Question

Consider a two-slit interference arrangement (Figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen falls at a distance D from the centre O.

Answer 1

Young's experiment to show interference of light passing through two slits. A pattern of bright and dark areas appears on the screen (as shown in figure (i)).

The condition for destructive interference is `Δx = S_2P - S_1P +- ((2n - 1)/2)lambda` where n = 1, 2, ...

For n^th minima to be formed on the screen path difference (Δx) between the rays coming from S₁ and S₂ must be `((2n - 1)/2)lambda`

The minima will occur when `Δx = S_2P - S_1P = ((2n - 1)/2)lambda` ......(i)

From the given figure,

`S_1P = sqrt((S_1T_1)^2 + (PT_1)^2) = sqrt(D^2 + (D - x)^2)`

And `S_2P = sqrt((S_2T_2)^2 + (T_2P^2)) = sqrt(D^2 + (D + x)^2)`

`T_2P = T_2O + OP = D + x`

And `T_1P = T_1O - OP = D - x`

Hence, `[D^2 + (D + x)^2]^(-1/2) - [D^2 + (D - x)^2]^(1/2) = lambda/2`  ......[For first minima n = 1]

If `x = D`

We can write, `[D^2 + 4D^2]^(-1/2) - [D^2 + 0]^(1/2) = lambda/2`

⇒ `[5D^2]^(1/2) - [D^2 + 0]^(1/2) = lambda/2`

⇒ `sqrt(5)D - D = lambda/2` or `D = lambda/(2sqrt(5) - 1)` = 0.404 λ