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Two converging glass lenses A and B have focal lengths in the ratio 2 : 1. The radius of curvature of first surface of lens A is 1/4th of the second surface where as the radius of curvature of first surface of lens B is twice that of second surface. Then the ratio between the radii of the first surfaces of A and B is

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5:6

1:2

5:3

3:5

answer is D.

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Detailed Solution

$For lens A \to R_{1 A} = \frac{R_{2 A}}{4}$

$For lens B \to R_{1 B} = 2 R_{2 B}$

$\frac{1}{f} \propto (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$\frac{f_{B}}{f_{A}} = \frac{\frac{1}{R_{1 A}} + \frac{1}{R_{2} A}}{\frac{1}{R_{1 B}} + \frac{1}{R_{2} A}}$

$\frac{1}{2} = \frac{\frac{1}{R_{1 A}} + \frac{1}{4 R_{1 A}}}{\frac{1}{R_{1 B}} + \frac{2}{R_{1 B}}}$

$\frac{R_{1 A}}{R_{1 B}} = \frac{10}{12} = \frac{5}{6}$

$R_{1 A} : R_{1 B} = 5 : 6$

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