(N/A) $(i)$ Let $n$ be the refractive index of the lens material. The time taken by light to travel from $S$ to $O$ through a point at distance $b$ from the axis is $T = \frac{SP_1}{c} + \frac{(n-1)w(b)}{c} + \frac{P_1O}{c}$.
Using the paraxial approximation: $SP_1 \approx u + \frac{b^2}{2u}$ and $P_1O \approx v + \frac{b^2}{2v}$.
$T = \frac{1}{c} \left[ u + \frac{b^2}{2u} + (n-1)(w_0 - \alpha b^2) + v + \frac{b^2}{2v} \right]$.
For $T$ to be an extremum, $\frac{dT}{db} = 0 \implies \frac{b}{u} + \frac{b}{v} - 2(n-1)\alpha b = 0$.
Since this must hold for all $b$, we have $\frac{1}{u} + \frac{1}{v} = 2(n-1)\alpha$. Comparing with the lens formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, the focal length is $f = \frac{1}{2(n-1)\alpha}$.
$(ii)$ For the gravitational lens, the optical path length is $L = \sqrt{u^2+b^2} + \sqrt{v^2+b^2} + (n-1)W(b)$.
Using paraxial approximation: $L \approx u + v + \frac{b^2}{2u} + \frac{b^2}{2v} + (n-1)K_1 \log \left( \frac{K_2}{b} \right)$.
Extremum condition $\frac{dL}{db} = 0 \implies \frac{b}{u} + \frac{b}{v} + (n-1)K_1 \left( -\frac{1}{b} \right) = 0$.
$b^2 \left( \frac{u+v}{uv} \right) = (n-1)K_1 \implies b = \sqrt{\frac{(n-1)K_1 uv}{u+v}}$.
The angular radius is $\beta \approx \frac{b}{v} = \sqrt{\frac{(n-1)K_1 u}{v(u+v)}}$.