Question

In Young’s double slit experiment intensity at a point is `(1/4)` of the maximum intensity. Angular position of this point is ______.

Options

• `sin^-1(λ/d)`

• `sin^-1(λ/(2d))`

• `sin^-1(λ/(3d))`

• `sin^-1(λ/(4d))`

Answer 1

In Young’s double slit experiment intensity at a point is `(1/4)` of the maximum intensity. Angular position of this point is `bbunderline(sin^-1(λ/(3d)))`.

Explanation:

In Young’s Double Slit Experiment (YDSE), the intensity I at any point is related to the maximum intensity I₀ and the phase difference Φ by:

`I = I_0 cos^2(phi/2)`

The problem states the intensity is `1/4` of the maximum:

`(I_0)/4 = I_0 cos^2 (phi/2)`

`1/4 = cos^2 (phi/2)`

`cos(phi/2) = 1/2`

∴ `phi/2 = 60°` 

∴ `pi/3 = phi = 2pi/3`

The relationship is `phi = (2pi)/λ Δ x`.

For YDSE, the path difference at an angular position θ is:

Δ x = d sin θ

`(2pi)/3 = (2pi)/λ (d sin θ)`

`1/3 = (d sin θ)/λ`

sin θ = `λ/(3d)`

θ = `sin^-1(λ/(3d))`