A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If the lenses are made of different material of refractive indices $μ_{1}$ and $μ_{2}$ and R is the radius of curvature of the curved surface of the lenses, then focal length of the combination is

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$(\frac{R}{μ_{1} - μ_{2}})$

$(\frac{2 R}{μ_{2} - μ_{1}})$

$\frac{R}{2 (μ_{2} - μ_{1})}$

$\frac{R}{2 - (μ_{1} + μ_{2})}$

answer is A.

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Detailed Solution

From the lens maker formula for plano convex lens, $\frac{1}{f_{1}} = (μ_{1} - 1) (\frac{1}{R})$
For plano cancave lens $\frac{1}{f_{2}} = - (μ_{2} - 1) (\frac{1}{R})$
For the combination, $\frac{1}{f_{e q}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = (μ_{1} - μ_{2}) (\frac{1}{R})$
This gives us $f_{e q} = \frac{R}{μ_{1} - μ_{2}}$