Hint: Use Fermat's principle.

Step 1: Find the condition that all paraxial rays starting from the source will converge at a point O on the axis.

(i) The time elapsed to travel from S to $P_1$ is
                         ${\mathrm{t}}_1=\frac{{\mathrm{SP}}_1}{\mathrm{c}}=\frac{\sqrt{{\mathrm{u}}^2+{\mathrm{b}}^2}}{\mathrm{c}}$
or                     ${\mathrm{t}}_1=\frac{\mathrm{u}}{\mathrm{c}}\left(1+\frac{1}{2}\frac{{\mathrm{b}}^2}{{\mathrm{u}}^2}\right) \mathrm{assuming} \mathrm{b}<<{\mathrm{u}}_0.$
The time required to travel from ${\mathrm{P}}_1 \mathrm{to} \mathrm{O}$ is,
                                                          ${\mathrm{t}}_2=\frac{{\mathrm{P}}_1\mathrm{O}}{\mathrm{c}}=\frac{\sqrt{{\mathrm{v}}^2+{\mathrm{b}}^2}}{\mathrm{c}}=\frac{\mathrm{v}}{\mathrm{c}} \left(1+\frac{1}{2}\frac{{\mathrm{b}}^2}{{\mathrm{v}}^2}\right)$
The time required to travel through the lens is
                                                            ${\mathrm{t}}_1=\frac{(\mathrm{n}-1)\mathrm{w}\left(\mathrm{b}\right)}{\mathrm{c}}$
where n is the refractive index.
Thus, the total time is
                                             $\mathrm{t}=\frac{1}{\mathrm{c}}\mathrm{u}+\mathrm{v}+\frac{1}{2}{\mathrm{b}}^2 \left(\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}\right)+(\mathrm{n}-1) \mathrm{w}\left(\mathrm{b}\right)$
Put                                         $\frac{1}{\mathrm{D}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$
Then,                                     $\mathrm{t}=\frac{1}{\mathrm{c}}\left(\mathrm{u}+\mathrm{v}+\frac{1}{2}\frac{{\mathrm{b}}^2}{\mathrm{D}}+(\mathrm{n}-1)\left({\mathrm{w}}_0+\frac{{\mathrm{b}}^2}{\mathrm{\alpha }}\right)\right)$
Fermat's principle gives the time taken should be minimum.
For that first derivation should be zero,
                                                       $\frac{\mathrm{dt}}{\mathrm{db}}=0=\frac{\mathrm{b}}{\mathrm{CD}}-\frac{2(\mathrm{n}-1)\mathrm{b}}{\mathrm{c\alpha }}$
$\mathrm{\alpha }=2(\mathrm{n}-1) \mathrm{D}$
Thus, a convergent lens is formed if $\alpha =2(n-1)D$. This is independent of B and hence, all paraxial rays from S will converge at O i.e., for rays
and                                             (b<<v).
Since,                                         $\frac{1}{\mathrm{D}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$, the focal length is D.

Step 2: Find the angular radius of the ring image.

(ii) In this case, differentiating expression of time taken t w.r.t. b;
                                      $\mathrm{t}=\frac{1}{\mathrm{c}}\left(\mathrm{u}+\mathrm{v}+\frac{1}{2}\frac{{\mathrm{b}}^2}{\mathrm{D}}+(\mathrm{n}-1){\mathrm{k}}_1\ln \left(\frac{{\mathrm{k}}_2}{\mathrm{b}}\right)\right)$
$\frac{\mathrm{dt}}{\mathrm{db}}=0=\frac{\mathrm{b}}{\mathrm{D}}-(\mathrm{n}-1)\frac{{\mathrm{k}}_1}{\mathrm{b}}$
$\Rightarrow$                                  ${\mathrm{b}}^2=(\mathrm{n}-1){\mathrm{k}}_1\mathrm{D}$
$\therefore$                                   $\mathrm{b}=\sqrt{(\mathrm{n}-1){\mathrm{k}}_1\mathrm{D}}$
Thus, all rays passing at a height b shall  contribute to the image, The ray paths make an angle.
                       $\mathrm{\beta } =\frac{\mathrm{b}}{\mathrm{v}}= \frac{\sqrt{(\mathrm{n}-1){\mathrm{k}}_1{\displaystyle \mathrm{D}}}}{{\mathrm{v}}^2}=\sqrt{\frac{(\mathrm{n}-1){\mathrm{k}}_1\mathrm{uv}}{{\mathrm{v}}^2(\mathrm{u}+\mathrm{v})}}=\sqrt{\frac{(\mathrm{n}-1){\mathrm{k}}_1\mathrm{u}}{(\mathrm{u}+\mathrm{v})\mathrm{v}}}$
This is the required expression.