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NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 9: Ray Optics and Optical Instruments (Part I)

Get NCERT Solutions for Class 12th Physics - Chapter 9: Ray Optics and Optical Instruments. Due to huge number of problems, we have divided solutions of this chapter in several parts. In this part, you will find solutions from question number 9.1 to 9.9. The solutions of remaining questions will be available in further parts. You can also download these solutions in PDF format.

Jun 2, 2017 09:15 IST
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NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 9: Ray Optics and Optical Instruments (Part I)

NCERT Solutions for Class 12th Physics - Chapter 9: Ray Optics and Optical Instruments are available here. These solutions are also available for download in PDF format. This chapter is one of the longest chapters of Class 12 Physics NCERT textbook and a large number of problems given in exercises at the end of this chapter. So, we have provided these solutions in 3 parts. This is Part I, in this part; solutions from question number 9.1 to 9.9 are available. The solutions of remaining questions will be available in further parts.

NCERT Solutions for Class 12 Physics ‒ Chapter 9: Ray Optics and Optical Instruments from question number 9.1 to 9.9 are given below:

Question 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?

Solution 9.1:

Given:

u = ‒27 cm

f = ‒18 cm

h = 2.5

From the formula

(1/v) + (1/u) = (1/ f)

(1/v) – (1/27) = ‒ 1/18

Gives v = ‒54 cm

The image is real, inverted and magnified.

Now

v / u = h’ / h

Substitution yields

h’ = 5 cm

As, uf, v → ∞, for u < f, image is virtual.

NCERT Exemplar: CBSE Class 12 Physics – Chapter 9

Question 9.2: 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.

Solution 9.2:

Given:

h = 4.5 cm

u = ‒12 cm

f = 15 cm

Putting the values in the formula used in previous question, v = 6.7 cm

Magnification, m = ‒v / u = 5 / 9

Therefore, h’ = 4.5 × (5/9) = 2.5 cm

As, μ → ∞, vf, m → 0

Question 9.3: 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?

Solution 9.3:

Given:

u = 12.5 cm

v = 9.4 cm

We know

Refractive index of the medium = Real depth / Apparent depth = 12.5 / 9.4 = 1.33.

Now, h’ = real depth / refractive index of medium = 12.5 / 1.63 = 7.67 cm

Microscope is situated at 9.4 cm, so adjustment shift = 9.4 – 7.67 = 1.7 cm.

Question 9.4: Figures 9.34(a) and (b) show 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 45º with the normal to a water-glass interface [Fig. 9.34(c)].

Solution 9.4:

For air-glass interface

nga = sin i / sin r = sin 600 / sin 350 = 1.51

For air-water interface

nwa = sin 60o/ sin 47o = 1.18

For glass-water interface

ngw = nga × naw = 1.51 × (1/1.18) = 1.28

ngw = sin 450 / sin r

Which gives, r = 33.54o.

Question 9.5: 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.)

Solution 9.5:

Given:

h = 80 cm

n = 1.33

By the relation

sin ic = 1 / n

ic = 48.75

Where, ic = critical angle

For rays to come out,

i < 48.75.

In triangle NSD

tan ic = ND / 0.8

Or r = ND = 0.8 x tan ic = 0.912 m

Thus,

Area = πr2 = 2.6 sq m.

Question 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°. 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.

Solution 9.6:

Given:

Dm = 40

A = 60

n(w) = 1.33

Refractive index of material of prism,

n(p) = [sin {(A + Dm)/2}] / sin(A / 2)

Substitution yields,

n(p) = 1.53

Now in water,

n(p) / n(w) = [sin {(A + Dm’) / 2}] / sin (A / 2)

Or

(1.53 / 1.33).sin 30 = sin [{A + Dm’} / 2]

Or

Dm = 10o.

CBSE Class 12 Chapter Notes

Question 9.7: 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 20 cm?

Solution 9.7:

Given:

n = 1.55

f = 20 cm

Using Lens Maker’s formula

n1 / f = (n2n1) × [1/R + 1/R]

Substituting values

1 / 20 = (1.55 ‒ 1) × [2 / R]

R = 22 cm.

Question 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?

Solution 9.8:

(a) Convex lens, f = 20 cm

u = 12 cm

(1/v) – (1/12) = (1/20)

Or

v = 7.5 cm.

(b) Concave f = 16 cm

u = 12 cm

(1/v) – (1/12) = ‒1 / 16

Or

v = 48 cm.

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

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