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

NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments: The chapter comes under the unit optics. The solutions of NCERT class 12 physics chapter 9 ray optics and the optical instrument is unavoidable as far as CBSE board exam is considered. Sometimes in such exams, questions are directly asked from NCERT book for which CBSE NCERT solutions for class 12 physics chapter 9 ray optics and the optical instrument can be helpful to score well. This chapter covers topics related to reflection, refraction, scattering and optical instruments. There are many problems related to refraction and reflection in the NCERT solutions for class 12 physics chapter 9 ray optics and optical instruments. Each and every topic of chapter 9 ray optics and optical instruments are important for exams and there is no scope of skipping any topic. NCERT solutions are one of the keys to success in CBSE board exams.

Knowledge of sign convention is mandatory to understand the solutions of NCERT class 12 physics chapter 9 ray optics and optical instruments. Some of the important formulas that will help in CBSE NCERT solutions for class 12 chapter 9 ray optics and optical instruments are listed below:

## NCERT solutions for class 12 physics chapter 9 ray optics and optical instruments- Important formulas:

 Formula Description Mirror equation: v is object distance from the pole of the mirror. u is the image distance from the pole of the mirror. F is the focal length of the mirror which is approximately equal to half the radius of curvature Magnification of a mirror $m= \frac{h_{i}}{h_{o}}= \frac{-v}{u}$ $h_{i}=$ height of the image from the principal axis $h_{o}=$ height of the object from the principal axis If a prism of refractive index n2is placed in a medium of refractive index n1then $\boldsymbol{\frac{n_2}{n_1} = \frac{\sin \left ( \frac{D _{m}+A}{2} \right )}{\sin \left ( A/2 \right )}}$ A is the angle of the prism Dm is the angle of  minimum deviation Thin lens formula $\frac{1}{v}-\frac{1}{u}= \frac{1}{f}$   Lens makers formula $\boldsymbol{\frac{1}{f}= \left ( \frac{\mu _{2}}{\mu _{1}}-1 \right )\left ( \frac{1}{R_{1}}- \frac{1}{R_{2}}\right )}$ v, u are image and object distance from the optical centre of the lens and f is the focal length $\mu _{1}=$ the refractive index of the medium of the object $\mu _{2}=$ the refractive index of the lens $R_{1}\, and \, R_{2}$ are the radius of curvature of two surface Power of a lens $\boldsymbol{P=\frac{1}{f}}$ Where P is the power and f is the focal length in meters The magnifying power of a simple microscope $m=1+ \frac{D}{f}$ For compound microscope magnification $m=m_e \times m_0$ D = 25 cm is the least distance of distinct vision and f is the focal length of the convex lens $m_e$ is magnification due to the eyepiece. $m_o$ is magnification produced by objective. For a telescope magnifying power is given by $m=\frac{\beta}{\alpha}=\frac{f_o}{f_e}$ Beta is the angle subtended by eyepiece at the image Alpha is the angle subtended at the eye by the object

## Q 9.1  A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm.  At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Given, size of the candle, h = 2.5 cm

Object distance, u = 27 cm

The radius of curvature of the concave mirror, R = -36 cm

focal length of a concave mirror = R/2 = -18 cm

let image distance = v

now, as we know

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

$\frac{1}{-27}+\frac{1}{v}=\frac{1}{-18}$

$\frac{1}{v}=\frac{1}{-18} + \frac{1}{27}$

$v= -54cm$

now, let the height of image be  $h'$

magnification of the image is given by

$m=\frac{h'}{h}=-\frac{v}{u}$

from here

${h'}=-\frac{-54}{-27}*2.5=-5cm$

Hence the size of the image will be -5cm. negative sign implies that the image is inverted and real

if the candle is moved closer to the mirror, we have to move the screen away from the mirror in order to obtain the image on the screen. if the image distance is less than the focal length image cannot be obtained on the screen and image will be virtual.

## Ray Optics and Optical Instruments Excercise:

### Question:

Given, the height of needle, h = 4.5 cm

distance of object = 12 cm

focal length of convex mirror =  15 cm.

Let the distance of the image be v

Now as we know

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

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

$\frac{1}{v}=\frac{1}{15} - \frac{1}{-12}$

$\frac{1}{v}=\frac{1}{15} + \frac{1}{12}$

v = 6.7 cm

Hence the distance of the image is 6.7 cm from the mirror and it is on the other side of the mirror.

Now, let the size of the image be h'

so.

$m=-\frac{v}{u} = \frac{h'}{h}$

$h'= -\frac{v}{u}*h$

$h'= -\frac{6.7}{-12}*4.5$

$h'= 2.5 cm$

Hence the size of the image is 2.5 cm. positive sign implies the image is erect, virtual and diminished.

magnification of the image =    $\frac{h'}{h}$     =     $\frac{2.5}{4.5}$     =   0.56

m = 0.56

The image will also move away from the mirror if we move the needle away from the mirror, and the size of the image will decrease gradually.

Given:

Actual height of the tank,h = 12.5 cm

Apparent height of tank,h' =  9.4 cm

let refrective index of the water be $\mu$

$\mu = \frac{h}{h'} =\frac{12.5}{9.4} = 1.33 (approx)$

so the refractive index of water is approximately 1.33.

Now, when water is replaced with a liquid having  $\mu = 1.63$

$\mu = \frac{h}{h'} =\frac{12.5}{h'_{new}} = 1.63$

$h'_{new}= \frac{12.5}{1.63}= 7.67 cm$

Hence the new apparent height of the needle is 7.67 cm.

Total distance we have to move in a microscope = 9.4 - 7.67  = 1.73 cm.

Since new apparent height is lesser than the previous apparent height we have to move UP the microscope in order to focus the needle.

As we know,by snell's law

$\mu_1sin\theta _1=\mu _2sin\theta _2$  where,

$\mu_1$ = refrective index of medium 1

$\theta _1$ = incident angle in medium 1

$\mu _2$ = refrective index of medium 2

$\theta _2$ = refraction angle in medium 2

NOW, APPLYING IT FOR fig (a)

$1sin60=\mu _{glass}sin35$

$\mu _{glass}=\frac{sin60}{sin35}=\frac{0.866025}{0.573576 } = 1.509$

Now applying for fig (b)

$1sin60=\mu _{water}sin47$

$\mu _{water}=\frac{sin60}{sin47}=\frac{.8660}{.7313} = 1.184$

Now in fig (c) Let refraction angle be $\theta$ so,

$\mu _{water}sin45=\mu _{glass}sin\theta$

$sin\theta =\frac{\mu _{water}*sin45}{\mu _{glass}}$

$sin\theta =\frac{1.184*0.707}{1.509} = 0.5546$

$\theta = sin^{-1}(0.5546) = 38.68$

Therefore the angle of refraction when ray goes from water to glass in fig(c) is 38.68.

Rays of light will emerge out in all direction and upto the angle when total internal reflection starts i.e. when the angle of refraction is 90 degree.

let the incident angle be i when refraction angle is 90 degree.

so, by snell's law

$\mu _{water}sini=1sin90$

from here, we get

$sini=\frac{1}{1.33}$

$i=sin^{-1}(\frac{1}{1.33})=48.75^0$

Now Let R be Radius of the circle of the area from which the rays are emerging out. and d be the depth of water which is = 80 cm.

From the figure:

$tani=\frac{R}{d}$

$R = tani*d=tan48.75^0*80cm$

R = 91 cm

So the area of water surface through which rays will be emerging out is

$\Pi R^2 = 3.14*(91)^2cm^2$

$= 2.61m^2$

therefore required area  $= 2.61m^2$.

## Q 9.6 A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be $40^{\circ}$. What is the refractive index of the material of the prism? The refracting angle of the prism is $60^{\circ}$. If the prism is placed in water, predict the new angle of minimum deviation of a parallel beam of light.

In Prism :

Prism angle ($A$) =  First Refraction Angle ( $r _1$) + Second refraction angle ( $r _2$)

also, Deviation angle ($\delta$) = incident angle($i$) + emerging angle($e$)  -  Prism angle ($A$) ..............(1)

the deviation angle is minimum when the incident angle($i$) and an emerging angle($e$) are the same. in other words

$i=e$ ...........(2)

from (1) and (2)

$\delta_{min} = 2i-A$

$i=\frac{\delta_{min} +A}{2}$..........................(3)

We also have

$r _1 = r_2 = r = \frac{A}{2}$   .................(4)

Now applying snells law using equation (3) and (4)

$\mu _1sini=\mu _2sinr$

$1sin(\frac{\delta _{min}+A}{2})=\mu _2sin\frac{A}{2}$

$\mu _2=\frac{sin(\frac{\delta _{min}+A}{2})}{sin\frac{A}{2}}$...................(5)

Given

$\\\delta _{min}= 40 \\A = 60$

putting those values in (5) we get

$\mu _2=\frac{sin(\frac{40+60}{2})}{sin\frac{60}{2}}=\frac{sin50}{sin30}=1.532$

Hence the refractive index of the prism is 1.532.

Now when the prism is in the water.

Applying Snell's law:

$\mu _1sin(\frac{\delta _{min}+A}{2})=\mu _2sin\frac{A}{2}$

$1.33sin(\frac{\delta _{min}+60}{2})=1.532sin\frac{60}{2}$

$sin(\frac{\delta _{min}+60}{2})=\frac{1.532*0.5}{1.33}$

$\frac{\delta _{min}+60}{2}=sin^{-1}\frac{1.532*0.5}{1.33}$

$\delta _{min} =2sin^{-1}0.5759 - 60$

$\delta _{min} =2*35.16- 60 =10.32^0$

Hence  minimum angle of deviation inside water is 10.32 degree.

As we know the lens makers formula

$\frac{1}{f}=(\mu _{21}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

[ This is derived by considering the case when the object is at infinity and image is at the focus]

Where $f$ = focal length of the lens

$\mu _{21}$ = refractive index of the glass of lens with the medium(here air)

$R_1$ and $R_2$ are the  Radius of curvature of faces of the lens.

Here,

Given, $f$ = 20cm,

$R_1$  = $R$ and $R_2$  =  $-R$

$\mu _{21}$ = 1.55

Putting these values in te equation,

$\frac{1}{20}=(1.55-1)(\frac{1}{R}-\frac{1}{-R})$

$\frac{2}{R}=\frac{1}{20}*\frac{1}{0.55}$

$R = 40*0.55$

$R = 22cm$

Hence Radius of curvature of the lens will be 22 cm.

## Q 9.8  A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?

In any Lens

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

$\\v=$ the distance of the image from the optical centre

$\\u=$ the distance of the object from the optical centre

$\\f=$ the focal length of the lens

a)

Here, The beam converges from the convex lens to point P. This image P  will now act as an object for the new lens which is placed 12 cm from it and focal length being 20 cm.

So

$\frac{1}{v} - \frac{1}{12}=\frac{1}{20}$

$\frac{1}{v}=\frac{1}{20} + \frac{1}{12}$

$\frac{1}{v}=\frac{8}{60}$

$v = 7.5 cm$

Hence distance of image is 7.5 cm and it will form towards the right as the positive sign suggests.

b)

HERE, Focal length $f$ = -16cm

so,

$\frac{1}{v} - \frac{1}{12}=\frac{1}{-16}$

$\frac{1}{v} =\frac{1}{-16} + \frac{1}{12} = \frac{1}{48}$

$v = 48 cm$

Hence image distance will be 48 cm in this case, and it will be in the right direction(as the positive sign suggests)

In any Lens

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

$\\v=$ the distance of the image from the optical centre

$\\u=$ the distance of the object from the optical centre

$\\f=$ the focal length of the lens

Here Given,

$\\u=$ -14 cm

$\\f=$ -21 cm

$\frac{1}{v} - \frac{1}{-14}=\frac{1}{-21}$

$\frac{1}{v} =\frac{1}{-21}- \frac{1}{14} = \frac{-5}{42}$

$v = -\frac{42}{5}= -8.4cm$

Hence image distance is -8.4 cm. the negative sign indicates the image is erect and virtual.

Also as we know,]

$m= -\frac{v}{u} = \frac{h'}{h}$

From Here

$h'= -\frac{v}{u}h$

$h'= -\frac{-8.4}{-12}*3= 1.8 cm$

Hence the height of the image is 1.8 cm.

As we move object further away from the lens, the image will shift toward the focus of the lens but will never go beyond that.size of the object will decrease as we move away from the lens.

When two lenses are in contact the equivalent is given by

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$

where $f_1$ and $f_2$ are the focal length of two individual lenses.

SO,Given,

$f_1 =$ 30 cm and $f_2 = -20cm$ (as focal length of the convex lens is positive and of the concave lens is negative by convention)

putting these values we get,

$\frac{1}{f}=\frac{1}{30}+\frac{1}{-20}$

$\frac{1}{f}=-\frac{1}{60}$

$f = -60cm$

Hence equivalent focal length will be -60 cm and since it is negative,equivalent is behaving as a concave lens which is also called diverging lens.

In a compound microscope, first, the image of an object is made by the objective lens and then this image acts as an object for eyepiece lens.

Given

the focal length of objective lens = $f_{objective}$ = 2 cm

focal length of eyepiece lense = $f_{eyepiece}$ = 6.25cm

Distance between the objective lens and eyepiece lens = 15 cm

a)

Now in Eyepiece lense

Image distance = $v_{final}$ = -25 cm (least distance of vision with sign convention)

focal length = $f_{eyepiece}$ = 6.25 cm

$\frac{1}{f_{eyepiece}} = \frac{1}{v_{final}} - \frac{1}{u}$

$\frac{1}{u} = \frac{1}{v_{final}} -\frac{1}{f_{eyepiece}}$

$\frac{1}{u} = \frac{1}{-25} -\frac{1}{6.25} = -\frac{1}{5}$

$u = -5 cm.$

Now, this object distance $u$ is from the eyepiece lens since the distance between lenses is given we can calculate this distance from the objective lens.

the distance of $u$ from objective lens =  $d+u=$ $15 - 5=10cm$. This length will serve as image distance for the objective lens.

$v = 10 cm$

so in the objective lens

$\frac{1}{f_{objective}} = \frac{1}{v} - \frac{1}{u_{initial}}$

$\frac{1}{u_{initial}} = \frac{1}{v} - \frac{1}{f_{objective}}$

$\frac{1}{u_{initial}} = \frac{1}{10} - \frac{1}{2} = -\frac{4}{10}=-\frac{2}{5}$

$u_{intial}$ = -2.5 cm

Hence the object distance required is -2.5 cm.

Now, the magnifying power of a microscope is given by

$m=\frac{v}{|u_{initial}|}(1+\frac{d}{f_{eyepiece}})$      where $d$ is the least distance of vision

so putting these values

$m=\frac{10}{2.5}(1+\frac{25}{6.25}) = 20$

Hence the lens can magnify the object to  20 times.

b) When image is formed at infinity

in eyepiece lens,

$\frac{1}{f_{eyepiece}} = \frac{1}{v_{final}} - \frac{1}{u}$

$\frac{1}{6.25} = \frac{1}{infinity} - \frac{1}{u}$

from here $u =$ - 6.25., this distance from objective lens = $d+u$ = 15 - 6.25 = 8.75 = $v$

in the optical lens:

$\frac{1}{f_{objective}} = \frac{1}{v} - \frac{1}{u_{initial}}$

$\frac{1}{2} = \frac{1}{-6.25} - \frac{1}{u_{initial}}$

$\frac{1}{u_{initial}} =- \frac{6.75}{17.5}$

$u_{initial}=-2.59cm$

Now,

$m=\frac{v}{|u_{initial}|}(1+\frac{d}{f_{eyepiece}})$      where $d$ is the least distance of vision

putting the values, we get,

$m=\frac{8.75}{2.59}(1+\frac{25}{6.25})= 13.51$

Hence magnifying power, in this case, is 13.51.

Inside  a microscope,

For the eyepiece lens,

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

we are given

$v_{eyepiece}= -25cm$

$f_{eyepiece}= 2.5cm$

$\frac{1}{2.5}=\frac{1}{-25} - \frac{1}{u_{eyepiece}}$

$\frac{1}{u_{eyepiece}}=\frac{1}{-25} - \frac{1}{2.5}=-\frac{11}{25}$

$u_{eyepiece}=- \frac{25}{11}=-2.27cm$

we can also find this value by finding image distance in the objective lens.

So, in objective lens

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

we are given

$f_{objective}= 0.8$

$u_{objective}= -0.9$

$\frac{1}{0.8}=\frac{1}{v_{objective}} - \frac{1}{-0.9}$

$\frac{1}{v_{objective}} = \frac{0.1}{0.72}$

$v_{objective} = 7.2cm$

Distance between object lens and eyepiece = $|u_{eyepiece}| + v_{objective}$ = 2.27 + 7.2 = 9.47 cm.

Now,

Magnifying power :

$m = \frac{v_{obejective}}{|u_{objective}|}(1+\frac{d}{f_{eyepiece}})$

$m = \frac{7.2}{0.9}(1+\frac{25}{2.5})= 88$

Hence magnifying power for this case will be 88.

## Q 9.13 A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

The magnifying power of the telescope is given by

$m=\frac{f_{objective}}{f_{eyepiece}}$

Here, given,

focal length of objective lens =  $f_{objective}=$   144 cm

focal length of eyepiece lens = $f_{eyepiece}=$  6 cm

$m=\frac{f_{objective}}{f_{eyepiece}}=\frac{144}{6}=24$

Hence magnifying power of the telescope is 24.

in the telescope distance between the objective and eyepiece, the lens is given by

$d= f_{objective}+ f_{eyepiece}$

$d = 144+6=150$

Therefore, the distance between the two lenses is 250 cm.

Angular magnification in the telescope is given by :

angular magnification =  $\alpha =$      $\frac{f_{objective} }{f_{eyepiece}}$

Here given,

focal length of objective length = 15m = 1500cm

the focal length of the eyepiece = 1 cm

so, angular magnification, $\alpha =$  $\frac{1500}{1}$

$\alpha = 1500$

Given,

The radius of the lunar orbit,r =  $3.8 \times 108m$.

The diameter of the moon,d =  $3.48 \times 106m$

focal length $f = 15m$

let  $d_1$  be the diameter of the image of the moon which is formed by the objective lens.

Now,

the angle subtended by diameter of the moon will be equal to the angle subtended by the image,

$\frac{d}{r}=\frac{d_1}{f}$

$\frac{3.48*10^6}{3.8*10^8}=\frac{d_1}{15}$

$d_1=13.74 cm$

Hence the required diameter is 13.74cm.

Q 9.15 (a) Use the mirror equation to deduce that:

an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

The equation we have for a mirror is:

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

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

Given condition  $f and $v>2f$

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

$-\frac{1}{2f}>-\frac{1}{u}>-\frac{1}{f}$

$\frac{1}{f}-\frac{1}{2f}>\frac{1}{f}-\frac{1}{u}>\frac{1}{f}-\frac{1}{f}$

$\frac{1}{2f}>\frac{1}{v}>0$

${2f}<{v}<0$

Here $f$ has to be negative in order to satisfy the equation and hence we conclude that our mirror is a concave Mirror. It also satisfies that $-v>-2f$(image lies beyond 2f)

Q. 9.15 (b)    Use the mirror equation to deduce that:

a convex mirror always produces a virtual image independent of the location of the object.

In a convex mirror focal length is positive conventionally.

so we have mirror equation

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

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

Here,  since object distance is always negative whenever we put our object in the left side of the convex mirror(which we always do, generally). So     $\frac{1}{v}$   is always the sum of two positive quantity(negative sign in the equation and negative sign of the $u$ will always make positive) and hence we conclude that $v$ is always greater than zero which means the image is always on the right side of the mirror which means it is a virtual image. Therefore, a convex lens will always produce a virtual image regardless of anything.

Q 9.15 (c)    Use the mirror equation to deduce that:

the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

In a convex mirror focal length is positive conventionally.

so we have mirror equation

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

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

here since $f$ is positive and $u$ is negative (conventionally) so we have,

$\frac{1}{v}>\frac{1}{f}$ that is '

$v

which means the image will always lie between pole and focus.

Now,

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

$magnification (m)=-\frac{v}{u} = \frac{f}{f-u}$

here since $u$ is always negative conventionally, it can be seen that magnification of the image will be always less than 1 and hence we conclude that image will always be diminished.

Q 9.15 (d) Use the mirror equation to deduce that:

an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

The focal length $f$ of concave mirror is always negative.

Also conventionally object distance $u$ is always negative.

So we have mirror equation:

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

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

Now in this equation whenever $u , $\frac{1}{v}$  will always be positive which means $v$ is always positive which means it lies on the right side of the mirror which means image is always virtual.

Now,

$m=-\frac{v}{u}=-\frac{f}{u-f}$

since the denominator is always less than the numerator, so the magnitude  magnification will always be greater than 1

Hence we conclude that image is always gonna be enlarged.

Hence an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

As we know,

Refractive index =

$\frac{actualdepth}{apparentdepth}$

Here actual depth = 15cm

let apparent depth be d'

And refractive index of the glass = 1.5

now putting these values, we get,

$1.5 = \frac{15}{d'}$

$d' =10$

the change in the apparent depth = 15 - 10 = 5 cm.

as long as we are not taking slab away from the line of sight of the pin, the apparent depth does not depend on the location of the slab.

## Q 9.17 (a)  In the following figure 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.

We are given,

Refractive index of glass($\mu _{glass}$) and outer covering($\mu _{outerlayer}$) is 1.68 and 1.44 respectively.

Now applying snell's law on upper glass - outer layer,

$\mu _{glass}sini'=\mu_{outerlayer}sin90$

$i' =$ the angle from where total Internal reflection starts

$sini'=\frac{\mu_{outerlayer}}{\mu_{glass}} = \frac{1.44}{1.68}=0.8571$

$i' = 59^0$

At this angle, in the air-glass interface

Refraction angle $r$ = 90 - 59 = 31 degree

let Incident Angle be  $i$ .

Applying Snell's law

$1sini=\mu_{glass}sinr$

$sini=1.68sin31=0.8652$

$i=60$(approx)

Hence total range of incident angle for which total internal reflection happen is $0

In the case when there is no outer layer,

Snell's law at glass-air interface(when the ray is emerging out from the pipe)

$\mu _{glass}sini'=1sin90$

$sini'=\frac{1}{\mu_{glass}} = \frac{1}{1.68}=0.595$

$i'=$ 36.5

refractive angle $r$ corresponding to this = 90 - 36.5 = 53.5.

the angle r is greater than the critical angle

So for all of the incident angles, the rays will get total internally reflected.in other words, rays won't bend in air-glass interference, it would rather hit the glass-air interference and get reflected

Q 9.18 (a)    Answer the following question:

You have learned that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.

If our object is virtual then Plane and convex mirrors can produce a real image. That is, when the light coming from infinity goes into the convex mirror, it creates a virtual object behind the convex mirror. the reflection of this virtual object in the convex mirror can be taken out on screen and hence convex mirror can make a real image.

Q 9.18 (b)    Answer the following question:

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?

No, there is no contradiction. A virtual image is formed whenever the light rays are diverging. We have a convex lens in our eye. This convex lens converges the diverging rays into our retina and forms a real image. In other words, the virtual image acts as an object to the convex lens of our eye to form a real image, which we see on the screen called retina.

Q 9.18 (c)  Answer the following question:

A diver underwater 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?

The diver is in denser medium (water) and fisherman is in lighter medium (air). As the diver is looking at the fisherman, rays of light will go from fisherman to divers eye, that is, from lighter medium to denser medium. Since rays deflect toward normal when it goes form lighter to a denser medium, the fisherman will look taller than actual to the diver.

Q 9.18 (d)  Answer the following question:

Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?

Yes, appearing depth of water will decrease when we view obliquely, this happens because of the fact that light bends from its direction whenever it goes from one medium to another medium.

Q 9.18 (e)  Answer the following question:

The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?

We use diamond as a cutter because it is very hard and sharp. The refractive index is high in diamond ensures that light goes through multiple total internal reflections so that light goes in all direction. This is the reason behind the shining of the diamond. Light entering  is totally reflected from faces before it getting out, hence  producing a sparkling effect

As we know for real image, the maximum focal length is given by

$f_{max}= \frac{d}{4}$

where d is the distance between the object and the lens.

So putting values we get,

$f_{max}= \frac{3}{4}=0.75$

Hence maximum focal length required is 0.75.

As we know that the relation between focal length $f$, the distance between screen $D$ and distance between two locations of the object $d$ is :

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

Given:$D$ = 90 cm.,$d$ = 20 cm ,

so

$f=\frac{90^2-20^2}{4*90}$

$f=\frac{90^2-20^2}{4*90}=\frac{770}{36}=21.39cm$

Hence the focal length of the convex lens is 21.39 cm.

Here there are two cases, first one is the one when we see it from convex side i.e. Light are coming form infinite and going into convex lens first and then goes to concave lens afterwords. The second case is a just reverse of the first case i.e. light rays are going in concave first.

1)When light is incident on convex lens first

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

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

$\frac{1}{v}=\frac{1}{30}+\frac{1}{infinite}$

$v= 30cm$

Now this will act as an object for the concave lens.

$\frac{1}{f_{concave}}=\frac{1}{v_{fromconcave}}-\frac{1}{u_{fromconcave}}$

$u_{fromconcave}= 30 - 8 = 22 cm$

$\frac{1}{-20}=\frac{1}{v_{fromconcave}}-\frac{1}{22}$

$\frac{1}{v}=-\frac{1}{220}$

$v = -220cm$

Hence parallel beam of rays will diverge from this point which is (220 - 4 = 216) cm away from the centre of the two lenses.

2) When rays fall on the concave lens first

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

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

$\frac{1}{v}=\frac{1}{-20}+\frac{1}{infinite}$

$v=-20cm$

Now this will act as an object for convex lens.

$\frac{1}{f_{convex}}=\frac{1}{v_{fromconvex}}-\frac{1}{u_{fromconvex}}$

$u_{fromconvex}= -20-8=-28cm$

$\frac{1}{-30}=\frac{1}{v_{fromconvex}}-\frac{1}{-28}$

$\frac{1}{v_{fromconvex}}=\frac{1}{30}-\frac{1}{28}=-\frac{1}{420}$

$v_{fromconvex}= -420cm$

Hence parallel beam will diverge from this point which is (420 - 4 = 146 cm ) away from the centre of two lenses.

As we have seen for both cases we have different answers so Yes, answer depend on the side of incidence when we talk about combining lenses. i .e. we can not use the effective focal length concept here.

Given

Object height = 1.5 cm

Object distance from convex lens = -40cm

According to lens formula

$\frac{1}{v_{fromconvex}}=\frac{1}{f_{convex}}+\frac{1}{u_{fromconvex}}$

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

$v_{fromconvex}=120$

Magnificatio due to convex lens:

$m_{convex}=-\frac{v}{u}=-\frac{120}{-40}=3$

The image of convex lens will act as an object for concave lens,

so,

$\frac{1}{v_{fromconcave}}=\frac{1}{f_{concave}}+\frac{1}{u_{fromconcave}}$

$u_{concave}= 120 - 8 = 112$

$\frac{1}{v_{fromconcave}}= \frac{1}{-20}+\frac{1}{112}$

$\frac{1}{v_{fromconcave}}= -\frac{92}{2240}$

$v_{fromconcave}= \frac{-2240}{92}$

Magnification due to concave lens :

$m_{concave}= \frac{2240}{92}*\frac{1}{112}=\frac{20}{92}$

The combined magnification:

$m_{combined}=m_{convex}*m_{concave}$

$m_{combined}=3*\frac{20}{92}=0.652$

Hence height of the image = $m_{combined}*h$

= 0.652 * 1.5 = 0.98cm

Hence height of image is 0.98cm.

Let prism be ABC ,

as emergent angle $e=90^0$,

$\mu_{glass}sinr_2 = 1sin90$

$sinr_2=\frac{1}{1.524} = 0.6562$

$r_2= 41 ^0$ (approx)

Now as we know in the prism

$r_1+r_2=A$

Hence, $r_1+=A-r_2=60-41=19^0$

Now applying snells law at surface AB

$1sini=\mu _{glass}sinr_1$

$sini=1.524sin19$

$sini=0.496$

$i = 29.75^0$

Hence the angle of incident is 29.75 degree.

(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?

(b) What is the angular magnification (magnifying power) of the lens?

(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.

Given,

Object distance u = -9cm

Focal length of convex lens = 10cm

According to the lens formula

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

$\frac{1}{10}=\frac{1}{v}-\frac{1}{-9}$

$\\\frac{1}{v}=\frac{1}{10}-\frac{1}{9}\\\Rightarrow v=-90\ cm$

a) Magnification

$m=\frac{v}{u}=\frac{-90}{9}=10 \ cm$

The area of each square in the virtual image

$=10\times10\times1=100mm^2=1cm^2$

b) Magnifying power

$=\frac{d}{|u|}=\frac{25}{9}=2.8$

c) No,

$magnification = \frac{v}{u}$

$magnifying\ power = \frac{d}{|u|}$.

Both the quantities will be equal only when image is located at the near point |v| = 25 cm

(b) What is the magnification in this case?

(c) Is the magnification equal to the magnifying power in this case?

a)

maximum magnifying is possible when our image distance will be equal to minimum vision point that is,

$v=-25$

$f = 10cm$     (Given)

Now according to the lens formula

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

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

$\frac{1}{u}=\frac{1}{-25}-\frac{1}{10}$

$\frac{1}{u}=-\frac{7}{50}$

$u=-\frac{50}{7}=-7.14cm$

Hence required object distance for viewing squares distinctly is 7.14 cm away from the lens.

b)

Magnification of the lens:

$m = \left | \frac{v}{u} \right |=\frac{25}{50}*7= 3.5$

c)

Magnifying power

$M = \frac{d}{u} =\frac{25}{50}*7= 3.5$

Since the image is forming at near point ( d = 25 cm ), both magnifying power and magnification are same.

Given

Virtual image area = 6.25 mm2

Actual ara = 1 mm2

We can calculate linear magnification as

$m=\sqrt{\frac{6.25}{1}}=2.5$

we also know

$m=\frac{v}{u}$

$v = mu$

Now, according to the lens formula

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

$\frac{1}{10}=\frac{1}{mu}-\frac{1}{u}$

$\frac{1}{u}(\frac{1}{2.5}-1)=\frac{1}{10}$

$u=-6cm$ and

$v=mu=2.5*(-6)=-15cm$

Since the image is forming at a distance which is less than 25 cm, it can not be seen by eye distinctively.

Angular magnification is the ratio of tangents of the angle formed by object and image from the centre point of the lens. In this question angle formed by the object and a virtual image is same but it provides magnification in a way that, whenever we have object place before 25cm, the lens magnifies it and make it in the vision range. By using magnification we can put the object closer to the eye and still can see it which we couldn't have without magnification.

Yes, angular magnification will change if we move our eye away from the lens. this is because then angle subtended by lens would be different than the angle subtended by eye. When we move our eye form lens, angular magnification decreases. Also, one more important point here is that object distance does not have any effect on angular magnification.

Firstly, grinding a lens with very small focal length is not easy and secondly and more importantly, when we reduce the focal length of a lens, spherical and chronic aberration becomes more noticeable. they both are defects of the image, resulting from the ways of rays of light.

We need more magnifying power and angular magnifying power in a microscope in order to use it effectively. Keeping both objective focal length and eyepiece focal length small makes the magnifying power greater and more effective.

When we view through a compound microscope, our eyes should be positioned a short distance away from the eyepiece lens for seeing a clearer image. The image of the objective lens in the eyepiece lens is the position for best viewing. It is also called "eye-ring" and all reflected rays from lens pass through it which makes it the ideal position for the eye for the best view.

When we put our eyes too close to the eyepiece lens, then we catch the lesser refracted rays from eyes, i.e. we reduce our field of view because of which the clarity of the image gets affected.

Given,

magnifying power = 30

objective lens focal length

$f_{objeective}$ = 1.25cm

eyepiece lens focal length

$f_{eyepiece}$ = 5 cm

Normally, image is formed at distance d = 25cm

Now, by the formula;

Angular magnification by eyepiece:

$m_{eyepiece}=1+\frac{d}{f_{eyepiece}}=1+\frac{25}{5}=6$

From here, magnification by the objective lens :

$m_{objective}=\frac{30}{6}=5$     since   ( $m_{objective}*m_{eyepiece}=m_{total}$)

$m_{objective}=-\frac{v}{u}=5$

$v=-5u$

According to the lens formula:

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

$\frac{1}{1.25}=\frac{1}{-5u}-\frac{1}{u}$

from here,

$u = -1.5cm$

hence object must be 1.5 cm away from the objective lens.

$v= -mu=(-1.5)(5)=7.5$

Now for the eyepiece lens:

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

$\frac{1}{5}=\frac{1}{-25}-\frac{1}{u}$

$\frac{1}{u}=-\frac{6}{25}$

$u = -4.17 cm$

Hence the object is 4.17 cm away from the eyepiece lens.

The separation between objective and eyepiece lens

$u_{eyepiece} +v_{objectivve}=4.17 + 5.7 = 11.67 cm$

Given,

the focal length of the objective lens $f_{objective}=140cm$

the focal length of the eyepiece lens $f_{eyepiece}=5cm$

normally, least distance of vision = 25cm

Now,

As we know magnifying power:

$m = \frac{f_{objective}}{f_{eyepiece}}=\frac{140}{5}=28$

Hence magnifying power is 28.

(b) the final image is formed at the least distance of distinct vision (25cm)?

Given,

the focal length of the objective lens $f_{objective}=140cm$

the focal length of the eyepiece lens $f_{eyepiece}=5cm$

normally, least distance of vision = 25cm

Now,

as we know magnifying power when the image is at d = 25 cm is

$m=\frac{f_{objective}}{f_{eyepiece}}(1+\frac{f_{eyepiece}}{d})=\frac{140}{5}(1+\frac{5}{25}) = 33.6$

Hence magnification, in this case, is 33.6.

a) Given,

focal length of the objective lens = $f_{objective}$= 140cm

focal length of the eyepiece lens = $f_{eyepiece}$ = 5 cm

The separation between the objective lens and eyepiece lens is given by:

$f_{eyepiece}+f_{objective}=140+5=145cm$

Hence, under normal adjustment separation between two lenses of the telescope is 145 cm.

Given,

focal length of the objectove lens = $f_{objective}$= 140cm

focal length of the eyepiece lens = $f_{eyepiece}$ = 5 cm

Height of tower $h_{tower}$ = 100m

Distance of object which is acting like a object $u$ = 3km = 3000m.

The angle subtended by the tower at the telescope

$tan\theta=\frac{h_{tower}}{u}=\frac{100}{3000}=\frac{1}{30}$

Now, let the height of the image of the tower by the objective lens is  $h_{image}$.

angle made by the image by the objective lens  :

$tan\theta'=\frac{h_{image}}{f_{objective}}=\frac{h_{image}}{140}$

Since both, the angles are the same we have,

$tan\theta=tan\theta'$

$\frac{1}{30}=\frac{h_{image}}{140}$

$h_{image}= \frac{140}{30}=4.7cm$

Hence the height of the image of the tower formed by the objective lens is 4.7 cm.

## Q 9.29 (c) What is the height of the final image of the tower if it is formed at 25cm

Given, image is formed at a distance $d$ = 25cm

As we know, magnification of eyepiece lens is given by :

$m=1+\frac{d}{f_{eyepiece}}$

$m=1+\frac{25}{5}=6$

Now,

Height of the final image is given by :

$h_{image}= mh_{object}= 6*4.7 = 28.2cm$

Therefore, the height of the final image will be 28.2 cm

Given,

Distance between the objective mirror and secondary mirror $d= 20mm$

The radius of curvature of the Objective Mirror

$R_{objective}=220mm$

So the focal length of the objective mirror

$f_{objective}=\frac{220}{2}=110mm$

The radius of curvature of the secondary mirror

$R_{secondary}=140mm$

so, the focal length of the secondary mirror

$f_{secondary}=\frac{140}{2}=70mm$

The image of an object which is placed at infinity, in the objective mirror, will behave like a virtual object for the secondary mirror.

So, virtual object distance for the secondary mirror

$u_{secondary}=f_{objective}-d=110-20=90mm$

Now, applying  the mirror formula in the secondary mirror:

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

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

$\frac{1}{v}=\frac{1}{70}-\frac{1}{90}$

$v=315mm$

Given

Angle of deflection $\delta = 3.5^0$

The distance of the screen from the mirror $D=1.5m$

The reflected rays will bet deflected by twice angle of deviation that is

$2\delta =3.5*2=7^0$

Now from the figure, it can be seen that

$tan2\delta =\frac{d}{1.5}$

$d=1.5*\tan2\delta = 2*\tan7=0.184m=18.4cm$

Hence displacement of the reflected spot of the light is $18.4cm$.

Given

The focal length of the convex lens $f_{convex}=30cm$

here liquid is acting like the mirror so,

the focal length of the liquid $=f_{liquid}$

the focal length of the system(convex + liquid) $f_{system}=45cm$

Equivalent focal length when two optical systems are in contact

$\frac{1}{f_{system}}=\frac{1}{f_{convex}}+\frac{1}{f_{liquid}}$

$\frac{1}{f_{liquid}}=\frac{1}{f_{system}}-\frac{1}{f_{convex}}$

$\frac{1}{f_{liquid}}=\frac{1}{45}-\frac{1}{30}=-\frac{1}{90}$

$f_{liquid}=-90cm$

Now, let us assume refractive index of the lens be $\mu _{lens}$

The radius of curvature are $R$ and $-R$.

As we know,

$\frac{1}{f_{convex}}=(\mu _{lens}-1)\left ( \frac{1}{R}-\frac{1}{-R} \right )$

$\frac{1}{f_{convex}}=(\mu _{lens}-1)\frac{2}{R}$

$R=2(\mu _{lens-1})f_{convex}=2(1.5-1)30=30cm$

Now, let refractive index of liquid be $\mu _{liquid}$

The radius of curvature of liquid in plane mirror side = infinite

Radius of curvature of liquid in lens side R  = -30cm

As we know,

$\frac{1}{f_{liquid}}=(\mu_{liquid-1})\left ( \frac{1}{R}-\frac{1}{infinite} \right )$

$-\frac{1}{90}=(\mu_{liquid}-1)(\frac{1}{30})$

$\mu_{liquid}= 1+\frac{1}{3}$

$\mu_{liquid}= 1.33$

Therefore the refractive index of the liquid is 1.33.

## NCERT solutions for class 12 physics- chapter wise

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions Solutions of NCERT class 12 physics chapter 11 Dual nature of radiation and matter CBSE NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei Solutions of NCERT class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

## Importance of NCERT solutions for class 12 physics chapter 9 ray optics and optical instruments:

The solutions of  NCERT class 12 physics chapter 9 ray optics and the optical instrument are important in the CBSE board exams since on an average 12% of questions are asked in the board exam from the unit optics, which includes chapter 9 and 10 of NCERT class 12 physics. For exams like NEET and JEE Mains, from the unit optics, 2 to 3 questions are expected. Learning the NCERT solutions for class 12 physics chapter 9 ray optics and optical instruments can be helpful to solve questions from other reference books also.