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 $P_n^{\prime}=P_f^{\prime}+4=54$

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

Any ray entering at an angle $i$ shall be guided along $A C$ if the angle ray makes with the face $A C(\phi)$ is greater than the critical angle as per the principle of total internal reflection $\phi+r=902$, therefore $\sin \phi=\cos r$

$$\begin{array}{lc} \Rightarrow & \sin \phi \geq \frac{1}{\alpha} \\ \Rightarrow & \cos r \geq \frac{1}{\alpha} \\ \text { or } & 1-\cos ^2 r \leq 1-\frac{1}{\alpha^2} \\ \text { i.e., } & \sin ^2 r \leq \frac{1}{\alpha^2} \\ \text { i.e., } & \sin ^2 r \leq 1-\frac{1}{\alpha^2} \end{array}$$

$$\begin{aligned} \text{since,}\quad\sin i & =\alpha \sin r \\ \frac{1}{\alpha_2} \sin ^2 i & \leq 1-\frac{1}{\alpha^2} \text { or } \sin ^2 i \leq \alpha^2-1 \end{aligned}$$

when $$i=\frac{\pi}{2}$$

Then, we have smallest angle $\phi$.

If that is greater than the critical angle, then all other angles of incidence shall be more than the critical angle.

Thus,

$$\begin{aligned} 1 & \leq \alpha^2-1 \\ \text{or}\quad\alpha^2 & \geq 2 \\ \Rightarrow\quad\alpha & \geq \sqrt{2} \end{aligned}$$

This is the required result.

Let us consider a portion of a ray between $x$ and $x+d x$ inside the liquid. Let the angle of incidence at $x$ be $\theta$ and let it enter the thin column at height $y$. Because of the bending it shall emerge at $x+d x$ with an angle $\theta+d \theta$ and at a height $y+d y$. From Snell's law,

$\alpha(y) \sin \theta=\alpha(y+d y) \sin (\theta+d \theta)$

$$\begin{gathered} \text{or}\quad\alpha(y) \sin \theta ;\left(\alpha(y)+\frac{d \alpha}{d y} d y\right)(\sin \theta \cos d \theta+\cos \theta \sin d \theta) \\ \text{or}\quad\alpha(y) \sin \theta+\alpha(y) \cos \theta d \theta+\frac{d \alpha}{d y} d y \sin \theta \\ \text{or}\quad\alpha(y) \cos \theta d \theta ; \frac{-d \alpha}{d y} d y \sin \theta \end{gathered}$$

$$\begin{gathered} d \theta ; \frac{-d \propto}{\alpha d y} d y \tan \theta \\ \text{But}\quad\tan \theta=\frac{d x}{d y}\text{(from the figure)} \end{gathered}$$

On solving, we have

$$\therefore \quad d \theta=\frac{-1 d \alpha}{\alpha d y} d x$$

Solving this variable separable form of differential equation.

$$\therefore \quad \theta=\frac{-1 d \alpha}{\alpha d y} \int_0^d d x=\frac{-1 d \propto}{\alpha d y} d$$