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A pole of height 4 $m$ is kept in front of a vertical plane mirror of length 2 $m$ . The lower end of the mirror is at a height of 6 $m$ from the ground. The horizontal distance between the mirror and the pole is 2 $m$ . Upto what minimum and maximum heights a man can see the image of top of the pole at a horizontal distance of 4 $m$ (from the mirror) standing on the same horizontal line which is passing through the base of the pole below the mirror?

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minimum height 2 $m$ and maximum height 4 $m$

minimum height 12 $m$ and maximum height 14 $m$

minimum height 10 $m$ and maximum height 16 $m$

minimum height 10 $m$ and maximum height 14 $m$

answer is C.

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Detailed Solution

From the question,

$P Q =$ Pole, $M N =$ Image of pole and AB =Man

$\frac{H G}{G N} = \frac{B D}{B N} B D = \frac{(H G) (B N)}{G N} = \frac{(2) (6)}{2} B D = 6 m$

$∴$ Minimum height required $= A D = B D + A B = 10 m$

Further, $\frac{I G}{G N} = \frac{B E}{B N} B E = \frac{(I G) (B N)}{G N} = \frac{(4) (6)}{2} B E = 12 m$

$∴$ Maximum height required $= A E = B E + A B = 16 m$

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