Question

A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Answer 1

Here, d = 2 mm = 2 × 10⁻³ m, D = l.2 m, λ₁ = 650 nm = 650 × 10⁻⁹ m, λ₂ = 520 nm = 520 × 10⁻⁹ m

At a linear distance 'y' from the center of the screen, the bright fringes due to both wavelengths coincide. Let n₁ number of bright fringes with wavelength λ₁ coinciding with n₂ number of bright fringe with wavelength λ₂· We can write:

y = n₁β₁ = n₂β₂

`n_1(λ_1D)/d = n_2(λ_2D)/d` or n₁λ₁ = n₂λ₂  ...(i)

Also at the first position of the coincidence, the n^th bright fringe of one will coincide with the (n + 1)^th bright fringe of the other.

If λ₂ < λ₁,

So, then n₂ > n₁

then n₂ = n₁ + 1 ...(ii)

Using equation (ii) in equation (i)

n₁λ₁ = (n₁ + 1)λ₂

n₁(650) × 10⁻⁹ = (n₁ + 1)520 × 10⁻⁹

65n₁ = 52n₁ + 52 or 12n₁ = 52 or n₁ = 4

Thus, y = n₁β₁ = `4[((6.5 xx 10^-7)(1.2))/(2 xx 10^-3)]`

= 1.56 × 10⁻³ m

= 1.56 mm

So, the fourth bright fringe of wavelength 520 nm coincides with the 5^th bright fringe of wavelength 650 nm.