(a) Derive an expression for path difference in Youngs double slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen. (b) The intensity at the central maxima in Youngs double slit experiment is I0. Find out the intensity at a point where the path difference is λ/6, λ/4 and λ/3.

(a)Path difference in Young's Double slit experiment at point P is given by:
Δx=S2P−S1P
S2P2−S1P2=[D2+(x+d/2)2]−[D2+(x−d/2)2]=2xd
(S2P−S1P)(S2P+S1P)=2xd
Assuming S2P+S1P≈2Das x<<D and d<<D
Δx≈xd/D
For constructive interference, Δx=nλ
Position of nth bright fringe is: xn=nλD/d
and for destructive interference, Δx=(2n+1)λ/2
Position of nth dark fringe is: xn=(2n+1)λD/2d
(b)Let intensity of light sources from slits be I.
Resultant intensity at a point is I′=I+I+2Icosφ
where φ is the phase difference at the point.
Path difference is given by: Δx=λφ/2π
Hence, I′=I+I+2Icos(2πΔx/λ)
Given intensity at central maximum is Io
Hence, Io=4I
At Δx=λ/6, I′=3I=3/4Io
At Δx=λ/4, I′=2I=1/2Io
At Δx=λ/3, I′=I=1/4Io