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In conjugate foci method the distance between source and screen is 80cm. When the lens is at a distance of 28cm from the source the big image is formed on the screen. When the lens is at a distance of 52cm from the source, a small image is formed on the screen then the focal length is _____ $\times 10^{- 1} cm .$

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Detailed Solution

$f = \frac{l^{2} - d^{2}}{4 l}$

$d_{1} = 28 cm$

$d_{2} = 52 cm$

$d = d_{2} - d_{1} l = 80 cm$

$After solving,$

$f = 18 .2 cm$

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