(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_0 - (b^2)/(prop)`, where `b` is the vertical distance from the pole. `w_0` is a constant. Using Fermat's principle, i.e., the time of transit for a ray between the source and observer is an exptremum, find the condition that al 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) = k_1 In ((k_2)/(b)) b_min lt b lt b_(max) w(b) = k_1 In ((k_2)/(b_(min))) b lt b_(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(((n - 1) k_1 (u)/(v))/(u + v))`.
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Explanation

Time required by light to travel in air from `S tp P_(1)` is
`t_(1) = (SP_(1))/(c ) = (sqrt(u^(2) + b^(2)))/(c )`
`= (u)/( c) [1 + (1)/(2)(b^(2))/(u^(2))]`, assuming `b lt lt u`
Similarly, time required by light to travel in air from `P_(1)` to `O , t_(2) = (v)/( c) (1 + (1)/(2)(b^(2))/(v^(2)))`
Now, thickness of lens varies as `w(b) = w_(0) - (b^(2))/(alpha)` ...(i)
`:.` Time required to travel through the lens is `t_(3) = (w(b))/(c//n) - (w(b))/(c ) = ((n - 1)w(b))/(c )`
Here, `n` is refractive index of the material of the lens.
Thus, the total time of travel from S to O is
`t = t_(1) + t_(2) + t_(3) = (u)/(c )(1 + (b^(2))/(2 u^(2))) + (v)/( c)(1 + (b^(2))/(2v^(2))) + ((n + 1) w(b))/(c )`
`t = (1)/(c) [u + v + (1)/(2)b^(2)((1)/(u) + (1)/(v)) + (n - 1)w(b)]`
Let us put `(1)/(u) + (1)/(v) = (1)/(D)` ...(ii)
`t = (1)/(c)[u + v + (b^(2))/(2 D) + (n - 1)(w_(0) - (b^(2))/(alpha))]` ...(iii)
According to Fermat's principle, `t = extermum`. Therefore `(dt)/(db) = 0`.
From (iii), `(dt)/(db) = (b)/(cD) - (2(n - 1)b)/(c alpha) = 0 or alpha = 2 (n - 1)D` ...(iv)
Hence, a convergent lens is formed if `alpha = 2 (n - 1) D`. This is independent of `b`. Hence, all paraxial rays from `S` will convergent at `O`.
From (iii), `(1)/(u) + (1)/(v) = (1)/(D)`. Therefore, focal length of lens `= D`.
(ii) Now, for a gravitational lens, `w(b) = k_(1) log_(e) ((k_(2))/(b)) , b_(min) lt b lt b_(max)`
`w_(b) = K_(1) log_(e) ((k_(2))/(b_(min))) , b lt b_(min)`
Proceeding as in the above case, from equ. (iii)
`t = (1)/( c) [u + v + (1)/(2)(b^(2))/(D) + (n - 1) k_(1) log_(e)((k_(2))/(b))]`
`(dt)/(db) = 0 = (b)/(D) - (n - 1)/(k_(1))/(b)`
`:. b^(2) = (n - 1) k_(1) D`
`b = sqrt((n - 1)k_(1) D)`
Hence, all rays passing through the lens at a height `b` shall form the image.
The paths of rays would make an angle
`beta = (b)/(v) = sqrt(((n - 1) k_(1)D)/(sqrtv_(2))) = sqrt(((n -1)k_(1)D)/(v^(2)(u + v)))` (as `(1)/(D) = (1)/(u) + (1)/(v))`
`beta = sqrt(((n - 1)k_(2)u//v)/((u + v)))`
Hence, the observer will see an image of a point object as a ring about the centre of the lens with angular radius `beta`.