equation (i), we get:

$\frac{1}{2f}<\frac{1}{v}<0$
$\frac{1}{v}$ is negative, i.e., v is negative.

$\frac{1}{2f}<\frac{1}{v}$

2f>v

-v>-2f
Therefore, the image lies beyond 2f.
(ii) For a convex mirror, the focal length (f) is positive.
Therefore f > 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
For image distance v, we have the mirror formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

Usig eqn (ii), we can conclude that:

$\frac{1}{v}<0$

v>0

Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object
distance.
(iii) For a convex mirror, the focal length (f) is positive.
Therefore f > 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative,
Therefore u < 0
For image distance v, we have the mirror formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

But we have u<0

Therefore $\frac{1}{v}>\frac{1}{f}$

v<f
Hence, the image formed is diminished and is located between the focus (f) and the
pole.
(iv) For a concave mirror, the focal length (f) is negative.

Therefore f < 0

When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
It is placed between the focus (f) and the pole.

Therefore f>u>0

$\frac{1}{f}<\frac{1}{u}<0$ $\frac{1}{f}-\frac{1}{u}<0$
For image distance v, we have the mirror formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

Therefore $\frac{1}{v}<0$

v>0
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write:

$\frac{1}{u}>\frac{1}{v}$

v>u
Magnification, m=$\frac{v}{u}$> 1
Hence, the formed image is enlarged.
Question 16:
A small pin fixed on a table top is viewed from above from a distance of 50 cm. By
what distance would the pin appear to be raised if it is viewed from the same point
through a 15 cm thick glass slab held parallel to the table? Refractive index of glass =
1.5. Does the answer depend on the location of the slab?

Actual depth of the pin, d = 15 cm
Apparent dept of the pin =d’
Refractive index of glass,$\mu$=1.5
Ratio of actual depth to the apparent depth is equal to the refractive index of glass,
i.e.

$\mu=\frac{d}{d’}$

Therefore d’=$\frac{d}{\mu}$ $\frac{15}{1.5}$=10cm
The distance at which the pin appears to be raised =d’-d=15-10=5cm
For a small angle of incidence, this distance does not depend upon the location of the
slab.

Question 17:
(i) Figure below shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive
index 1.68. The outer covering of the pipe is made of a material of refractive index
1.44. What is the range of the angles of the incident rays with the axis of the pipe
for which total reflections inside the pipe take place, as shown in the figure.
(ii) What is the answer if there is no outer covering of the pipe? (i) Refractive index of the glass fibre,

Refractive index of the outer covering of the pipe, $\mu_{2}$= 1.44
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i’
The refractive index ($\mu$) of the inner core − outer core interface is given as:

$\mu=\frac{\mu_{2}}{\mu_{1}}=\frac{1}{sin i’}$ $sin i’=\frac{\mu_{1}}{\mu_{2}}$ $=\frac{1.44}{1.68}=0.8571$ $Therefore\;i’=59°$
For the critical angle, total internal reflection (TIR) takes place only when i>i’, i.e., i
> $59°$
Maximum angle of reflection,           $r_{max}=90°-i’=90°-59°=31°$
Let, $i_{max}$ be the maximum angle of incidence.
The refractive index at the air − glass interface, $\mu_{1}=1.68$ $\mu_{1}=\frac{sin i_{max}}{sin r_{max}}$

1.68 sin 31°

$1.68\times 0.5150$=0.8652

Therefore $i_{max}=sin^{-1}0.8652=60°$

Thus, all the rays incident at angles lying in the range 0 < i <$60°$ will suffer total internal reflection.

(ii) If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe, $\mu_{1}$=Refractive index of air=1
For the angle of incidence i =$90°$, we can write Snell’s law at the air − pipe interface
as:
$\frac{sin i}{sin r}=\mu_{2}=1.68$ $sin r=\frac{sin 90°}{1.68}=\frac{1}{1.68}$ $r=sin^{-1}(0.5932)$ $=36.5°$

Therefore $i’=90°-36.5°=53.5°$

Since i’>r, all incident rays will suffer total internal reflection.

Question 18:
(i) You have learnt that plane and convex mirrors produce virtual images of objects.
Can they produce real images under some circumstances? Explain.
(ii) A virtual image, we always say, cannot be caught on a screen.
Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’
(i.e., the retina) of our eye. Is there a contradiction?
(iii) A diver under water, looks obliquely at a fisherman standing on the bank of a lake.
Would the fisherman look taller or shorter to the diver than what he actually is?
(iv) Does the apparent depth of a tank of water change if viewed obliquely? If so,
does the apparent depth increase or decrease?
(v) The refractive index of diamond is much greater than that of ordinary glass.
Is this fact of some use to a diamond cutter?

(i) Yes

Plane and convex mirrors can produce real images as well. If the object is virtual,
i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror)
are reflected to a point on a screen placed in front of the mirror, then a real image will
be formed.
(ii) No
A virtual image is formed when light rays diverge. The convex lens of the eye causes
these divergent rays to converge at the retina. In this case, the virtual image serves
as an object for the lens to produce a real image.
(iii) The diver is in the water and the fisherman is on land (i.e., in air). Water is a
denser medium than air. It is given that the diver is viewing the fisherman. This
indicates that the light rays are travelling from a denser medium to a rarer
medium. Hence, the refracted rays will move away from the normal. As a result,
the fisherman will appear to be taller.
(iv) Yes; Decrease
The apparent depth of a tank of water changes when viewed obliquely. This is because
light bends on travelling from one medium to another. The apparent depth of the tank
when viewed obliquely is less than the near-normal viewing.
(v) Yes
The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The
critical angle for diamond is less than that for glass. A diamond cutter uses a large
angle of incidence to ensure that the light entering the diamond is totally reflected
from its faces. This is the reason for the sparkling effect of a diamond.
Question 19:
The image of a small electric bulb fixed on the wall of a room is to be obtained on the
opposite wall 3 m away by means of a large convex lens. What is the maximum
possible focal length of the lens required for the purpose?

Distance between the object and the image, d = 3 m

Maximum focal length of the convex lens =$f_{max}$
For real images, the maximum focal length is given as:

$f_{max}=\frac{d}{4}$ $\frac{3}{4}=0.75m$
Hence, for the required purpose, the maximum possible focal length of the convex lens
is 0.75 m.
Question 20:
A screen is placed 90 cm from an object. The image of the object on the screen is
formed by a convex lens at two different locations separated by 20 cm. Determine the
focal length of the lens.

Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:

$f=\frac{D^{2}-d^{2}}{4D}$

=$\frac{90^{2}-(20^{2})}{4\times90}=\frac{770}{36}=21.39cm$
Therefore, the focal length of the convex lens is 21.39 cm.

Question 21:
(i) Determine the ‘effective focal length’ of the combination of the two lenses in
Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does
the answer depend on which side of the combination a beam of parallel light is incident?
Is the notion of effective focal length of this system useful at all?

(ii) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement
(a) above. The distance between the object and the convex lens is 40 cm. Determine
the magnification produced by the two-lens system, and the size of the image.

Focal length of the convex lens, $f_{1}$= 30 cm
Focal length of the concave lens, $f_{2}$ = −20 cm
Distance between the two lenses, d = 8.0 cm
(i) When the parallel beam of light is incident on the convex lens first:
According to the lens formula, we have:

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
Where,
$u_{1}$= Object distance
$v_{1}$= Image distance

$\frac{1}{v_{1}}=\frac{1}{30}-\frac{1}{\infty}=\frac{1}{30}$

Therefore $v_{1}=30cm$
The image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$

Where,
$u_{2}$=Object distance
= (30 − d) = 30 − 8 = 22 cm
$v_{2}$= Image distance

$\frac{1}{v_{2}}=\frac{1}{22}-\frac{1}{20}=\frac{10-11}{220}=\frac{-1}{220}$

Therefore $v_{2}=-220cm$

The parallel incident beam appears to diverge from a point that
is $(220-\frac{d}{2}=220-4)216cm$ from the centre of the combination of the two lenses.
When the parallel beam of light is incident, from the left, on the concave lens
first:
According to the lens formula, we have:

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$ $\frac{1}{v_{2}}=\frac{1}{f_{2}}+\frac{1}{u_{2}}$
Where,
$u_{2}$=Object distance = −∞

$v_{2}$=Image distance

$\frac{1}{v_{2}}=\frac{1}{-20}+\frac{1}{-\infty}=-\frac{1}{20}$

Therefore $v_{2}=-20cm$
The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
Where,
$u_{1}$= Object distance
= −(20 + d) = −(20 + 8) = −28 cm
$v_{1}$= Image distance

$\frac{1}{v_{1}}=\frac{1}{30}+\frac{1}{-28}=\frac{14-15}{420}=\frac{-1}{420}$

Therefore $v_{2}=-420cm$

Hence, the parallel incident beam appear to diverge from a point that is (420 − 4) 416
cm from the left of the centre of the combination of the two lenses.
The answer does depend on the side of the combination at which the parallel beam of
light is incident. The notion of effective focal length does not seem to be useful for this
combination.
(ii) Height of the image, $h_{1}$ = 1.5 cm
Object distance from the side of the convex lens, $u_{1}$ = -40 cm
$|u_{1}|=40cm$

According to the lens formula:

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
Where,

$v_{1}$= Image distance

$\frac{1}{v_{1}}=\frac{1}{30}+\frac{1}{-40}=\frac{4-3}{120}=\frac{1}{120}$

Therefore $v_{1}=120cm$

Magnification, m=$\frac{v_{1}}{|u_{1}|}=\frac{120}{40}=3$
Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens.
According to the lens formula:

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$
Where,
$u_{2}$= Object distance
= +(120 − 8) = 112 cm.

$v_{2}$=Image distance

$\frac{1}{v_{2}}=\frac{1}{-20}+\frac{1}{112}=\frac{-112+20}{2240}=\frac{-92}{2240}$

Therefore $v_{2}=\frac{-2240}{92}$

Magnification, m’=$|\frac{v_{2}}{u_{2}}|=\frac{2240}{92}\times \frac{1}{112}=\frac{20}{92}$

Hence, the magnification due to the concave lens is $\frac{20}{92}$.
The magnification produced by the combination of the two lenses is calculated as:

$m \times m’$
The magnification of the combination is given as:

$\frac{h_{2}}{h_{1}}=0.652$ $h_{2}=h_{1}\times 0.652$
Where,
$h_{1}$= Object size = 1.5 cm
$h_{2}$ = Size of the image
Therefore $h_{2}=0.652\times1.5=0.98cm$

Hence, the height of the image is 0.98 cm.

Question 22:
At what angle should a ray of light be incident on the face of a prism of refracting angle
$60°$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

The incident, refracted, and emergent rays associated with a glass prism ABC are
shown in the given figure. Angle of prism, Therefore A = $60°$
Refractive index of the prism, $\mu$= 1.524
$i_{1}$= Incident angle
$r_{1}$= Refracted angle
$r_{2}$= Angle of incidence at the face AC
e= Emergent angle = $90°$
According to Snell’s law, for face AC, we can have:

$\frac{sin e}{sin r_{2}}=\mu$ $sin r_{2}=\frac{1}{\mu}\times sin\;90°$

=$\frac{1}{1.524}=0.6562$

Therefore $r_{2}=sin^{-1}0.6562=41°$

Therefore ${r}_{1}=A-{}_{}$