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A source of light is placed at a distance of 75 cm from a screen. Two sharp images are formed on screen when a lens is placed between source and screen at two positions. If one image is four times larger than the other, then focal length of the lens used, is

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$\frac{35}{2} c m$

25 cm

15 cm

$\frac{50}{3} c m$

answer is D.

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Detailed Solution

$m_{2} = 4 m_{1} and m_{1} m_{2} = 1 \Rightarrow m_{1} = \frac{1}{2}, m_{2} = 2$
For $m_{2} = 2$
$u = 25 c m, v = 50 c m$
$∴ \frac{1}{f} = \frac{1}{25} + \frac{1}{50} = \frac{3}{50}$
$∴ f = \frac{50}{3} c m$

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