(a) $\mathit{O}\mathit{P}$ is the incidence ray on the prism and $\mathit{Q}\mathit{R}$ emergent ray.





For minimum deviation, $\angle {\mathit{r}}_1=\angle {\mathit{r}}_2=\angle \mathit{r}$
$\mathit{A}=\angle {\mathit{r}}_1+\angle {\mathit{r}}_2$
So, $\mathit{A}=\angle \mathit{r}+\angle \mathit{r}=\angle 2\mathit{r}$
$\angle \mathit{r}=2\mathit{A}$
Also, $\angle {\mathit{i}}_1+\angle {\mathit{i}}_2=\angle \mathit{i}$
$\mathit{A}+{\mathit{\delta }}_{\mathit{m}}=\angle {\mathit{i}}_1+\angle {\mathit{i}}_2$
So, $\mathit{i}=\frac{\left(\mathit{A}+{\mathit{\delta }}_{\mathit{m}}\right)}{2}$
Now, from snell's law,
$\propto =\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{i}}{\mathit{s}\mathit{i}\mathit{n}\mathit{r}}$
$\mathit{\mu }=\frac{\mathit{s}\mathit{i}\mathit{n}\frac{\mathrm{A}+{\mathit{\delta }}_{\mathit{m}}}{2}}{\mathit{s}\mathit{i}\mathit{n}\frac{\mathrm{A}}{2}}$
(b)(i)The critical angle $=\mathit{s}\mathit{i}\mathit{n}{}^{-1}\frac{1}{1.5}={41.3}^{\circ }$.
So, the ray which is incident on $\mathit{A}\mathit{B}$ surface will be reflected making an angle ${45}^{\circ }$.
The angle of incidence on $\mathrm{AC}$ 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 $\mathit{\mu }=1.4$, then the critical angle $=$ $\mathit{s}\mathit{i}\mathit{n}{}^{-1}\frac{1}{1.4}={45.23}^{\circ }$.
So, the ray will be refracted out from the $\mathit{A}\mathit{B}$ face.
Angle of incidence $={45}^{\circ }$
$\therefore \mathit{\mu }=\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{i}}{\mathit{s}\mathit{i}\mathit{n}\mathit{r}}$



So, the ray will be refracted out making an angle of refraction ${81.9}^{\circ }$.
OR
(a) Consider two thin lenses ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$ of focal length ${\mathit{f}}_1$ and ${\mathit{f}}_2$ are placed coaxially in contact with each other.
The lenses are so thin that their optical centers are assumed to coincide at point $\mathit{P}$.

An object is placed at $\mathrm{O}$ on the common principal axis. The lens ${\mathrm{L}}_1$ produces an image at ${\mathit{I}}_1$ and this image acts as the object for the second lens ${\mathit{L}}_2$. The final image is produced at $\mathit{l}$ as shown in figure.
$\mathrm{PO}=\mathit{u}$, object distance for the first lens $\left({\mathrm{L}}_1\right)$,
$\mathrm{PI}=\mathit{v}$, final image distance and
$\mathit{P}{\mathit{I}}_1={\mathit{v}}_1$, image distance for the first lens $\left({\mathrm{L}}_1\right)=$ object distance for second lens $\left({\mathrm{L}}_2\right)$.
For the image ${\mathit{I}}_1$ produced by the first lens ${\mathit{L}}_1$
$\frac{1}{{\mathit{v}}_1}-\frac{1}{\mathit{u}}=\frac{1}{{\mathit{f}}_1}$ ......(1)
For the final image I, produced by the second lens ${\mathrm{L}}_2$,
$\frac{1}{\mathit{v}}-\frac{1}{{\mathit{v}}_1}=\frac{1}{{\mathit{f}}_2}$ .......(2)
Adding equations (1) and (2),
$\frac{1}{\mathit{v}}-\frac{1}{\mathit{u}}=\frac{1}{{\mathit{f}}_1}+\frac{1}{{\mathit{f}}_2}$ .......(3)
If the combination is replaced by a single lens of focal length $\mathit{f}$ such that it forms the image of $\mathrm{O}$ at the same position $\mathit{I}$, then
$\frac{1}{\mathit{v}}-\frac{1}{\mathit{u}}=\frac{1}{\mathit{f}}$ ......(4)
Comparing equations (3) and (4),
$\frac{1}{{\mathit{f}}_1}+\frac{1}{{\mathit{f}}_2}=\frac{1}{\mathit{f}}$
(b) Refractive index of the medium of lens $=1.5$
Power of the lens $=-10\mathrm{D}$
Focal length of the lens
In liquid, its focal length
According to lens makers' formula

Or, $\frac{1}{10}=\left[\frac{1.5}{1}-1\right]\left[\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right]$
$\therefore \frac{1}{10}=0.5\times \left[\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right]$ .......(1)
In the liquid,

Or, $\frac{1}{-50}=\left[\frac{1.5}{{\mathit{n}}_1}-1\right]\left[\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right]$ .......(2)
Dividing eqn (1) by eqn (2),
$-\frac{\frac{1}{10}}{\frac{1}{50}}=\frac{0.5}{\frac{1.5}{{\mathit{n}}_1}-1}$
Or, $\frac{1.5}{{\mathit{n}}_1}-1=\frac{1}{10}$
$\therefore {\mathit{n}}_1=$ Refractive index of the liquid medium $=1.36$