An opaque ball of diameter `D = 40 mm` is placed between a source of light with wavelength `lambda = 0.55 mu m` and a photographic plate. The distance between the source and the ball is equal to `a = 12 m` and that between the ball and the photographic plate is equal to `b = 18m`. find:
(a) the image dimension `y'` on the plate if the transverse dimension of the source is `y = 6.0 mm`,
(b) the minimum height of irregularities, covering the surface of the ball at random, at which the ball obstructs light.
Note: As calculations and experience shown, that happens when the height of irregularities is comparable with the width of the Fresnel zone along which the edge of an opaque screen passes.

Explanation

If a point source is placed before an opaque ball, the diffraction pattern consists of a bright spot inside a dark disc followed by fringes. The bright spot is on the line joining the point source and the cnetre of the ball. When the object is a finite source of transverse diamension `y`, every point of teh source has its corresponding image on the line joining that point and the centre of the ball. Thus the trasverse dimension of the image is given by
`y' =(b)/(a)y = 9mm`.
(b) The minimum height of the irregularities covering the surface of the ball at random, at which the ball obstructs light is according to the note at the end of the problem, compareble with the width of the Fresnel zone along which the edge of opaque screen passes.
So `h_(min) ~~ Delta r`
To find `Delta r` we note that
`r^(2) = (k lambda ab)/(a + b)`
or `2r Delta r = D Delta r = (lambda ab)/(a+b) Deltak`
Where `D =` diameter of disc `(=`diameter of the last Fresnel zone) and `Delta k = 1`
Thus `h_(min) = (lambda ab)/(D(a+b)) = 0.099 mm`.