Question

Using analytical method, obtain an expression for the fringe width of two interfering waves.

Answer 1

Let the distance between the two slits S₁ and S₂ be d and that between O and O' be D.

The point O' on the screen is equidistant from S₁ and S₂. Thus, the distances travelled by the wavelets starting from S₁ and S₂ to reach O' will be equal. The two waves will be in phase at O', resulting in constructive interference. Thus, there will be a bright spot at O' and a bright fringe at the centre of the screen.

Now, consider any point P on the screen. The two wavelets from S₁ and S₂ travel different distances to reach P, and so the phases of the waves reaching P will not be the same. If the path difference (Δl) between S₁P and S₂P is an integral multiple of λ, the two waves arriving there will interfere constructively, producing a bright fringe at P. If the path difference between S₁P and S₂P is a half-integer multiple of λ, there will be destructive interference, and a dark fringe will be located at P.

Considering triangles `S_1S_1^'P` and `S_2S_2^'P`, we can write:

(S₂P)² − (S₁P)² = `{D^2 + (y + d/2)^2} - {D^2 + (y - d/2)^2}`

= 2 yd

∴ Δl = S₂P − S₁P

= `(2 yd)/(S_2P + S_1P)`

For `d/D` << 1, we can write S₂P + S₁P ≈ 2 D

∴ Δl = `(2 yd)/(2 D)`

= `y d/D`

Thus, the condition for constructive interference at P can be written as:

Δl = `y_n d/D`

= n λ

y_n being the position (y-coordinate) of the n^th bright fringe (n = 0, ±1, ±2, ...). It is given by:

y_n = `(nlambdaD)/d`

Similarly, the position of n^th (n = +1, +2, ...) dark fringe (destructive interference) is given by:

Δl = `y_n d/D`

= `(n - 1/2) lambda`

y_n = `(n - 1/2) (lambda D)/d`

The distance between any two successive dark or any two successive bright fringes (y_n+1 − y_n) is equal. This is called the fringe width and is given by:

Fringe width (W) = Δy

= y_n+1 − y_n

= `(n + 1) (lambda D)/d - n (lambda D)/d`

= `(lambda D)/d`

This is the required expression for the fringe width.