Consider a thin lens placed between a source $(S)$ and an observer $(O)$ (See figure). Let the thickness of the lens vary as $w(b) = w_0 - \alpha b^2$, where $b$ is the vertical distance from the pole. $w_0$ and $\alpha$ are constants. 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 $O$ on the axis. Find the focal length.
$(ii)$ $A$ gravitational lens may be assumed to have a varying width of the form $W = K_1 \log \left( \frac{K_2}{b} \right)$ (where $b_{\min} < b < b_{\max}$). 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 $\beta = \sqrt{\frac{(n - 1)K_1 u}{v(u + v)}}$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(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 $\beta \approx \frac{b}{v} = \sqrt{\frac{(n-1)K_1 u}{v(u+v)}}$.

Explore More

Similar Questions

Derive the thin lens equation.

In the displacement method,we use a lens of focal length $f$ and the distance between the object and the screen is $60 \, cm$. $A$ possible value for the focal length is ........ $cm$.

An object is placed at a point distant $x$ from the focus of a convex lens and its image is formed at $I$ as shown in the figure. The distances $x$ and $x'$ satisfy the relation:

$A$ lens behaves as a converging lens in air and a diverging lens in water. The refractive index of the material of the lens is:

The focal length of a convex lens is $f$. An object is placed at a distance $x$ from its first focal point. The ratio of the size of the real image to that of the object is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo