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A plano convex lens fits exactly in to a plano concave lens. Their plane surfaces are parallel to each other. If the lenses are made of different materials of refractive indices $μ$ ₁ and $μ$ ₂ and ‘R’ is the radius of curvature of the curved surfaces of the lenses then the focal length of the combination is

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$\frac{R}{μ_{1} - μ_{2}}$

$\frac{2 R}{μ_{1} - μ_{2}}$

$\frac{R}{2 (μ_{1} - μ_{2})}$

$\frac{R}{(μ_{1} + μ_{2})}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{μ_{1} - 1}{R} - \frac{μ_{2} - 1}{R} \\ f = \frac{R}{μ_{1} - μ_{2}} \end{array}$

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