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 : 3

3 : 5

5 : 6

1 : 2

answer is D.

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Detailed Solution

$F o r A \frac{1}{f_{A}} = (μ - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}}) R_{1} = \frac{R_{2}}{4} \Rightarrow 4 R_{1} \frac{1}{f_{A}} = (μ - 1) (\frac{1}{R_{1}} + \frac{1}{4 R_{1}}); f_{A} = \frac{4 R_{1}}{5 (μ - 1)} F o r B \frac{1}{f_{B}} = (μ - 1) (\frac{1}{R_{1}^{1}} + \frac{1}{R_{2}^{1}}) R_{1}^{1} = 2 R_{2}^{1} \Rightarrow R_{2}^{1} = \frac{R_{1}^{1}}{2} \frac{1}{f_{B}} = (μ - 1) (\frac{1}{R_{1}^{1}} + \frac{2}{R_{1}^{1}}) f_{B} = \frac{R_{1}^{1}}{3 (μ - 1)} \frac{f_{A}}{f_{B}} = \frac{4 R_{1}}{5} \times \frac{3}{R_{1}^{1}} \frac{2}{1} = \frac{12}{5} \times \frac{R_{1}}{R_{1}^{1}}; \frac{R_{1}}{R_{1}^{1}} = \frac{5}{6}$