A convex lens is in contact with concave lens. The magnitude of the ratio of their focal length is $\frac{2}{3}$ . Their equivalent focal length is 30 cm. Their individual focal lengths are

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– 75 cm, 50 cm

– 15 cm, 10 cm

– 10 cm, 15 cm

– 50 cm, 75 cm

answer is D.

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Detailed Solution

From equivalent focal length

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \Rightarrow \frac{1}{2} [\frac{f_{2}}{f_{1}} + 1] \to 1$

From achromatic doublet

$\frac{w_{2}}{w_{1}} = - \frac{f_{2}}{f_{1}} \Rightarrow \frac{f_{2}}{f_{1}} = - \frac{2}{3} \to 2$

From 2 and 1

$f_{1} = - 15 c m, f_{2} = 10 c m$

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