A thin plano-convex lens fits exactly into a plano-concave lens with their plane surface parallel to each other as shown in the figure. The radius of curvature of the curved surface $R = 30 c m .$ The lens is made of a different material having refractive index $μ_{1} = \frac{3}{2}$ and $μ_{2} = \frac{5}{4}$ as shown in the figure. If the plane surface of the plano-convex lens is silvered, then calculate the equivalent focal length of this system (in cm).

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answer is 60.

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Detailed Solution

$P_{e q} = 2 P_{l 1} + 2 P_{l 2} + P_{m} ........ (1)$
For plano-cancavelens : $\frac{1}{f_{l 1}} = (\frac{3}{2} - 1) (\frac{1}{\infty} - \frac{1}{30})$
$f_{l 1} = - 60 c m \Rightarrow P_{l 1} = \frac{100}{f_{l 1}} \Rightarrow P_{l 1} = - \frac{5}{3} D$
For plano-canvexlens : $\frac{1}{f_{l 2}} = (\frac{5}{4} - 1) (\frac{1}{30} - \frac{1}{\infty}) \Rightarrow \frac{1}{f_{l 2}} = \frac{1}{120}$
For plane mirror : $P_{m} = 0$ since $f_{m} = \infty$
Effective power of the combination is
$P_{e q} = 2 P_{l 1} + 2 P_{l 2} + P_{m}$
$P_{e q} = - \frac{2 \times 5}{3} + \frac{2 \times 5}{6} + 0$
$\Rightarrow P_{e q} = - \frac{10}{6} D ∵ P_{e q} = - \frac{1}{f_{e q}}$
$\Rightarrow f_{e q} = 60 c m$