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

$$\begin{array}{rlr}& \frac{1}{f_1}=\frac{1}{v_1}-\frac{1}{u}& \frac{1}{f_2}=\frac{1}{v}-\frac{1}{v_1}\end{array}$$

Adding, we get

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

$$\begin{array}{rlrllllll}{f}_1& \frac{1}{f_2}& =\frac{1}{v}-\frac{1}{u}=\frac{1}{f}& \therefore & \frac{1}{f}& =\frac{1}{f_1}+\frac{1}{f_2}& \therefore & P& =P_1+P_2\end{array}$$

(ii) At minimum deviation

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

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

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

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

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

[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 $n_2(n_2>n_1)$. 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 $\mathbf{1}\cdot \mathbf{6}$. 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

$$\begin{array}{rlrl}\tan \mathrm{\angle }NOM& =\frac{MN}{OM}\tan \mathrm{\angle }NCM& =\frac{MN}{MC}\tan \mathrm{\angle }NIM& =\frac{MN}{MI}\end{array}$$

For $\mathrm{△}NOC,i$ is exterior angle, therefore

$$\begin{array}{rlr}i& =\mathrm{\angle }NOM+\mathrm{\angle }NCM& =\frac{MN}{OM}+\frac{MN}{MC}\end{array}$$

Similarly,

$$r=\frac{MN}{MC}-\frac{MN}{MI}$$

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

$$\begin{array}{rlrlr}& n_{1}i=n_{2}r& \therefore \frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2-n_1}{MC}& \therefore OM=-u,MI=+v& MC=+R\end{array}$$

(using sign convention)

$\therefore \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

$1/2$

(ii) Lens maker’s formula is

$$\frac{1}{f}=(\frac{n_2}{n_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$$

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

$$\begin{array}{rlr}\frac{1}{20}& =(1\cdot 6-1)(\frac{1}{R_1}-\frac{1}{R_2})\therefore (\frac{1}{R_1}-\frac{1}{R_2})& =\frac{1}{20\times 0\cdot 6}=\frac{1}{12}\end{array}$$

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

$$\begin{array}{rlr}& \therefore \frac{1}{f^{\prime }}=\left(\frac{1\cdot 6-1\cdot 3}{1\cdot 3}\right)(\frac{1}{R_1}-\frac{1}{R_2})& =\frac{0\cdot 3}{12\times 1\cdot 3}\end{array}$$

Or

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

[CBSE Marking Scheme 2017]