A plane monochromatic light wave falls normally on an opaque half-plane. A screen is located at a distance `b = 100 cm` behind the half-plane. Making use of the cornu spiral (Fig.) find :
(a) the ratio of intensities of the first maximum and the neighbouring minimum,
(b) the wavelength of light if the first two maxima are separated by a distance `Deltax = 0.63mm`.

Explanation

From th Cornu's spiral, the intensity of the first maximum is given as
`I_(max,1) = 1.37 I_(0)`
and the intensity of the first minimum is given by
`I_(min) = 0.78I_(0)`
so the required ratio is
`(I_(max))/(I_(min)) = 1.76`
(b) The value of the distance `x` is related to the parameter `v` in Fresnel's intergral by
`v = x sqrt((2)/(b lambda))`.
For the first two maxima the distance `x_(1),x_(2)` are related to the parameters `v_(1), v_(2)` by
`x_(1) = sqrt((b lambda)/(2))v_(1), x_(2) = sqrt((b lambda)/(2))v_(2)`
Thus `(v_(2) - v_(1)) sqrt((b lambda)/(2)) = x_(2) - x_(1) = Deltax`
or `lambda = (2)/(b) ((Deltax)/(v_(2) - v_(1)))^(2)`
From the Cornu's spiral the positions of the maxima are
`v_(1) = 1.22, v_(2) = 2.34, v_(3) = 3.08 etc`
Thus `lambda = (2)/(b) ((Deltax)/(1.12))^(2) = 0.63 mu m`.