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Illustrations

Example-1

An object of height 4 cm is placed at 30 cm from a concave mirror of focal length 20 cm. Find the position of the image. Comment on the nature and size of the image.

Solution

Given, ho = 4 cm, u = +30 cm, f = + 20 cm

 

To find: v, hi and nature of image.

 

From mirror formula, we have

 

Description: 84551.png

 

Description: 84562.png

 

Description: 84572.png

 

Description: 84582.png

 

Magnification, Description: 84592.png

 

or Description: 84600.png

 

Thus, image is formed at a distance of 60 cm in front of the mirror. Height of the image is 8 cm. As v is positive, the image is real. Since hi is negative, it is inverted.
 

 

Example-2

A point source of light is kept in front of a convex mirror of radius of curvature 40 cm. The image is formed at 10 cm behind the mirror. Calculate the object distance.

Solution

Given: For a convex mirror, R = –40 cm.

 

v = –10 cm (image is virtual).

 

From mirror formula, we have

 

Description: 84621.png

 

Description: 84628.png

 

Description: 84637.png

 

= +20 cm.

 

Thus, object is placed 20 cm in front of the mirror.
 

 

Example-3

The refractive index of glass is 1.5. The speed of light in vacuum is 3  108 m s–1. Calculate the speed of light in glass.

Solution

Given, refractive index of glass, ng = 1.5

 

Velocity of light in vacuum, c = 3  108 m s–1

 

Velocity of light in glass, vg = ?

 

Refractive index of glass with respect to air or vacuum is
Description: 84654.png

 

i.e., vg = 2  108 m s–1.
 

 

Example-4

A ray of light travels from water to glass. The angle of incidence is 40°. Calculate the angle of refraction and the deviation produced. Given ng = 1.5, nw = 1.3.

Solution

i = 40°, r = ?, ng = 1.5, nw = 1.3

 

Since, light is travelling from water to glass, we have

 

Description: 84672.png

 

i.e. Description: 84681.png

 

ng sin r = nw sin i

 

sin r = Description: 84692.png

 

Description: 84700.png

 

∴ r = sin–1 (0.5570) = 33° 51′

 

Thus, angle of refraction is 33° 51′.

 

Deviation, d = i – r = 40° − 33° 51′

 

∴ d = 6° 9′.
 

 

Example-5

At what distance should an object be placed in front of a convex lens of focal length 0.4 m so that the image is thrice the size of the object?

Solution

Given f = 0.4 m, hi = 3ho, u = ?

 

We know that, Description: 84726.png

 

Description: 84735.png

 

Also from lens formula, we have

 

Description: 84743.png

 

Description: 84752.png

 

Description: 84759.png

 

Description: 84770.png

 

i.e., object distance, u = 0.267 m.

 

The object must be placed at 0.267 m from the lens.
 

 

Example-6

The eye lens has a focal length of 2.5 cm. Find the range of accommodation while changing the focus from infinity to D = 25 cm.

Solution

For the object at infinity, u1 = ∞,

 

v1 = f1 = 2.5 cm

 

Description: 84779.png

 

Description: 84786.png

 

For the object at 25 cm, u2 = 25 cm,

 

v2 = 2.5 cm, f2 = ?

 

Description: 84796.png

 

Description: 84805.png

 

= 44 dioptre.

 

Thus, the change in power, i.e., the range of accommodation is p2 − p1 = 4 dioptre.
 





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