Question

Two light waves of wavelengths 800 and $600 nm$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7 m$ away from plane of slits If the two slits are separated by $0.35 mm$, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ___$mm$

Hint:

The shortest distance between two coinciding bright fringes for two wavelengths can be found using the condition for fringe coincidence.

Answer 1

Correct Answer: 48

To solve the problem, we need to find the shortest distance from the central bright maximum where the bright fringes of two wavelengths coincide in Young's double-slit experiment. The path difference for bright fringes is given by:

y_n = nλD/d

where y_n is the fringe position, n is the fringe order, λ is the wavelength, D is the distance to the screen, and d is the slit separation.

Given: λ₁ = 800 nm, λ₂ = 600 nm, D = 7 m, and d = 0.35 mm.

The bright fringes of the two wavelengths will coincide for the smallest y when:

nλ₁ = mλ₂, n, m are integers.

This implies:

n/m = λ₂/λ₁ = 600/800 = 3/4 => n = 3 and m = 4 satisfies this.

Using n = 3, we calculate y:

y = (nλ₁D)/d = (3×800×10⁻⁹×7)/(0.35×10⁻³)

Calculate: y = 48 mm.

This calculated distance of 48 mm matches the expected range [48,48]. Thus, the shortest distance is 48 mm.