Question

(i) Consider a thin lens placed between a source (S) and an observer (O) (Fig.). Let the thickness of the lens vary as w(b) = w_o − \(\frac{b^2}{α}\), where b is the verticle distance from the pole. w_o is a constant. Using Fermat’s principle i.e., the time of transit for 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(b) = k₁ ln(k₂/b) b_min<b <b_max

= k₁ ln(\(\frac{k_2}{b_{min}}\)) b<b_min

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 β =

Answer 1

(i) The time required to travel from S to P₁ is t₁ = SP₁/c= \(\sqrt {u^2+b^2}\over{c}\); u/c (1 + 1/2 \(\frac{b^2}{u^2}\)) assuming b <u₀

The time required to travel from P1 to O is

The time required to travel through the lens is t1 = \(\frac{(n-1)w(b)}{c}\) where is the refractive index.

Thus, the total time is

Fermet’s principles gives

\(\frac{dt}{db}=∩=\frac{b}{cd}\frac{2(n-1)b}{cα}\)

α = 2(n-1)D

Thus, a convergent lens is formed if α = 2(n – 1) D. This is independent of b and hence all paraxial rays from S will converge at O (i.e. for rays b < < n and b < < v).

Since \(\frac{1}{D}=\frac{1}{u}+\frac{1}{v}\), the focal length is D.

(ii) In this case

Thus, all rays passing at a height b shall contribute to the image. The ray paths make an angle