Question

A plano-convex lens fits exactly into plano-concave lens as shown in figure. Their plane surfaces are parallel to each other. If the lens are made of different materials of refractive indices $1.6$ and $1.5$ respectively. If $R$ is the radius of curvature of curved surfaces of lenses. Then the focal length of the combination

• A. $\frac{R}{0.2}$
• B. $\frac{R}{6.2}$
• C. $\frac{R}{3.1}$
• D. $\frac{R}{0.1}$

Answer 1

The correct option is D $\frac{R}{0.1}$

According to formula
$\frac{1}{f_1}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{f_1}=(1.6-1)(\frac{1}{R}-\frac{1}{\infty })$
$\frac{1}{f_1}=0.6\times \frac{1}{R}=\frac{6}{10R}$
Given, $n=1.6$
$R_1=R$
$R_2=\infty$ (Because surface is plane)
$f_1=\frac{10R}{6}$ ....(i)
and $\frac{1}{f_2}=(1.5-1)(\frac{1}{-R}-\frac{1}{\infty })$
$\frac{1}{f_2}=-0.5\times \frac{1}{R}$
$f_2=-\frac{10R}{5}$
So, the focal length of the combination
$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$
$\frac{1}{F}=\frac{6}{10R}-\frac{5}{10R}$
$F=\frac{10R}{1}=\frac{R}{1/10}$
$F=\frac{R}{0.1}$