A gravitational lens may be assumed to have a varying width of the form
$\text{w}(\text{b})=\text{k}_1\text{ In }\Big(\frac{\text{k}_2}{\text{b}}\Big)\ \ \text{b}_\text{min}<\text{b}<\text{b}_\text{max}$
$=\text{k}_1\text{ In }\Big(\frac{\text{k}_2}{\text{b}_\text{min}}\Big)\ \ \text{b}<\text{b}_\text{min}$
Show that an observer will see an image of a point object as a ring about the center of the lens with an angular radius
$\beta=\sqrt{\frac{(\text{n}-1)\text{k}_1\frac{\text{u}}{\text{v}}}{\text{u}+\text{v}}}$
$\text{w}(\text{b})=\text{k}_1\text{ In }\Big(\frac{\text{k}_2}{\text{b}}\Big)\ \ \text{b}_\text{min}<\text{b}<\text{b}_\text{max}$
$=\text{k}_1\text{ In }\Big(\frac{\text{k}_2}{\text{b}_\text{min}}\Big)\ \ \text{b}<\text{b}_\text{min}$
Show that an observer will see an image of a point object as a ring about the center of the lens with an angular radius
$\beta=\sqrt{\frac{(\text{n}-1)\text{k}_1\frac{\text{u}}{\text{v}}}{\text{u}+\text{v}}}$
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It this case, differentiating expression of time taken t w.r.t.b.
$\text{t}=\frac{1}{\text{c}}\bigg(\text{u}+\text{v}+\frac{1}{2}\frac{\text{b}^2}{\text{D}}+(\text{n}-1)\text{k}_1\text{ In }\Big(\frac{\text{k}_2}{\text{b}}\Big)\bigg)$
$\frac{\text{dt}}{\text{db}}=0=\frac{\text{b}}{\text{D}}-(\text{n}-1)\frac{\text{k}_1}{\text{b}}$
$\Rightarrow\ \text{b}^2=(\text{n}-1)\text{k}_1\text{D}$
$\therefore\ \text{b}=\sqrt{(\text{n}-1)\text{k}_1\text{D}}$
Thus, all rays passing at a height b shall contribute to the image. They ray paths make an angle.
$\beta=\frac{\text{b}}{\text{v}}=\frac{\sqrt{(\text{n}-1)\text{k}_1\text{D}}}{\text{v}}=\sqrt{\frac{(\text{n}-1)\text{k}_1\text{uv}}{\text{v}^2(\text{u}+\text{v})}}=\sqrt{\frac{(\text{n}-1)\text{k}_1\text{u}}{(\text{u}+\text{v})\text{v}}}$. This is the required expression.
$\text{t}=\frac{1}{\text{c}}\bigg(\text{u}+\text{v}+\frac{1}{2}\frac{\text{b}^2}{\text{D}}+(\text{n}-1)\text{k}_1\text{ In }\Big(\frac{\text{k}_2}{\text{b}}\Big)\bigg)$
$\frac{\text{dt}}{\text{db}}=0=\frac{\text{b}}{\text{D}}-(\text{n}-1)\frac{\text{k}_1}{\text{b}}$
$\Rightarrow\ \text{b}^2=(\text{n}-1)\text{k}_1\text{D}$
$\therefore\ \text{b}=\sqrt{(\text{n}-1)\text{k}_1\text{D}}$
Thus, all rays passing at a height b shall contribute to the image. They ray paths make an angle.
$\beta=\frac{\text{b}}{\text{v}}=\frac{\sqrt{(\text{n}-1)\text{k}_1\text{D}}}{\text{v}}=\sqrt{\frac{(\text{n}-1)\text{k}_1\text{uv}}{\text{v}^2(\text{u}+\text{v})}}=\sqrt{\frac{(\text{n}-1)\text{k}_1\text{u}}{(\text{u}+\text{v})\text{v}}}$. This is the required expression.
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