A point sources `S` emitting light of wavelength `600nm` is placed at a very small height `h` above the flat reflecting surface `AB`(see figure).The intensity of the reflected light is`36%` of the intensity.interference firnges are observed on a screen placed parallel to the reflecting surface a very large distance `D` from it.
(A)What is the shape of the interference fringes on the screen?

(B)Calculate the ratio of the minimum to the maximum to the maximum intensities in the interference fringes fromed near the point `P` (shown in the figure) (c) if the intenstities at point `P` corresponds to a maximum,calculate the minimum distance through which the reflecting surface `AB` should be shifted so that the intensity at `P` again becomes maximum.

Answer

(a) circular (b)`(1)/(16)` (C )=`3000Å`

Explanation

(A)circualr

(b)`(I_(min))/(I_(max))=(sqrt(I_(1))-sqrt(I_(2)))^(2)/(sqrt(I_(1))+sqrt(I_(2)))^(2)`
`I_(1)`=intensity of light from source `rArr I_(2)`=intensity of reflected light `rArr I_(2)=0.36 I_(1)`
`I_min/I_max=((sqrtI_1-sqrt(0.36I_1))/(sqrtI_1+sqrt(0.36I_1)))^2implies I_min/I_max=((sqrtI_1-sqrt(0.6I_1))/(sqrtI_1+sqrt(0.6I_1)))^2`
`(I_(min))/(I_(max))=((0.4)/(0.6))^(2) rArr (I_(min))/(I_(max))=(1)/(16)`
(c) path difference `=nlambda`
`2h+(lambda)/(2)=nlambda rArr h=((2n-1)lambda)/(4)....(1)`
for next maxima
`2(h+Deltax)+(lambda)/(2)=(n+1)lambda rArr 2(h+Deltax)=nlambda+(lambda)/(2)`
`h+Deltax=((2n+1)lambda)/(4)...(2)`
From(1)&(2)
`(2n-1)(lambda)/(4)+Deltax=(2n+1)(lambda)/(4) rArrDeltax=(lambda)/(2) rArr Deltax=(600)/(2)nm`
`Deltax=300nm`