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

From lens maker's equation,
$P=\frac{1}{f}=\left(\right.\mu -1\left.\right)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)$
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,
$P_{2}=2P_{1}$
$\therefore \frac{- \left(\mu _{2} - 1\right)}{R}=\frac{2 \left(\mu _{1} - 1\right)}{R}$
$\therefore -\mu _{2}+1=2\mu _{1}-2$
$\therefore 2\mu _{1}+\mu _{2}=3$
$\therefore \mu _{1}+\frac{\mu _{2}}{2}=1.5=\frac{3}{2}$