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.

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

Let us consider two planes at $r$ and $r+d r$. Let the light be incident at an angle $\theta$ at the plane at $r$ and leave $r+d r$ at an angle $\theta+d \theta$. Then from Snell's law,

$$\begin{gathered} n(r) \sin \theta=n(r+d r) \sin (\theta+d \theta) \\ \Rightarrow \quad n(r) \sin \theta ;\left(n(r)+\frac{d n}{d r} d r\right)(\sin \theta \cos d \theta+\cos \theta \sin d \theta) \\ ;\left(n(r)+\frac{d n}{d r} d r\right)(\sin \theta+\cos \theta d \theta) \end{gathered}$$

Ignoring the product of differentials or we have,

$$\begin{aligned} n(r) \sin \theta ; n(r) \sin \theta & +\frac{d n}{d r} d r \sin \theta+n(r) \cos \theta d \theta \\ -\frac{d n}{d r} \tan \theta & =n(r) \frac{d \theta}{d r} \\ \frac{2 G M}{r^2 c^2} \tan \theta & =\left(1+\frac{2 G M}{r c^2}\right) \frac{d \theta}{d r} \approx \frac{d \theta}{d r} \\ \int_0^{\theta_0} d \theta & =\frac{2 G M}{c^2} \int_{-\infty}^{\infty} \frac{\tan \theta d r}{r^2} \end{aligned}$$

Now substitution for integrals, we have

Now, $r^2=x^2+R^2$ and $\tan \theta=\frac{R}{x}$

$$\begin{aligned} 2 r d r & =2 x d x \\ \int_0^{\theta_0} d \theta & =\frac{2 G M}{c^2} \int_{-\infty}^{\infty} \frac{R}{x} \frac{x d x}{\left(x^2+R^2\right)^{\frac{3}{2}}} \end{aligned}$$

$$\begin{aligned} \text{Put}\quad x & =R \tan \phi \\ d x & =R \sec ^2 \phi d \phi \\ \therefore\quad\theta_0 & =\frac{2 G M R}{c^2} \int_{-\pi / 2}^{\pi / 2} \frac{R \sec ^2 \phi d \phi}{R^3 \sec ^3 \phi} \\ & =\frac{2 G M}{R c^2} \int_{-\pi / 2}^{\pi / 2} \cos \phi d \phi=\frac{4 G M}{R c^2} \end{aligned}$$

This is the required proof.