Question

A plano-convex lens with $R.I.$ ${\mu }_1$ and power $P_1$ is placed in contact with a plano concave lens with $R.I.$ ${\mu }_2$ and power $P_2$ .

Let both lenses have same radius of curvature $R$ each and $P_2=2P_1$ . Then the refractive indices are related to each other as ${\mu }_1+\frac{{\mu }_2}{2}=\frac{n}{2}$ . What is the value of $n$ ?

Answer 1

Answer: 3

From lens maker's equation,
$P=\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
For plano - convex lens,
$R_1=\infty ,R_2=-R$
$\therefore P_1=\left({\mu }_1-1\right)\left(\frac{1}{\infty }-\frac{1}{-R}\right)=\frac{\left({\mu }_1-1\right)}{R}$
For plano-concave lens,
$R_1=-R,R_2=\infty$
$\therefore P_2=\left({\mu }_2-1\right)\left(\frac{1}{-R}-\frac{1}{\infty }\right)=\frac{-\left({\mu }_2-1\right)}{R}$
Given that,
$P_2=2P_1$
$\therefore \frac{-\left({\mu }_2-1\right)}{R}=\frac{2\left({\mu }_1-1\right)}{R}$
$\therefore -{\mu }_2+1=2{\mu }_1-2$
$\therefore 2{\mu }_1+{\mu }_2=3$
$\therefore {\mu }_1+\frac{{\mu }_2}{2}=1.5=\frac{3}{2}$