JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 26)
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.
_26th_February_Evening_Shift_en_26_1.png)
Answer
150
Explanation
Given, length of mirror, m = 50 cm = 50 $$\times$$ 10$$-$$2 m
Distance of source from mirror, d = 60 cm = 60 $$\times$$ 10$$-$$2 m
Distance of man from mirror, dm = 1.2 m
By using the concept of ray diagram of plane mirror shown below
_26th_February_Evening_Shift_en_26_2.png)
Now, using the concept of similar triangle,
$$\Delta$$HAI $$\sim$$ $$\Delta$$GAE and $$\Delta$$BAI $$\sim$$ $$\Delta$$CAE
$$\therefore$$ $${{AI} \over {AE}} = {{HI} \over {EG}}$$
$$ \Rightarrow {{0.60} \over {1.8}} = {{0.25} \over {EG}}$$ ($$\because$$ AI = IS)
$$ \Rightarrow EG = 0.25 \times {{1.8} \over {0.6}} = 0.25 \times 3 = 0.75$$ m
As, $$CG = 2EG$$
$$ \Rightarrow CG = 0.75 \times 2 = 1.50$$ m
Hence, distance between the extreme points, where he can see image of light source in mirror is 150 cm.
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