The figure shows an equiconvex lens (of refractive index $1.50$) in contact with a liquid layer on top of a plane mirror. $A$ small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be $45.0 \; cm$. The liquid is removed and the experiment is repeated. The new distance is measured to be $30.0 \; cm$. What is the refractive index of the liquid?

(1.33) Focal length of the convex lens,$f_{1} = 30 \; cm$.
The liquid layer acts as a plano-concave lens. Let its focal length be $f_{2}$.
The combination of the convex lens and the liquid layer acts as a system with focal length $f = 45 \; cm$.
For a pair of optical systems placed in contact,the equivalent focal length is given by:
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\frac{1}{f_{2}} = \frac{1}{f} - \frac{1}{f_{1}} = \frac{1}{45} - \frac{1}{30} = \frac{2 - 3}{90} = -\frac{1}{90}$
$\therefore f_{2} = -90 \; cm$.
For the equiconvex lens,using the lens maker's formula:
$\frac{1}{f_{1}} = (\mu_{1} - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (1.5 - 1) \left( \frac{2}{R} \right) = \frac{1}{R}$
Since $f_{1} = 30 \; cm$,we have $R = 30 \; cm$.
For the plano-concave liquid lens,the radius of curvature of the upper surface is $R = -30 \; cm$ and the lower surface is $\infty$.
Using the lens maker's formula:
$\frac{1}{f_{2}} = (\mu_{2} - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right)$
$-\frac{1}{90} = (\mu_{2} - 1) \left( \frac{1}{-30} - 0 \right)$
$\mu_{2} - 1 = \frac{30}{90} = \frac{1}{3}$
$\mu_{2} = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.33$.
Thus,the refractive index of the liquid is $1.33$.