Answer –
According to the question –
Focal length of the given convex lens is f1 = 30 cm
Since, the liquid acts as a mirror. Focal length of the liquid is denoted by f2
Total focal length of the system (convex lens + liquid) is f = 45 cm
The equivalent focal length of the pair of optical systems which are in contact is expressed 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,we have
f2 = -90 cm
Assume that $ \mu_{1}$ is the refractive index and R is the radius of curvature of one surface. Therefore, -R will be the radius of curvature of the other surface.
R can be determined using the relation :
$\frac{1}{f_{1}}=(\mu_{1} -1)(\frac{1}{R}+\frac{1}{-R})$
$\frac{1}{30}$ = (1.5 – 1) $\left ( \frac{2}{R} \right )$
Therefore,
R = $\frac{30}{1}$
R = 30 cm
Consider $\mu _{2}$ as the refractive index of the liquid.
Radius of curvature of the liquid on the side of the plane mirror = ∞ Radius of curvature of the liquid on the side of the lens,
R = −30 cm
The value of $\mu _{2}$ can be obtained by using the relation:
$\frac{1}{f_{2}}=\left ( \mu _{2} -1\right )\left [ \frac{1}{-R}-\frac{1}{infinity} \right ]$
$\mu _{2}$ – 1 = $\frac{1}{3}$
Therefore, we have
$\mu _{2}$ = 1.33
Therefore, the refractive index of the liquid is 1.33 .