Q89 / 90 · JEE Main 2023 (31 Jan Shift 2) · Physics

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.

Correct answer: 48

Solution

Fringe width for both cases can be written as.ω1=λ1Dd & ω2=λ2Dd.

Using values of wavelength, we get ω1=16 mm & ω2=12 mm.

Let y be the common distance of the bright fringes by the both wavelength, then

y=n1ω1=n2ω2.

As LCM ω1,ω2=48 mm, therefore at y=48 mm distance both bright fringes will be found.

Hence, the correct answer is 48.

Relevant Questions