The relationship between refractive index, prism angle $A$ and angle of minimum deviation is given by

$$\propto=\frac{\sin \left[\frac{\left(A+D_m\right)}{2}\right]}{\sin \left(\frac{A}{2}\right)}$$

Here,

$\therefore$ Given, $$D_m=A$$

Substituting the value, we have

$$\begin{aligned} & \therefore \quad \alpha=\frac{\sin A}{\sin \frac{A}{2}} \\ & \text { On solving, we have } \quad =\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2} \end{aligned}$$

$$\propto=\frac{\sin A}{\sin \frac{A}{2}}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2}$$

For the given value of refractive index, we have

$$\begin{array}{llr} \therefore & \cos \frac{A}{2}=\frac{\sqrt{3}}{2} \\ \text { or } & \frac{A}{2} =30 \Upsilon \\ \therefore & A =60 \Upsilon \end{array}$$

This is the required value of prism angle.

Since, the object distance is $u$. Let us consider the two ends of the object be at distance $u_1=u-L / 2$ and $u_2=u+L / 2$, respectively so that $\left|u_1-u_2\right|=L$. Let the image of the two ends be formed at $v_1$ and $v_2$, respectively so that the image length would be

$L^{\prime}=\left|v_1-v_2\right|\quad\text{.... (i)}$

Applying mirror formula, we have

$$\frac{1}{u}+\frac{1}{v}=\frac{1}{f} \text { or } v=\frac{f u}{u-f}$$

On solving, the positions of two images are given by

$$v_1=\frac{f(u-L / 2)}{u-f-L / 2}, v_2=\frac{f(u+L / 2)}{u-f+L / 2}$$

For length, substituting the value in (i), we have

$$L^{\prime}=\left|v_1-v_2\right|=\frac{f^2 L}{(u-f)^2 \times L^2 / 4}$$

Since, the object is short and kept away from focus, we have

$$\begin{gathered} L^2 / 4<<(u-f)^2 \\ \text{Hence, finally}\quad L^{\prime}=\frac{f^2}{(u-f)^2} L \end{gathered}$$

This is the required expression of length of image.

Refering to the figure, $A M$ is the direction of incidence ray before liquid is filled. After liquid is filled in , $B M$ is the direction of the incident ray. Refracted ray in both cases is same as that along $A M$.

Let the disc is separated by $O$ at a distance $d$ as shown in figure. Also, considering angle

$$N=90 \Upsilon, O M=a, C B=N B=a-R, A N=a+R$$

Here, in figure

$$\begin{aligned} \sin t & =\frac{a-R}{\sqrt{d^2+(a-R)^2}} \\ \text{and}\quad\sin \alpha & =\cos (90-\alpha)=\frac{a+R}{\sqrt{d^2+(a+R)^2}} \end{aligned}$$

But on applying Snell's law,

$$\frac{1}{\alpha}=\frac{\sin t}{\sin r}=\frac{\sin t}{\sin \alpha}$$

On substituting the values, we have the separation

$$d=\frac{\alpha\left(a^2-b^2\right)}{\sqrt{(a+r)^2-\alpha(a-r)^2}}$$

This is the required expression.

If there was no cut, then the object would have been at a height of 0.5 cm from the principal axis OO'.

Applying lens formula, we have

$$\begin{aligned} \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \\ \frac{1}{v} & =\frac{1}{u}+\frac{1}{f}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50} \\ v & =50 \mathrm{~cm} \\ \text { Mangnification is } m & =\frac{v}{u}=-\frac{50}{50}=-1 \end{aligned}$$

Thus, the image would have been formed at 50 cm from the pole and 0.5 cm below the principal axis. Hence, with respect to the $X$-axis passing through the edge of the cut lens, the coordinates of the image are $(50 \mathrm{~cm},-1 \mathrm{~cm})$.

Principal of reversibility is states that the position of object and image are interchangeable. So, by the versibility of $u$ and $v$, as seen from the formula for lens.

$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$

It is clear that there are two positions for which there shall be an image.

On the screen, let the first position be when the lens is at $O$. Finding $u$ and $v$ and substituting in lens formula.

$$\begin{aligned} \text{Given,}\quad -u+v & =D \\ \Rightarrow\quad u & =-(D-v) \end{aligned}$$

Placing it in the lens formula

$$\frac{1}{D-v}+\frac{1}{v}=\frac{1}{f}$$

On solving, we have

$$\begin{aligned} & \Rightarrow \quad \frac{v+D-v}{(D-v) v}=\frac{1}{f} \\ & \Rightarrow \quad v^2-D v+D f=0 \\ & \Rightarrow \quad v=\frac{D}{2} \pm \frac{\sqrt{D^2-4 D f}}{2} \end{aligned}$$

$$\begin{aligned} &\text { Hence, finding } u\\ &u=-(D-v)=-\left(\frac{D}{2} \pm \frac{\sqrt{D^2-4 D f}}{2}\right) \end{aligned}$$

$$\begin{aligned} \text{When, the object distance is}\quad & \frac{D}{2}+\frac{\sqrt{D^2-4 D f}}{2} \\ \text{the image forms at}\quad& \frac{D}{2}-\frac{\sqrt{D^2-4 D f}}{2} \end{aligned}$$

Similarly, when the object distance is

$$\begin{aligned} & \frac{D}{2}-\frac{\sqrt{D^2-4 D f}}{2} \\ \text{The image forms at}\quad & \frac{D}{2}+\frac{\sqrt{D^2-4 D f}}{2} \end{aligned}$$

The distance between the poles for these two object distance is

$$\frac{D}{2}+\frac{\sqrt{D^2-4 D f}}{2}-\left(\frac{D}{2}-\frac{\sqrt{D^2-4 D f}}{2}\right)=\sqrt{D^2-4 D f}$$

Let $$d=\sqrt{D^2-4 D f}$$

If $u=\frac{D}{2}+\frac{d}{2}$, then the image is at $v=\frac{D}{2}-\frac{d}{2}$.

$\therefore \quad$ The magnification $m_1=\frac{D-d}{D+d}$

If $u=\frac{D-d}{2}$, then $v=\frac{D+d}{2}$

$\therefore$ The magnification $m_2=\frac{D+d}{D-d}$

Thus, $$\frac{m_2}{m_1}=\left(\frac{D+d}{D-d}\right)^2$$

This is the required expression of magnification.