Question

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 itslelf. 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.

Answer 1

Given, refractive index of lens, ${\mu }_g$ = 1.5

The distance of the needle from the lens in the first case = The focal length of the combination of convex lens and plano- concave lens formed by the liquid, f =x

And, the distance measured in the second case = Focal length of the convex lens, $f_1=y$

If the focal length of plano-concave lens formed by the liquid be $f_2$, then

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$

$\Rightarrow \frac{1}{f_2}=\frac{1}{f}-\frac{1}{f_1}=\frac{1}{x}-\frac{1}{y}$

=$\frac{y-x}{xy}$

or $f_2=\frac{xy}{y-x}$

Now, refractive index of lens, $\mu =1.5$, radius of curvature of one surface= R and radius of curvature of other surface = - R

$\therefore \frac{1}{f_1}=({\mu }_g-1)(\frac{1}{R}-\frac{1}{-R})$

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

And, if refractive index of liquid is ${\mu }_l$,
Radius of curvature on the side of plane mirror =$\infty$
Radis of curvature on the side of lens =- R

$\therefore \frac{y-x}{xy}=({\mu }_l-1)(\frac{1}{-R}-\frac{1}{\infty })=({\mu }_l-1)(-\frac{1}{y})$

$\Rightarrow (\mu -1)=-\left(\frac{y-x}{x}\right)$

or ${\mu }_l=\frac{x-y}{x}+1$

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