Solution:

Ans. (i)

Two thin lenses, of focal length $f_{1}$ and $f_{2}$ are kept in contact. Let $O$ be the position of object and let $u$ be the object distance. The distance of the image (which is at $I_{1}$ ), for the first lens is $v_{1}$.

This image serves as object for the second lens. $1 / 2$ Let the final image be at I. We then have

$$ \begin{aligned} & \frac{1}{f_{1}}=\frac{1}{v_{1}}-\frac{1}{u} \ & \frac{1}{f_{2}}=\frac{1}{v}-\frac{1}{v_{1}} \end{aligned} $$

Adding, we get

$$ \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}} $$

$$ \begin{array}{rlrl} f_{1} & \frac{1}{f_{2}} & =\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \ & \therefore & \frac{1}{f} & =\frac{1}{f_{1}}+\frac{1}{f_{2}} \ & \therefore & P & =P_{1}+P_{2} \end{array} $$

(ii) At minimum deviation

$$ r=\frac{A}{2}=30^{\circ} $$

We are given that, $\quad i=\frac{3}{4} A=45^{\circ}$

$$ \therefore \quad \mu=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\sqrt{2} $$

$\therefore$ Speed of light in the prism $=\frac{c}{\sqrt{2}}$

$$ \left(\cong 2.1 \times 10^{8} \mathrm{~ms}^{-1}\right) $$

[Award $1 / 2$ mark if the student writes the formula :

$$ \mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)} $$

but does not do any calculations.]

[CBSE Marking Scheme 2017] AI Q. 2. (i) A point object is placed on the principal axis of a convex spherical surface of radius of curvature $R$, which separates the two media of refractive indices $n_{1}$ and $n_{2}\left(n_{2}>n_{1}\right)$. Draw the ray diagram and deduce the relation between the object distance $(u)$, image distance $(v)$ and the radius of curvature $(R)$ for refraction to take place at the convex spherical surface from rarer to denser medium.

(ii) A converging lens has a focal length of $20 \mathrm{~cm}$ in air. It is made of a material of refractive index $\mathbf{1 \cdot 6}$. If it is immersed in a liquid of refractive index $1 \cdot 3$, find its new focal length

U] [Foreign 2017]

Ans. (i)

For small angles

$$ \begin{aligned} \tan \angle N O M & =\frac{M N}{O M} \ \tan \angle N C M & =\frac{M N}{M C} \ \tan \angle N I M & =\frac{M N}{M I} \end{aligned} $$

For $\triangle N O C, i$ is exterior angle, therefore

$$ \begin{aligned} i & =\angle N O M+\angle N C M \ & =\frac{M N}{O M}+\frac{M N}{M C} \end{aligned} $$

Similarly,

$$ r=\frac{M N}{M C}-\frac{M N}{M I} $$

For small angles, Snell’s law can be written as

$$ \begin{aligned} & n_{1} i=n_{2} r \ & \therefore \quad \frac{n_{1}}{O M}+\frac{n_{2}}{M I}=\frac{n_{2}-n_{1}}{M C} \ & \therefore \quad O M=-u, M I=+v \ & M C=+R \end{aligned} $$

(using sign convention)

$\therefore \quad \frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}$

$1 / 2$

(ii) Lens maker’s formula is

$$ \frac{1}{f}=\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) $$

Focal length of lens is $=20 \mathrm{~cm}$

$$ \begin{aligned} \frac{1}{20} & =(1 \cdot 6-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \ \therefore \quad\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) & =\frac{1}{20 \times 0 \cdot 6}=\frac{1}{12} \end{aligned} $$

Let $f^{\prime}$ be the focal length of the lens in water

$$ \begin{aligned} & \therefore \quad \frac{1}{f^{\prime}}=\left(\frac{1 \cdot 6-1 \cdot 3}{1 \cdot 3}\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \ & =\frac{0 \cdot 3}{12 \times 1 \cdot 3} \end{aligned} $$

Or

$$ f^{\prime}=\frac{120 \times 1 \cdot 3}{3}=52 \mathrm{~cm} $$

[CBSE Marking Scheme 2017]