NCERT · Class 12 · Physics

Chapter 9: Ray Optics and Optical Instruments

Practice each question, then open the answer or worked solution.

Questions

38 questions

Q1Written answer

A small candle, 2.5 cm in size, is placed at 27 cm in front of a concave mirror of a 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?

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Worked solution

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,

\(\begin{array}{l}f=\frac{R}{2}=-18 cm\end{array} \)

Image distance= v

\(\begin{array}{l}\frac{1}{u}\end{array} \)

\(\begin{array}{l}\frac{1}{v}\end{array} \)

\(\begin{array}{l}\frac{1}{f}\end{array} \)

\(\begin{array}{l}\frac{1}{v}\end{array} \)

\(\begin{array}{l}\frac{1}{f}\end{array} \)

\(\begin{array}{l}\frac{1}{u}\end{array} \)

\(\begin{array}{l}\frac{1}{-18}\end{array} \)

\(\begin{array}{l}\frac{1}{-27}\end{array} \)

\(\begin{array}{l}\frac{-3+2}{54}\end{array} \)

\(\begin{array}{l}-\frac{1}{54}\end{array} \)

\(\begin{array}{l}∴\end{array} \)

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

Image magnification is

\(\begin{array}{l}m=\frac{h’}{h}=-\frac{v}{u}\end{array} \)

\(\begin{array}{l}∴ h’=\frac{-v}{u}\times h\end{array} \)

\(\begin{array}{l}=-(\frac{-54}{-27})\times2.5\end{array} \)

5 cm 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.

Q2Written answer

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.

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Worked solution

Size of the needle, h₁=4.5 cm

Object distance, u=-12 cm

The focal length of the convex mirror, f= 15 cm

Image distance= v

\(\begin{array}{l}\frac{1}{u}\end{array} \)

\(\begin{array}{l}\frac{1}{v}\end{array} \)

\(\begin{array}{l}\frac{1}{f}\end{array} \)

\(\begin{array}{l}\frac{1}{v}\end{array} \)

\(\begin{array}{l}\frac{1}{f}\end{array} \)

\(\begin{array}{l}\frac{1}{u}\end{array} \)

\(\begin{array}{l}\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}\end{array} \)

∴

\(\begin{array}{l}\frac{60}{9}\end{array} \)

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

Using the magnification formula,

\(\begin{array}{l}m=\frac{h_{2}}{h_{1}}=-\frac{v}{u}\end{array} \)

\(\begin{array}{l} h_{2}=\frac{-v}{u}\times h_{1}\end{array} \)

\(\begin{array}{l}=\frac{-6.7}{-12}\times 4.5\end{array} \)

\(\begin{array}{l}m=\frac{h_{2}}{h_{1}}=\frac{2.5}{4.5}\end{array} \)

2.5 cm 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.

Q3Written answer

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?

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Worked solution

The actual depth of the needle in water, h₁=12.5 cm

The apparent depth of the needle in water, h₂ =9.4 cm

The refractive index of water =

\(\begin{array}{l}\mu\end{array} \)

The value of

\(\begin{array}{l}\mu\end{array} \)

\(\begin{array}{l}\mu=\frac{h_{1}}{h_{2}}\end{array} \)

\(\begin{array}{l}=\frac{12.5}{9.4}\end{array} \)

Hence, 1.33 is the refractive index of water.

Now, water is replaced by a liquid with a refractive index of 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.

\(\begin{array}{l}\mu ‘=\frac{h_{1}}{x}\end{array} \)

Therefore y=

\(\begin{array}{l}\frac{h_{1}}{\mu ‘}\end{array} \)

\(\begin{array}{l}=\frac{12.5}{1.63}\end{array} \)

The new apparent depth of the needle is 7.67cm. It is observed that the value is less than

\(\begin{array}{l}h_{2}\end{array} \)

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

Q4Written answer

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 a glass when the angle of incidence in water is 45° with the normal to a water-glass interface.

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Worked solution

For the glass-air interface:

The angle of incidence, i =60°

The 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

\(\begin{array}{l}\mu_{g}^{a}=\frac{sin\;i}{sin\;r}\\ =\frac{sin\;60°}{sin\;35°}=\frac{0.8660}{0.5736}=1.51\end{array} \)

For the air-water interface:

The angle of incidence, i=

\(\begin{array}{l}60°\end{array} \)

The angle of refraction, r=

\(\begin{array}{l}47°\end{array} \)

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

\(\begin{array}{l}\mu_{w}^{a}=\frac{sin\;i}{sin\;r}\\ =\frac{sin\;60°}{sin\;47°}=\frac{0.8660}{0.7314}=1.184\end{array} \)

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

\(\begin{array}{l}\mu_{g}^{w}=\frac{\mu_{g}^{a}}{\mu_{w}^{a}}\\ =\frac{1.51}{1.184}=1.275\end{array} \)

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

The angle of incidence, i=

\(\begin{array}{l}45°\end{array} \)

The angle of refraction= r

r can be calculated from Snells’s law as

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

Therefore,

\(\begin{array}{l}38.68°\end{array} \)

Q5Written answer

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

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Worked solution

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

The refractive index of water,

\(\begin{array}{l}\mu\end{array} \)

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.

\(\begin{array}{l}\frac{AC}{2}\end{array} \)

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

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

The relation:

tan i=

\(\begin{array}{l}\frac{OC}{OB}=\frac{R}{d_{1}}\end{array} \)

Therefore, R=

\(\begin{array}{l}tan\;48.75° \times 0.8=0.91m\end{array} \)

Area of the surface of water=

\(\begin{array}{l}nR^{2}=n(0.91)^{2}=2.61m^{2}\end{array} \)

\(\begin{array}{l}2.61m^{2}\end{array} \)

Q6Written answer

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.

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Worked solution

The angle of minimum deviation,

\(\begin{array}{l}\delta_{m}=40°\end{array} \)

The angle of the prism, A =

\(\begin{array}{l}60°\end{array} \)

The refractive index of water,

\(\begin{array}{l}\mu=1.33\end{array} \)

The refractive index of the material=

\(\begin{array}{l}\mu ‘\end{array} \)

The angle of deviation is related to refractive index (

\(\begin{array}{l}\mu ‘\end{array} \)

\(\begin{array}{l}\frac{sin\frac{A+\delta_{m}}{2}}{sin\frac{A}{2}}\end{array} \)

\(\begin{array}{l}\frac{sin\frac{60°+40°}{2}}{sin\frac{60°}{2}}=\frac{si n 50°}{sin 30°}\end{array} \)

1.532 is the refractive index of the material of the prism.

\(\begin{array}{l}\delta_{m}\end{array} \)

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

\(\begin{array}{l}\delta_{g}^{w}=\frac{\mu ‘}{\mu}=\frac{sin\frac{(A+\delta_{m}’)}{2}}{sin\frac{A}{2}} \\ sin\frac{(A+\delta_{m}’)}{2}=\frac{\mu ‘}{\mu}sin\frac{A}{2}\\ sin\frac{(A+\delta_{m}’)}{2}=\frac{1.532}{1.33}\times sin\frac{60°}{2}=0.5759\\ \frac{(A+\delta_{m}’)}{2}=sin^{-1}0.5759=35.16°\\ 60°+\delta_{m}’=70.32°\\ Therefore\;\delta_{m}’=70.32°-60°=10.32°\end{array} \)

\(\begin{array}{l}10.32°\end{array} \)

Q7Written answer

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?

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Worked solution

The refractive index of glass,

\(\begin{array}{l}\mu=1.55\end{array} \)

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

The radius of curvature of one face of the lens=R₁

The radius of curvature of the other face of the lens=R₂

The radius of curvature of the double-convex lens=R

Therefore,

\(\begin{array}{l}R_{1}= R\;and \;R_{2}= – R\end{array} \)

The value of R is calculated as

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

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

Q8Written answer

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?

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Worked solution

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

\(\begin{array}{l}\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \frac{1}{v}-\frac{1}{12}=\frac{1}{20}\\ \frac{1}{v}=\frac{1}{20}+\frac{1}{12}=\frac{3+5}{60}=\frac{8}{60}\\ Therefore\;v=\frac{60}{8}=7.5cm\end{array} \)

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

\(\begin{array}{l}\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \frac{1}{v}=-\frac{1}{16}+\frac{1}{12}\\ =\frac{-3+4}{48}=\frac{1}{48}\\\end{array} \)

∴ v= 48cm.

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

Q9Written answer

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?

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Worked solution

Size of the object,

\(\begin{array}{l}h_{1} = 3 cm\end{array} \)

Object’s distance, u = -14 cm

The focal length of the concave lens, f = -21 cm

Image distance = v

\(\begin{array}{l}\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \frac{1}{v}=-\frac{1}{21}-\frac{1}{14}\\ =\frac{-2-3}{42}=\frac{-5}{42}\\ Therefore\;v=-\frac{42}{5}=-8.4cm\end{array} \)

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

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

The height of the image is 1.8 cm.

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.

Q10Written answer

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

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Worked solution

The focal length of the convex lens, f₁=30 cm

The focal length of the concave lens, f₂= -20 cm

The focal length of the system of lenses= f

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

\(\begin{array}{l}\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\end{array} \)

\(\begin{array}{l}\frac{1}{f}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}\end{array} \)

Therefore, f=-60 cm

Therefore, 60 cm is the focal length of the combination of lenses. Also, the negative sign is an indication that the system of lenses used acts as a diverging lens.

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Chapter 9: Ray Optics and Optical Instruments

Questions