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

NCERT Solutions for Class 12 Physics Chapter 9 – Free PDF Download

The NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments given here, is one of the best study materials that you will come across to score good marks in the CBSE Class 12 term – II examination. Chapter 9 of NCERT Solutions for Class 12 Physics PDF provided here includes 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 an important chapter of the Unit-Optics and is categorized under the latest term – II CBSE Syllabus 2021-22. Opting for the right reference material for this chapter is made easy with the NCERT Solutions for Class 12 provided here. It helps students understand key concepts of the chapter in an easy and interesting way.

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Class 12 Physics NCERT Solutions Ray Optics Optical Instruments Important Questions


Question 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?

Answer

Height of the candle, h =2.5 cm

Image size=h’

Object distance, u= -27 cm

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

Focal length of the concave mirror, f=R2=18cmf=\frac{R}{2}=-18cm

Image distance=v

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

 

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

=118\frac{1}{-18}- 127\frac{1}{-27}=3+254\frac{-3+2}{54}=154-\frac{1}{54}

 

v=-54cm

Therefore, to obtain a sharp image, the distance between the screen and the mirror should be 54cm.

Image magnification is:

m=hh=vum=\frac{h’}{h}=-\frac{v}{u}

 

h=vu×h∴ h’=\frac{-v}{u}\times h

 

=(5427)×2.5=-(\frac{-54}{-27})\times2.5=-5cm

5cm is the height of the image of the candle. As there is a negative sign, the image is inverted and virtual.

If the candle is moved closer to the mirror, then the screen needs to be moved away from the mirror so as to obtain the image.

Question 2:

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Answer :

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

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

 

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

 

115+112=4+560=960\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}

 

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

The distance between the needle image and the mirror is 6.7cm and the image is obtained on the other side of the mirror.

Using magnification formula:

m=h2h1=vum=\frac{h_{2}}{h_{1}}=-\frac{v}{u}

 

h2=vu×h1 h_{2}=\frac{-v}{u}\times h_{1}

 

=6.712×4.5=\frac{-6.7}{-12}\times 4.5=+2.5cm

 

m=h2h1=2.54.5m=\frac{h_{2}}{h_{1}}=\frac{2.5}{4.5}=0.56

2.5cm is the height of the image. As it has a positive sign, the image is erect, virtual and diminished.

As the needle is moved farther from the mirror, the image also moves away from the mirror resulting in a reduction in the size of the image.

Question 3:

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If
water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Answer:

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 =μ\mu

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

μ=h1h2\mu=\frac{h_{1}}{h_{2}}

 

=12.59.4=\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.

μ=h1x\mu ‘=\frac{h_{1}}{x}

Therefore y=h1μ\frac{h_{1}}{\mu ‘}

 

=12.51.63=\frac{12.5}{1.63}=7.67cm

The new apparent depth of the needle is 7.67cm. It is observed that the value is less than h2h_{2}, therefore the needle needs to be moved up to the focus again.

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

Question 4:

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Question 4

The figures above show the refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence
in water is 45° with the normal to a water-glass interface.

For the glass-air interface:

Angle of incidence, i =60°

Angle of refraction, r=35°

From Snell’s law we know that the refractive index of the glass with respect to air is given as:

μga=sin  isin  r=sin  60°sin  35°=0.86600.5736=1.51\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°60°

Angle of refraction, r=47°47°

From Snell’s law we know that the refractive index of the water with respect to air is given as:

μwa=sin  isin  r=sin  60°sin  47°=0.86600.7314=1.184\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:

μgw=μgaμwa=1.511.184=1.275\mu_{g}^{w}=\frac{\mu_{g}^{a}}{\mu_{w}^{a}}\\ =\frac{1.51}{1.184}=1.275

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Question 4 Solution

Angle of incidence, i= 45°45°

Angle of refraction= r

r can be calculated from Snells’s law as

sin  isin  r=μgwsin45°sin  r=1.275sin  r=121.275=0.5546Therefore  r=sin1(0.5546)=38.68°\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, 38.68°38.68° is the angle of refraction at the water-glass interface.

Question 5:

A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Answer :

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

Refractive Index of water, μ\mu= 1.33

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Question 5 Solution

I = Angle of incidence

r =Angle of refraction – 90°

As the bulb is used as a point source, the emergent light is considered to be a circle.

R=AC2\frac{AC}{2}=AO=OB

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

μ=sin  rsin  i1.33=sin  90°sin  iTherefore  i=sin1(11.33)=48.75°\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=OCOB=Rd1\frac{OC}{OB}=\frac{R}{d_{1}}

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

Area of the surface of water=nR2=n(0.91)2=2.61m2nR^{2}=n(0.91)^{2}=2.61m^{2}

 

2.61m22.61m^{2} is found to be the area of water through which the light from the water can emerge out.

Question 6:

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on the face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the
material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Answer:

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

Angle of the prism, A = 60°60°

Refractive Index of water, μ=1.33\mu=1.33

Refractive Index of the material= μ\mu ‘

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

μ\mu ‘=sinA+δm2sinA2\frac{sin\frac{A+\delta_{m}}{2}}{sin\frac{A}{2}}

=sin60°+40°2sin60°2=sin50°sin30°\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

δm\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:

δgw=μμ=sin(A+δm)2sinA2sin(A+δm)2=μμsinA2sin(A+δm)2=1.5321.33×sin60°2=0.5759(A+δm)2=sin10.5759=35.16°60°+δm=70.32°Therefore  δm=70.32°60°=10.32°\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°10.32° is the new minimum angle of deviation.

Question 7:

Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?

Answer:

Refractive Index of glass, μ=1.55\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, R1=R  and  R2=RR_{1}= R\;and \;R_{2}= – R

The value of R is calculated as:

1f=(μ1)[1R11R2]120=(1.551)[1R+1R]120=0.55×2RTherefore  R=0.55×2×20=22cm\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

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

 Question 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?

Answer :

The object is virtual and the image formed is real.

Object distance, u= +12 cm

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

Image distance=v

1v1u=1f1v112=1201v=120+112=3+560=860Therefore  v=608=7.5cm\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

1v1u=1f1v=116+112=3+448=148\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}\\

∴ v=48cm.

Therefore, the image obtained is 48cm away from the lens towards the right.

Question 9:

An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens? 

Answer:

Size of the object, h1=3cmh_{1} = 3 cm

Object distance, u = -14 cm

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

Image distance = v

1v1u=1f1v=121114=2342=542Therefore  v=425=8.4cm\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \frac{1}{v}=-\frac{1}{21}-\frac{1}{14}\\ =\frac{-2-3}{42}=\frac{-5}{42}\\ Therefore\;v=-\frac{42}{5}=-8.4cm

Therefore, the image obtained is 8.4cm away and is obtained on the other side of the lens. Since there is negative sign, it is understood that the image is virtual and erect.

m=Image  height(h2)Object  height(h1)=vuTherefore  h2=8.414×3=0.6×3=1.8cmm=\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

The height of the image is 1.8cm.

If the object is moved further away from the lens, then the virtual image will move towards the mirror. As the object distance increases, the image size decreases.

Question 10:

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

Answer:

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:

1f=1f1+1f2\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}

 

1f=130120=2360=160\frac{1}{f}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}

Therefore f=-60cm

Therefore, 60cm is the focal length of the combination of lenses. Also, the negative sign is an indication that the system of lenses used act 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?

Answer:

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, v2v_{2}= -25cm

Object distance for the eyepiece= u2

According to the lens formula, we have the relation:

1v21u2=1f2\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}

 

1u2=1v21f2\frac{1}{u_{2}}=\frac{1}{v_{2}}-\frac{1}{f_{2}}

 

12516.25=1425=525\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}

 

u2=5cmu_{2}=-5cm

Image distance for the objective lens, v1=d+u2=155=10cmv_{1}=d+u_{2}=15 – 5=10 cm

Object distance for the objective lens = u1u_{1}

 

1v11u1=1f1\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}

 

1u1=1v11f1\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}

 

11012=1510=410\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}

 

u1=2.5cmu_{1}=-2.5cm

Magnifying power

m=v0u0(1+dfe)=102.5(1+256.25)=20\\m=\frac{v_{0}}{|u_{0}|}(1+\frac{d}{f_{e}}) \\ \\=\frac{10}{2.5}(1+\frac{25}{6.25}) \\ \\=20

Hence, 20 is the magnifying power of the microscope.

(b) The final Image is formed at infinity.

Therefore image distance for the eyepiece, v2=v_{2}=\infty

Object distance for the eyepiece=u2 u_{2}

According to the lens formula, we have the relation:

1v21u2=1f2\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}

 

11u2=16.25\frac{1}{\infty}-\frac{1}{u_{2}}=\frac{1}{6.25}

 

u2=6.25cmu_{2}=-6.25cm

Image distance for the objective lens, v1=d+u2=156.25=8.75cmv_{1} = d + u_{2}=15 -6.25 = 8.75 cm

Object distance for the objective lens=u1

Following relation is obtained from the lens formula:

1v11u1=1f1\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}

 

1u1=1v11f1\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}

 

18.7512.0=28.7517.5\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}

 

u1=17.56.75=2.59cmu_{1}=-\frac{17.5}{6.75}=-2.59cm

Magnitude of the object distance, u1|u_{1}|=2.59 cm

Following is the relation explaining the magnifying power of a compound microscope:

m=v1u1(du2)\frac{v_{1}}{|u_{1}|}(\frac{d’}{|u_{2}|})

=8.752.59×256.25\frac{8.75}{2.59}\times\frac{25}{6.25}=13.51

Hence, 13.51 is the magnifying power of the microscope.

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,

Answer:

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

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

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

Least distance of distant vision, d=25 cm

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

Object distance for the eyepiece =ueu_{e}.

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

1ve1ue=1fe\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}

 

1ue=1ve1fe\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}

 

12512.5=11025\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}

 

ue=2511=2.27cmu_{e}=-\frac{25}{11}=-2.27cm

With the help of lens formula, the value of the image distance for the objective (v) lens is obtained:

1vo1uo=1fo\frac{1}{v_{o}}-\frac{1}{u_{o}}=\frac{1}{f_{o}}

 

1vo=1fo+1uo\frac{1}{v_{o}}=\frac{1}{f_{o}}+\frac{1}{u_{o}}

 

=10.810.9=\frac{1}{0.8}-\frac{1}{0.9}

 

=17.2=\frac{1}{7.2}

vo = 7.2 cm

The separation between two lenses are determined as follows:

= v0 + |ue|

= 7.2 + 2.27

= 9.47 cm

The magnifying power of the microscope is calculated as follows:

Magnifying power, M = Mo × Me

M=vouo(1+Dfe)M=7.20.9(1+2525)\\M = \frac{v_{o}}{u_{o}}(1+\frac{D}{f_{e}}) \\ \\M = \frac{7.2}{0.9}(1+\frac{25}{25})

= 8 × 11

= 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?
Answer:

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

=1446\frac{144}{6}=24

The separation between the objective lens and the eyepiece is calculated as: fo+fef_{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 3.48×106m3.48 \times 10^{6} m,and the radius of lunar orbit is 3.8×108m3.8 \times 10^{8}m.

Answer:
Focal length of the objective lens, fof_{o}=15m =15×102cm15 \times 10^{2} cm
Focal length of the eyepiece, fef_{e}= 1.0 cm
(i) α=fofe\alpha =\frac{f_{o}}{f_{e}} is the angular magnification of a telescope.

=15×1021\frac{15\times10^{2}}{1}=1500
Hence, the angular magnification of the given refracting telescope is 1500.
(ii) Diameter of the moon, d = 3.48×106m3.48 \times 10^{6} m
Radius of the lunar orbit, ror_{o}= 3.8×108m3.8 \times 10^{8} m
Consider d’ to 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.

dro=dfo\frac{d}{r_{o}}=\frac{d’}{f_{o}}

 

3.48×1063.8×108=d15\frac{3.48\times10^{6}}{3.8\times10^{8}}=\frac{d’}{15}

Therefore d’=3.483.8×102×15\frac{3.48}{3.8}\times10^{-2}\times15

 

13.74×102m=13.74cm13.74\times10^{-2}m=13.74cm

Therefore 13.74cm is the diameter of the image of the moon 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.]
Answer:

(i) The focal length is negative for a concave mirror.
Therefore f < 0
The object distance (u) is negative when the object is placed on the left side of the mirror.
Therefore u < 0
Lens formula for image distance v is written as:

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

 

1v=1f1u\frac{1}{v}=\frac{1}{f}-\frac{1}{u}——————(i)

For a concave mirror, f is negative, i.e., f < 0

For a real object, i.e., which is on the left side of the mirror.

For u between f and 2f implies that 1/u lies between 1/f and 1/2f

12f>1u>1f(asu,farenegative)1f12f<1f1u<012f<1v<0\\\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}(as \, u, \, f \, are \, negative) \\ \\\frac{1}{f}-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}<0 \\ \\\frac{1}{2f}<\frac{1}{v}<0

1/v is negative

which implies v is negative and greater than 2f. Hence, image lies beyond 2f and it is real.

(ii) The focal length is positive for a convex mirror.
Therefore f > 0
The object distance (u) will be negative if the object is placed on the left side of the mirror.
Therefore u < 0
Using mirror formula, we can calculate the image distance v:

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

 

1v=1f1u\frac{1}{v}=\frac{1}{f}-\frac{1}{u}

Using eqn (ii), we can conclude that:

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

v>0

Thus, the image is obtained on the back side of the mirror, Therefore, it can be concluded that a convex mirror always produces a virtual image irrespective of the object distance.

(iii) The focal length is positive for a convex mirror.
Therefore f > 0
The object distance (u) is negative when the object is placed on the left side of the mirror
Therefore u < 0
For image distance v, we have the mirror formula:

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

 

1v=1f1u\frac{1}{v}=\frac{1}{f}-\frac{1}{u}

But we have u<0

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

v<f

Therefore, the image is formed between the focus and the pole and is diminished.
(iv) The focal length of the concave mirror is negative.

Therefore f < 0

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

Therefore f>u>0

1f<1u<0\frac{1}{f}<\frac{1}{u}<0

 

1f1u<0\frac{1}{f}-\frac{1}{u}<0
For image distance v, we have the mirror formula:

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

 

1v=1f1u\frac{1}{v}=\frac{1}{f}-\frac{1}{u}

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

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

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

v>u
Magnification, m=vu\frac{v}{u}> 1
Hence, the formed image is enlarged.

Question 16:
A small pin fixed on a tabletop 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?
Answer:

The actual depth of the pin, d = 15 cm
Apparent depth of the pin =d’
Refractive index of glass,μ\mu=1.5
The ratio of actual depth to the apparent depth and the refractive index of the glass are equal.
i.e.

μ=dd\mu=\frac{d}{d’}

Therefore d’=dμ\frac{d}{\mu}

= 151.5\frac{15}{1.5}=10cm
The distance at which the pin appears to be raised =d’-d=15-10=5cm
When the angle of incidence is small, the distance is independent of 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?

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Question 17

Answer:
(i) Refractive index of the glass – fibre, μ2\mu_{2}= 1.68

Refractive index of the outer covering, μ1\mu_{1}= 1.44
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i’
The refractive index (μ\mu) of the inner core − outer core interface is given as:

μ=μ2μ1=1sini\mu=\frac{\mu_{2}}{\mu_{1}}=\frac{1}{sin i’}

 

sini=μ1μ2sin i’=\frac{\mu_{1}}{\mu_{2}}

 

=1.441.68=0.8571=\frac{1.44}{1.68}=0.8571

 

Therefore  i=59°Therefore\;i’=59°
For the critical angle, total internal reflection (TIR) takes place only when i>i’, i.e., i> 59°59°
Maximum angle of reflection,           rmax=90°i=90°59°=31°r_{max}=90°-i’=90°-59°=31°
Let, imaxi_{max} be the maximum angle of incidence.
The refractive index at the air − glass interface, μ1=1.68\mu_{1}=1.68

We know that

μ1=sinimaxsinrmax\mu_{1}=\frac{sin i_{max}}{sin r_{max}}

Rearranging the equation, we get:

sin imax = 1.68 × sin rmax

Substituting the values in the above equation, we get:

= 1.68 sin 31°

= 1.68×0.51501.68\times 0.5150=0.8652

Therefore imax=sin10.8652=60°i_{max}=sin^{-1}0.8652=60°

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

(ii) If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe, μ1\mu_{1}=Refractive index of air=1
For the angle of incidence i =90°90°,

We can write Snell’s law at the air − pipe interface as:
sinisinr=μ2=1.68\frac{sin i}{sin r}=\mu_{2}=1.68

 

sinr=sin90°1.68=11.68sin r=\frac{sin 90°}{1.68}=\frac{1}{1.68}

 

r=sin1(0.5932)r=sin^{-1}(0.5932)

 

=36.5°=36.5°

Therefore i=90°36.5°=53.5°i’=90°-36.5°=53.5°

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

Question 18:
Answer the following questions:
(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 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?
(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?
Answer:

(i) Yes.

Real images can be obtained from plane and convex mirrors too. When the light rays converge at a point behind the plane or convex mirror, the object is said to be virtual. The real image of this object is obtained on the screen which is placed in the front of the mirror. This is where the real image is obtained.
(ii) No

To obtain a virtual image, the light rays must diverge. In the eyes, the convex lens help in converging these divergent rays at the retina. This is an example where the virtual image acts as an object for the lens to produce a real image.

(iii) The diver is in the water while the fisherman is on the land. Air is less dense when compared to the water as a medium. It is mentioned that the diver is viewing the fisherman. This explains that the light rays are travelling from the denser medium to the rarer medium. This means that the refracted rays move away from the normal making fisherman appear taller.

(iv) Yes; Decrease

The reason behind the change in depth of the tank when viewed obliquely is because the light rays bend when they travel from one medium to another. This also means that the apparent depth is less than the near-normal viewing.

(v) Yes

2.42 is the refractive index of the diamond while 1.5 is the refractive index of the ordinary glass. Also, the critical angle of the diamond is less than the glass. The sparkling effect of a diamond is possible because of the large cuts in the angle of incidence. Larger cuts ensures that the light entering the diamond is totally reflected from all the faces.

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?
Answer:

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

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

fmax=d4f_{max}=\frac{d}{4}

=34=0.75m\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.
Answer:

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=D2d24Df=\frac{D^{2}-d^{2}}{4D}

=902(202)4×90=77036=21.39cm\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.
Answer:

Focal length of the convex lens, f1f_{1}= 30 cm
Focal length of the concave lens, f2f_{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:
Using lens formula, we can write:

1v11u1=1f1\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}
Where,
u1u_{1}= Object distance
v1v_{1}= Image distance

1v1=1301=130\frac{1}{v_{1}}=\frac{1}{30}-\frac{1}{\infty}=\frac{1}{30}

Therefore v1=30cmv_{1}=30cm
For a concave lens, the image acts as a virtual object
Using lens formula for the concave lens, we can write:

1v21u2=1f2\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}

Where,
u2u_{2}=Object distance
= (30 − d) = 30 − 8 = 22 cm
v2v_{2}= Image distance

1v2=122120=1011220=1220\frac{1}{v_{2}}=\frac{1}{22}-\frac{1}{20}=\frac{10-11}{220}=\frac{-1}{220}

Therefore v2=220cmv_{2}=-220cm

 The parallel incident beam appears to diverge from a point that is  (220d2=2204)216cm(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:
Using the lens formula, we can write:

1v21u2=1f2\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}

 

1v2=1f2+1u2\frac{1}{v_{2}}=\frac{1}{f_{2}}+\frac{1}{u_{2}}
Where,
u2u_{2}=Object distance = −∞

 

v2v_{2}=Image distance

 

1v2=120+1=120\frac{1}{v_{2}}=\frac{1}{-20}+\frac{1}{-\infty}=-\frac{1}{20}

Therefore v2=20cmv_{2}=-20cm

For a convex lens, the image will act as a real object.
Applying lens formula to the convex lens, we have:

1v11u1=1f1\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}
Where,
u1