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

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$\frac{R}{(μ_{1} - μ_{2})}$

$\frac{2 R}{(μ_{2} - μ_{1})}$

$\frac{R}{2 (μ_{1} + μ_{2})}$

$\frac{R}{2 (μ_{1} - μ_{2})}$

answer is A.

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Detailed Solution

The combination of two lenses 1 and 2 is as shown in figure.

$∴ \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$

According to lens maker's formula

$\frac{1}{f_{1}} = (μ_{1} - 1) (\frac{1}{\infty} - \frac{1}{R}) = \frac{(μ_{1} - 1)}{R} \frac{1}{f_{2}} = (μ_{2} - 1) (\frac{1}{- R} - \frac{1}{\infty}) = (μ_{2} - 1) (- \frac{1}{R}) = - \frac{(μ_{2} - 1)}{R} ∴ \frac{1}{f} = \frac{(μ_{1} - 1)}{R} - \frac{(μ_{2} - 1)}{R} \frac{1}{f} = \frac{(μ_{1} - μ_{2})}{R} f = \frac{R}{(μ_{1} - μ_{2})}$