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Two thin symmetrical lenses of different nature have equal radii of curvature of all faces R = 20 cm. The lenses are put close together and immersed in water. The focal length of the system is 24 cm. The difference between refractive indices of the lenses is $(\frac{5}{n}) .$ Find $n$ . The refractive index of water is $\frac{4}{3} .$

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answer is 9.

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Detailed Solution

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \Rightarrow \frac{1}{24} = [(\frac{μ_{1}}{\frac{4}{3}} - 1) (\frac{1}{20} + \frac{1}{20}) + (\frac{μ_{2}}{\frac{4}{3}} - 1) (\frac{1}{- 20} - \frac{1}{20})]$

$(μ_{1} - μ_{2}) = (\frac{5}{9})$

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