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

Answer: 60

(i) $P_{eq}=2P_{l1}+2P_{l2}+P_m...\left(1\right)$
For plano-concave lens : $\frac{1}{f_{l1}}=\left(\frac{3}{2}-1\right)\left(\frac{1}{\infty }-\frac{1}{30}\right)$
$f_{l1}=-60cm\Rightarrow P_{l1}=\frac{100}{f_{l1}}\Rightarrow P_{l1}=-\frac{5}{3}D$
For plano-convex lens : ​ $\frac{1}{f_{l2}}=\left(\frac{5}{4}-1\right)\left(\frac{1}{30}-\frac{1}{\infty }\right)$ $\Rightarrow$ $\frac{1}{f_{l2}}=\frac{1}{120}$
$f_{l2}=+120cm\Rightarrow P_{l2}=\frac{100}{f_{l2}}\Rightarrow P_{l2}=+\frac{5}{6}D$
For plane mirror : $P_m=0\text{since}{f}_m=\infty$
Effective power of the combination is
$P_{eq}=2P_{l1}+2P_{l2}+P_m$
$P_{eq}=-\frac{2\times 5}{3}+\frac{2\times 5}{6}+0$
$\Rightarrow P_{eq}=-\frac{10}{6}D$ $\because P_{eq}=-\frac{1}{f_{eq}}$
⇒ $f_{eq}=60cm$