(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.