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 ${\mu }_1=\frac{3}{2}$ and ${\mu }_2=\frac{5}{4}$ 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) $P_{eq}=2P_{\ell 1}+2P_{\ell 2}+P_m\dots (1)$
For plano-convex lens: $\frac{1}{(f{)}_t}=\left(\frac{3}{2}-1\right)\left(\frac{1}{\infty }-\frac{1}{30}\right)$
$f_{\ell 1}=-60cm$
$\Rightarrow P_{\ell 1}=\frac{100}{f_{\ell 1}}$
$\Rightarrow P_{\ell 1}=-\frac{5}{3}D$
For plano-convex lens: $\frac{1}{(f{)}_P}=\left(\frac{5}{4}-1\right)\left(\frac{1}{30}-\frac{1}{\infty }\right)$
$\Rightarrow \frac{1}{f_{C2}}=\frac{1}{120}$
$f_{\ell 2}=120cm\Rightarrow P_{\ell 2}=\frac{100}{f_{\ell 2}}$
$\Rightarrow P_{\ell 2}=\frac{5}{6}D$
For plane mirror: $P_m=0$ sincef ${}_m=\infty$
$\Rightarrow P_{eq}=2P_{\ell 1}+2P_{\ell 2}+P_m$
$P_{\text{eq }}=-\frac{10}{3}+\frac{10}{6}+0$
$\Rightarrow P_{\text{eq }}=-\frac{10}{6}D$
$\because P_{\text{eq }}=-\frac{1}{F_{\text{eq }}}$
$\Rightarrow F_{\text{eq }}=60cm$