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A converging beam of light forms a sharp image on a screen. A lens is placed in the path of the beam, the lens being 10cm from the screen. It is found that the screen has to be moved 8 cm further away from the lens to obtain a sharp image. The focal length of lens is

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$- 22.5 c m$

$40 c m$

$- 40 c m$

$22.5 c m$

answer is A.

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Detailed Solution

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\begin{array}{l} \frac{1}{18} - \frac{1}{+ 10} = \frac{1}{f} \\ \frac{1}{f} = \frac{5 - 9}{90} = \frac{- 4}{90} \\ f = - 22.5 c m \end{array}$

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