(a) $O P$ is the incidence ray on the prism and $Q R$ emergent ray. $\angle i_1 = \text{ angle if incidence }$ $\angle i_2 = \text{ angle of emergence }$ $A = \text{ angle of the prism }$ $\mu = \text{ refractive index of the material of the prism. }$ $\delta = \text{ angle of deviation }$For minimum deviation, $\angle r_1 = \angle r_2 = \angle r$ $A = \angle r_1 + \angle r_2$So, $A = \angle r + \angle r = \angle 2 r$ $\angle r = 2A$ Also, $\angle i_1 + \angle i_2 = \angle i$ $A + \delta_m = \angle i_1 + \angle i_2$So, $i = \frac{A + \delta_m}{2}$Now, from snell's law, $\propto = \frac{\sin i}{\sin r}$ $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ (b)(i)The critical angle $= \sin^{-1}\left(\frac{1}{1.5}\right) = 41.3^{\circ}$.So, the ray which is incident on $A B$ surface will be reflected making an angle $45^{\circ}$.The angle of incidence on $AC$ surface is also $45^{\circ}$; so, the ray will be reflected making an angle $45^{\circ}$.The ray is incident normally on the surface BC. So, there will be no deviation due to refraction.(ii) If $\mu = 1.4$, then the critical angle $=$ $\sin^{-1}\left(\frac{1}{1.4}\right) = 45.23^{\circ}$.So, the ray will be refracted out from the $A B$ face. Angle of incidence $= 45^{\circ}$ $\therefore \mu = \frac{\sin i}{\sin r}$ $\text{ Or, } \frac{1}{1.4} = \frac{\sin 45^{\circ}}{\sin r}$ $\text{ Or, } \sin r = 1.4 \times \sin 45^{\circ} = 0.99$ $\therefore r = \text{ angle of refraction } = 81.9^{\circ}$ So, the ray will be refracted out making an angle of refraction $81.9^{\circ}$. OR (a) Consider two thin lenses $L_1$ and $L_2$ of focal length $f_1$ and $f_2$ are placed coaxially in contact with each other. The lenses are so thin that their optical centers are assumed to coincide at point $P$. An object is placed at $O$ on the common principal axis. The lens $L_1$ produces an image at $I_1$ and this image acts as the object for the second lens $L_2$. The final image is produced at $l$ as shown in figure.$PO = u$, object distance for the first lens $(L_1)$,$PI = v$, final image distance and$P I_1 = v_1$, image distance for the first lens $(L_1) =$ object distance for second lens $(L_2)$.For the image $I_1$ produced by the first lens $L_1$ $\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1}$ ......(1) For the final image I, produced by the second lens $L_2$, $\frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2}$ .......(2) Adding equations (1) and (2), $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2}$ .......(3) If the combination is replaced by a single lens of focal length $f$ such that it forms the image of $O$ at the same position $I$, then $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ ......(4) Comparing equations (3) and (4), $\frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{f}$ (b) Refractive index of the medium of lens $= 1.5$Power of the lens $= -10 \mathrm{D}$Focal length of the lens $= f_{\text{air }} = \frac{1}{10} = 0.1 \mathrm{m} = 10 \mathrm{cm}$In liquid, its focal length $= f_{\text{liquid }} = -50 \mathrm{cm}$According to lens makers' formula $\frac{1}{f_{\text{air }}} = \left[\frac{n_2}{n_1} - 1\right] \left[\frac{1}{R_1} - \frac{1}{R_2}\right]$Or, $\frac{1}{10} = \left[\frac{1.5}{1} - 1\right] \left[\frac{1}{R_1} - \frac{1}{R_2}\right]$ $\therefore \frac{1}{10} = 0.5 \times \left[\frac{1}{R_1} - \frac{1}{R_2}\right]$ .......(1)In the liquid, $\frac{1}{f_{\text{liquid }}} = \left[\frac{n_2}{n_1} - 1\right] \left[\frac{1}{R_1} - \frac{1}{R_2}\right]$Or, $\frac{1}{-50} = \left[\frac{1.5}{n_1} - 1\right] \left[\frac{1}{R_1} - \frac{1}{R_2}\right]$ .......(2) Dividing eqn (1) by eqn (2), $-\frac{\frac{1}{10}}{\frac{1}{50}} = \frac{0.5}{\frac{1.5}{n_1} - 1}$Or, $\frac{1.5}{n_1} - 1 = \frac{1}{10}$$\therefore n_1 =$ Refractive index of the liquid medium $= 1.36$