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.

Let $d$ be the diameter of the disc. The spot shall be invisible if the incident rays from the dot at $O$ to the surface at $d / 2$ at the critical angle.

Let $i$ be the angle of incidence.

Using relationship between refractive index and critical angle, then,

$$\sin t=\frac{1}{\propto}$$

Using geometry and trigonometry.

Now,

$$\begin{aligned} \frac{d / 2}{h} & =\tan i \\ \Rightarrow\quad\frac{d}{2} & =h \tan i=h\left[\sqrt{\alpha^2-1}\right]^{-1} \\ \therefore\quad d & =\frac{2 h}{\sqrt{\alpha^2-1}} \end{aligned}$$

This is the required expression of $d$.

(i) Let the power at the far point be $P_f$ for the normal relaxed eye of an average person. The required power

$$P_f=\frac{1}{f}=\frac{1}{0.1}+\frac{1}{0.02}=60 \mathrm{D}$$

By the corrective lens the object distance at the far point is $\infty$.

The power required is

$$P_f^{\prime}=\frac{1}{f^{\prime}}=\frac{1}{\infty}+\frac{1}{0.02}=50 \mathrm{D}$$

So for eye + lens system,

we have the sum of the eye and that of the glasses $P_g$

$$\begin{array}{lr} \therefore & P_f^{\prime}=P_f+P_g \\ \therefore & P_g=-10 \mathrm{D} \end{array}$$

(ii) His power of accomodatlon is 4 D for the normal eye. Let the power of the normal eye for near vision be $P_n$.

Then,

$$4=P_n-P_f \text { or } P_n=64 \mathrm{D}$$

Let his near point be $x_n$, then

$$\begin{aligned} \frac{1}{x_n}+\frac{1}{0.02} & =64 \text { or } \frac{1}{x_n}+50=64 \\ \frac{1}{x_n} & =14 \end{aligned}$$

$$\therefore \quad x_n=\frac{1}{14} ; 0.07 \mathrm{~m}$$

(iii) With glasses $P_n^{\prime}=P_f^{\prime}+4=54$

$$\begin{aligned} & 54=\frac{1}{x_n^{\prime}}+\frac{1}{0.02}=\frac{1}{x_n^{\prime}}+50 \\ & \frac{1}{x_n^{\prime}}=4 \\ \therefore\quad & x_n^{\prime}=\frac{1}{4}=0.25 \mathrm{~m} \end{aligned}$$