If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refractive index of the medium given by
$$n(r)=1+2 G M / r c^2$$
where $r$ is the distance of the point of consideration from the centre of the mass of the massive body, $G$ is the universal gravitational constant, $M$ the mass of the body and $c$ the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object.
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.
An infinitely long cylinder of radius $R$ is made of an unusual exotic material with refractive index-1 (figure). The cylinder is placed between two planes whose normals are along the $y$-direction. The centre of the cylinder 0 lies along the $y$-axis. A narrow laser beam is directed along the $y$-direction from the lower plate. The laser source is at a horizontal distance $x$ from the diameter in the $y$ direction. Find the range of $x$ such that light emitted from the lower plane does not reach the upper plane.
Since, the material is of refractive index $-1, \theta_r$, is negative and $\theta_r^{\prime}$ positive.
Now, $$\left|\theta_t\right|=\left|\theta_r\right|=\left|\theta_r^{\prime}\right|$$
The total deviation of the outcoming ray from the incoming ray is $4 \theta_t$. Rays shall not reach the recieving plate if
$$\frac{\pi}{2} \leq 4 \theta_t \leq \frac{3 \pi}{2}$$ [angles measured clockwise from the $y$-axis]
On solving, $$\frac{\pi}{8} \leq \theta_t \leq \frac{3 \pi}{8}$$
$$\begin{aligned} \text{Now,}\quad\sin \theta_t & =\frac{x}{R} \\ \frac{\pi}{8} & \leq \sin ^{-1} \frac{x}{R} \leq \frac{3 \pi}{8} \\ \text{or}\quad\frac{\pi}{8} & \leq \frac{x}{R} \leq \frac{3 \pi}{8} \end{aligned}$$
Thus, for light emitted from the source shall not reach the receiving plate. If $\frac{R \pi}{8} \leq x \leq \frac{R 3 \pi}{8}$.
(i) Consider a thin lens placed between a source $(S)$ and an observer $(0)$ (Figure). Let the thickness of the lens vary as $w(b)=w_0-\frac{b^2}{\alpha}$, where $b$ is the verticle distance from the pole, $w_0$ is a constant. Using Fermat's principle i.e., the time of transit for a ray between the source and observer is an extremum find the condition that all paraxial rays starting from the source will converge at a point 0 on the axis. Find the focal length.
(ii) A gravitational lens may be assumed to have a varying width of the form
$$\begin{aligned} w(b) & =k_1 \ln \left(\frac{k_2}{b}\right) b_{\min }< b < b_{\max } \\ & =k_1 \ln \left(\frac{k_2}{b_{\min }}\right) b< b_{\min } \end{aligned}$$
Show that an observer will see an image of a point object as a ring about the centre of the lens with an angular radius
$$\beta=\sqrt{\frac{(n-1) k_1 \frac{u}{v}}{u+v}}$$
(i) The time elapsed to travel from $S$ to $P_1$ is
$$t_1=\frac{S P_1}{c}=\frac{\sqrt{u^2+b^2}}{c}$$
or $$\frac{u}{c}\left(1+\frac{1}{2} \frac{b^2}{u^2}\right) \text { assuming } b < < u_0$$
The time required to travel from $P_1$ to $O$ is
$$t_2=\frac{P_1 O}{c}=\frac{\sqrt{v^2+b^2}}{c} ; \frac{v}{c}\left(1+\frac{1}{2} \frac{b^2}{v^2}\right)$$
The time required to travel through the lens is
$$t_1=\frac{(n-1) w(b)}{c}$$
where $n$ is the refractive index.
Thus, the total time is
$$t=\frac{1}{c} u+v+\frac{1}{2} b^2\left(\frac{1}{u}+\frac{1}{v}\right)+(n-1) w(b)$$
Put $$\frac{1}{D}=\frac{1}{u}+\frac{1}{v}$$
Then, $$t=\frac{1}{c}\left(u+v+\frac{1}{2} \frac{b^2}{D}+(n-1)\left(w_0+\frac{b^2}{\alpha}\right)\right)$$
Fermet's principle gives the time taken should be minimum.
For that first derivative should be zero.
$$\begin{aligned} \frac{d t}{d b} & =0=\frac{b}{C D}-\frac{2(n-1) b}{C \alpha} \\ \alpha & =2(n-1) D \end{aligned}$$
Thus, a convergent lens is formed if $\alpha=2(n-1) D$. This is independant of and hence, all paraxial rays from $S$ will converge at $O$ i.e., for rays
and $$(b < < v )$$
Since, $\frac{1}{D}=\frac{1}{u}+\frac{1}{v}$, the focal length is $D$.
(ii) In this case, differentiating expression of time taken $t$ w.r.t. $b$
$$\begin{aligned} t & =\frac{1}{c}\left(u+v+\frac{1}{2} \frac{b^2}{D}+(n-1) k_1 \ln \left(\frac{k_2}{b}\right)\right) \\ \frac{d t}{d b} & =0=\frac{b}{D}-(n-1) \frac{k_1}{b} \\ \Rightarrow\quad b^2 & =(n-1) k_1 D \\ \therefore\quad b & =\sqrt{(n-1) k_1 D} \end{aligned}$$
Thus, all rays passing at a height $b$ shall contribute to the image. The ray paths make an angle.
$$\beta ; \frac{b}{v}=\frac{\sqrt{(n-1) k_1 D}}{v^2}=\sqrt{\frac{(n-1) k_1 u v}{v^2(u+v)}}=\sqrt{\frac{(n-1) k_1 u}{(u+v) v}}$$
This is the required expression.