Question

In Young's double slit experiment, derive the condition for

(i) constructive interference and

(ii) destructive interference at a point on the screen.

Answer 1

Young’s double slit experiment: Consider two narrow rectangular slits S₁ and S₂ placed perpendicular to the plane of paper. Slit S is placed on the perpendicular bisector of S₁S₂ and is illuminated with monochromatic light.

The slits are separated by a small distance d. A screen is placed at a distance D from S₁, S₂.

Consider a point P on the screen at distance x from O.

The path difference between the waves reaching P from S₁ and S₂ is:

P = S₂P − S₁P

Draw S₁N perpendicular to S₂P. Then,

P = S₂P − S₁P = S₂P − NP = S₂N

From right-angled 

`DeltaS_1S^2N=(S_2N)/(S_2S_1) = sintheta`

`therefore P =S_2N=S_2S_1sintheta = d sintheta`

From ΔCOP,

When θ is small,

`sintheta≈theta≈tantheta = x/D`

`therefore P=(xd)/D`

For constructive interference,

`(xd)/D =nlambda,n=0,1,2,3,.....`

Position of n^th bright fringe, `x_n = (nDlambda)/d =0,(Dlambda)/d,(2Dlambda)/d,(3Dlambda)/d,.......`

When n = 0, x_n = 0, central bright fringe is formed at O.

For destructive interference,

`(xd)/D = (2n +1)lambda/2`

`or x_n = (2_n +1) (lambdaD)/(2d) = 1/2(lambdaD)/d,3/2(lambdaD)/d,5/2(lambdaD)/d,......`

Thus, alternate bright and dark fringes are formed on the screen.