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 $P_{2}$.

If both lenses have same radius of curvature $R$ each and $P_{2}=2 P_{1}$, then the refractive indices are related to each other as $\mu_{1}+\frac{\mu_{2}}{2}=$______.

Answer 1

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, $P_{2}=2 P_{1}$
$\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$