Solution:

Ans. Lens maker’s formula $1 / 2$ Formula for ‘combination of lenses’ $1 / 2$ Obtaining the expression for $\mu$

(a) Let $\mu_{1}$ denote the refractive index of the liquid. When the image of the needle coincides with the lens itself; its distance from the lens, equals the relevant focal length.

With liquid layer present, the given set up, is equivalent to a combination of the given (convex) lens and a concavo plane/plano concave ’liquid lens’.

We have $\quad \frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

and

$$ \frac{1}{f}=\left(\frac{1}{f_{1}}+\frac{1}{f_{2}}\right) $$

as per the given data,

$$ \begin{aligned} \frac{1}{f_{2}} & =\frac{1}{y}=(1 \cdot 5-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right) \quad 1 / 2 \ & =\frac{1}{R} \end{aligned} $$

$$ \therefore \quad \frac{1}{x}=\left(\mu_{l}-1\right)\left(-\frac{1}{R}\right)+\frac{1}{y}=\frac{-\mu_{l}}{y}+\frac{2}{y} \quad 1 / 2 $$

$$ \begin{aligned} & \therefore \quad \frac{\mu_{1}}{y}=\frac{2}{y}-\frac{1}{x}=\left(\frac{2 x-y}{x y}\right) \ & \text { or } \quad \mu_{l}=\left(\frac{2 x-y}{x}\right) \end{aligned} $$

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Detailed Answer :

Given, Radius of curvature of convex lens is $R$ and $\mu$ $=1.5$

When lens is placed on a layer of water then focal length of combination is $f$. Thus the first measurement gives the focal length of combination i.e. $f=x$.

In second measurement, we get the focal length $f_{1}$ of convex lens i.e., $f_{1}=y$, therefore the focal length of plano convex lens of water $f_{2}$ is given as

$$ \begin{align*} \Rightarrow \quad \frac{1}{f_{1}}+\frac{1}{f_{2}} & =\frac{1}{f} \ \Rightarrow \quad & \frac{1}{f_{2}}=\frac{1}{f}-\frac{1}{f_{1}}=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{x y} \ \Rightarrow \quad & f_{2}=\frac{x y}{y-x} \tag{i} \end{align*} $$

For plano convex lens of liquid of refractive index $\mu_{l}$

$\frac{1}{f_{2}}=\left(\mu_{l}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=\left(\mu_{l}-1\right)\left(-\frac{1}{R}-\frac{1}{\infty}\right)$

$\Rightarrow \frac{1}{f_{2}}=-\frac{\left(\mu_{l}-1\right)}{R}$

$\Rightarrow \mu_{l}=1-\frac{R}{f_{2}}$

Now from eqn. (i) and (ii),

$\mu_{l}=1-\frac{R}{x y /(y-x)}=1-\frac{R(y-x)}{x y}$

Now again for plano convex lens of air, if image coincide at $2 f_{1}$ then for case II we have

$$ \begin{array}{ll} \Rightarrow & 2 f_{1}=R \ \Rightarrow & f_{1}=\frac{R}{2} \ \text { again } & y=2 f_{1}=2 \times \frac{R}{2} \ \Rightarrow & y=R \tag{iv} \end{array} $$

Now from eqn. (iii) and (iv),

$$ \begin{aligned} & \mu_{l}=1-\frac{y(y-x)}{x y}=1-\frac{y-x}{x} \ & \mu_{l}=\frac{2 x-y}{x} \end{aligned} $$

This is the required expression for the refractive index of liquid.