Two thin symmetrical lenses of different nature and of different material have equal radii of curvature R = 15 cm. The lenses are put close together and immersed in water $(μ_{w} = \frac{4}{3})$ . The focal length of the system in water is 30 cm. The difference between refractive of the two lenses is

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answer is C.

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Detailed Solution

$\begin{array}{l} Let f_{1} and f_{2} be the focal length in water . \\ Then \frac{1}{f_{1}} = (\frac{μ_{1}}{μ_{w}} - 1) = (\frac{1}{R} + \frac{1}{R}) = (\frac{μ_{1}}{μ_{w}} - 1) (\frac{2}{R}) . . . . (1) \\ \frac{1}{f_{1}} = (\frac{μ_{2}}{μ_{w}} - 1) (- \frac{1}{R} - \frac{1}{R}) = (\frac{μ_{2}}{μ_{w}} - 1) (- \frac{2}{R}) . . . . . (2) \\ Adding (1) and (2), we get \\ \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{2 (μ_{1} - μ_{2})}{μ_{w} R} \\ or \frac{1}{30} = \frac{2 (μ_{1} - μ_{2})}{μ_{w} R} \\ ∴ (μ_{1} - μ_{2}) = \frac{μ_{w} R}{60} \\ Substitutiong the values \\ (μ_{1} - μ_{2}) = \frac{4 \times 15}{3 \times 60} = \frac{1}{3} \end{array}$