Figure 9.37 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror.

Solution:

Focal length of the convex lens, f1 = 30 cm

The liquid acts as a mirror. Focal length of the liquid = f2

Focal length of the system (convex lens + liquid), f = 45 cm

For a pair of optical systems placed in contact, the equivalent focal length is given as:

$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$

$\frac{1}{f_{2}}=\frac{1}{f}-\frac{1}{f_{1}}$

$=\frac{1}{45}-\frac{1}{30}=-\frac{1}{90}$

$\therefore f_{2}=-90 \mathrm{~cm}$

Let the refractive index of the lens be $\mu_{1}$ and the radius of curvature of one surface be $R$. Hence, the radius of curvature of the other surface is $-R$.

R can be obtained using the relation:

$\frac{1}{f_{1}}=\left(\mu_{1}-1\right)\left(\frac{1}{R}+\frac{1}{-R}\right)$

$\frac{1}{30}=(1.5-1)\left(\frac{2}{R}\right)$

$\therefore R=\frac{30}{0.5 \times 2}=30 \mathrm{~cm}$

Let $\mu_{2}$ be the refractive index of the liquid.

Radius of curvature of the liquid on the side of the plane mirror $=\infty$

Radius of curvature of the liquid on the side of the lens, R = −30 cm

The value of $\mu_{2}$ can be calculated using the relation:

$\frac{1}{f_{2}}=\left(\mu_{2}-1\right)\left[\frac{1}{-R}-\frac{1}{\infty}\right]$

$\frac{-1}{90}=\left(\mu_{2}-1\right)\left[\frac{1}{+30}-0\right]$

$\mu_{2}-1=\frac{1}{3}$

$\therefore \mu_{2}=\frac{4}{3}=1.33$

Hence, the refractive index of the liquid is 1.33.