A point source S emitting light of wavelength 600 nm is placed at a very small height h above a flat figure. The intensity of the reflected light id 36% of the incident intensity. Interference fringe are observed on a screen placed parallel to the reflecting surface at a very large distance D from it.
a. What is the shape of the interference fringe on the screen?
b. Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near point P (shown in the figure).
c. If the intensity at point P corresponds to a maximum, calculate the mininum distance through which the reflecting surface. AB should at P again becomes maxinum.

Answer

(i). Circular
(ii). `(1)/(16)`
(iii). 300 nm.

Explanation

(i). S is a point source, fringes formed will be circular
(ii). Ratio of minimum and maximum intensities intensity of light direct from source `I_(1)=I_(0)`(say)
Intensity after reflection `I_(2)=0.36I_(0)`
`therefore(I_(min))/(I_(max))=((sqrt(I_(1))sqrt(I_(2)))/(sqrt(1))+sqrt(I_(2)))^(2)=((0.4)/(1.6))^(2)=(1)/(16)`
(iii). Shift of AB for same intensity
if intensity at P correspionds to maximum it means that constructive interference occurs at P.
`therefore` Path difference between direct waves from S and reflected waves, from reflector AB, is `nlamda`
Let AB is shifted by x (towards P or away from P)
`therefore` Additional path difference introduced =2x
For minimum value of `x_(1),n=1`
`therefore` Path difference `=1xxlamda=600nm`
`2x=600nmimpliesx=300nm`