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 .