# A symmetric biconvex lens of radius of curvature R and made of glass of refractive index 1·5, is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its tip on the principal axis of the lens is moved along the axis until its real, inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be x. On removing the liquid layer and repeating the experiment, the distance is found to be y. Obtain the expression for the refractive index of the liquid in terms of x and y.

Given,
Refractive index of lens $\mu _{1}= 1\cdot 5$
Distance of needle from the lens = focal length of systems = x = f
Distance after removing liquid layer = focal length of lens = y = f1
Let, focal length of liquid = f2
refractive index of liquid = $\mu _{2}$
The equivalente focal length f is
$\frac{1}{f}= \frac{1}{f_{1}}+\frac{1}{f_{2}}$
$\frac{1}{f_{2}}= \frac{1}{f}-\frac{1}{f_{1}},\; so \; f_{2}= \frac{f_{1}\times f}{f_{1}-f}$
$\frac{1}{f_{2}}= \frac{xy}{y-x}$
Given that radius of Curvature of other surfaces of the lens  is -R
from lens maker formula
$\frac{1}{f_{1}}= \left ( \mu _{1}-1 \right )\left ( \frac{1}{R} -\frac{1}{-R}\right )$
$\frac{1}{y}= \left ( 1\cdot 5-1 \right )\times \frac{2}{R}$
$R= \frac{y}{0\cdot 5\times 2}= y\, cm$
Since the liquid act as a plane mirror, therefore
Applying lens maker formula for liquid, we get
$\frac{1}{f_{2}}= \left ( \mu _{2}-1 \right )\left ( \frac{1}{R} -\frac{1}{\infty }\right )$
$\frac{1}{-\frac{xy}{y-x}}= \left ( \mu _{2} -1\right )\frac{1}{y};\; \; \mu _{2}-1= \frac{x-y}{x}$
$\mu _{2}= \frac{x-y}{x}+1\, = 2-\frac{y}{x}$
Therefore, the expression for the refractive index of the liquid is

$2-\frac{y}{x}$

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