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Q.

A plano convex lens fits exactly into a plano 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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a

$\frac{R}{2 (μ_{1} - μ_{2})}$

b

$\frac{R}{(μ_{1} - μ_{2})}$

c

$\frac{2 R}{(μ_{2} - μ_{1})}$

d

$\frac{R}{2 (μ_{1} + μ_{2})}$

answer is B.

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Detailed Solution

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

Hence, focal length of the combination is $\frac{R}{μ_{1} - μ_{2}}$

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A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices μ1 and μ2 and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is.