A lens of diameter `5.0cm` and focal length `f = 25.0 cm` was cut along the diameter into two identical halves. In the process, the layer of the lens `alpha = 1.00 mm` in thieckness was lost. Then the halves were put together to form a composite lens. In this focal plane a narrow slit was placed, emitting monochromatic light with wavelength `lambda = 0.60 mu m`. Behind the lens a screen was located at a distance `b = 50 cm` from it. Find:
(a) the which of a fringe on the screen and the number of possible maxima,
(b) the maximum width of the slit `del_(max)` at which the fringes on the screen will be still observed sufficiently sharp.

Explanation

(a) We show the upper half o fthe lens. The emergent light is at an angle `(a)/(2f)` from the axis.
Thus the divergence angle of the two incident light beams is
`psi = (a)/(f)`
When they interfere the fringes produced have a width
`Delta x = (lambda)/(psi) = (f lambda)/(a) = 0.15 mm`.
The patch on the screen illuminated by both light has s width `b psi` and this contains
`(b psi)/(Delta x) = (ba^(2))/(f^(2) lambda)` fringes `= 13` fringes
(if we ignore `1` in comparies on to `(b psi)/(Delta x)` (if 5.71(a))
(b) We follow the logic of. From one edge of the slit to the edge the distance is of magnitude `del` (i.e `(a)/(2)` to `(a)/(2) + del)`.
If we imagine teh edge to shift by this distance, the angle `psi//2` will increase by `(Delta psi)/(2) = (del)/(2f)`
and the light will shift `+- b(del)/(2f)`
The fringe patten will therefore shift by `(del.b)/(f)`
Equating this to `(deltax)/(2) = (f lambda)/(2a)` we get `del_(max) = (f^(2) lambda)/(2ab) = 37.5 mum `.