Question 271: Ray Optics

Question

A point source of light S, placed at a distance 60cm in front  of the centre of a plane mirror of width 50 cm, hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 1.2 m from it (see in the figure). The distance between the extreme points where he can see the image of the light source in the mirror is __________ cm.

Accepted Answer

Given, length of mirror, m = 50 cm = 50 $$	imes$$ 10
$$-$$2
m
Distance of source from mirror, d = 60 cm = 60 $$	imes$$ 10
$$-$$2
m
Distance of man from mirror, d
m
= 1.2 m
By using the concept of ray diagram of plane mirror shown below
Now, using the concept of similar triangle,
$$\Delta$$HAI $$\sim$$ $$\Delta$$GAE and $$\Delta$$BAI $$\sim$$ $$\Delta$$CAE
$$	herefore$$ $${{AI} \over {AE}} = {{HI} \over {EG}}$$
$$ \Rightarrow {{0.60} \over {1.8}} = {{0.25} \over {EG}}$$ ($$\because$$ AI = IS)
$$ \Rightarrow EG = 0.25 	imes {{1.8} \over {0.6}} = 0.25 	imes 3 = 0.75$$ m
As, $$CG = 2EG$$
$$ \Rightarrow CG = 0.75 	imes 2 = 1.50$$ m
Hence, distance between the extreme points, where he can see image of light source in mirror is 150 cm.

Ray Optics

Integer / Numeric Answer Type