Question

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 $R=30cm$ . The lens is made of a different material having refractive index $\text{μ}_{1} = \frac{3}{2} \text{ and } \text{μ}_{2} = \frac{5}{4}$ 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 $cm$ ).

Answer 1

(i) $P_{eq}=2P_{l 1}+2P_{l 2}+P_{m }...\left(1\right)$
For plano-concave lens : $\frac{1}{f_{l 1}}=\left(\frac{3}{2} - 1\right)\left(\frac{1}{\infty } - \frac{1}{3 0}\right)$
$f_{l 1}=-60cm\Rightarrow P_{l 1}=\frac{1 0 0}{f_{l 1}}\Rightarrow P_{l 1}=-\frac{5}{3}D$
For plano-convex lens : ​ $\frac{1}{f_{l 2}}=\left(\frac{5}{4} - 1\right)\left(\frac{1}{30} - \frac{1}{\infty }\right)$ $⇒$ $\frac{1}{f_{l 2}}=\frac{1}{120}$
$f_{l 2}=+120cm\Rightarrow P_{l 2}=\frac{100}{f_{l 2}}\Rightarrow P_{l 2}=+\frac{5}{6}D$
For plane mirror : $P_{m}=0\text{since}f_{m}=\infty $
Effective power of the combination is
$P_{eq}=2P_{l 1}+2P_{l 2}+P_{m}$
$P_{eq}=-\frac{2 \times 5}{3}+\frac{2 \times 5}{6}+0$
$\Rightarrow P_{eq}=-\frac{10}{6}D$ $\because P_{eq}=-\frac{1}{f_{eq}}$
⇒ $f_{e q}=60cm$