Optics Question 156

Question: In a Young’s double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength λ0=750nm and λ=900nm . The minimum distance from the common central bright fringe on a screen 2m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is

Options:

A) 1.5 mm

B) 3 mm

C) 4.5 mm

D) 6 mm

Show Answer

Answer:

Correct Answer: C

Solution:

From the given data, note that the fringe width (b1) for λ1=900nm is greater than fringe width (b2) for λ2=750nm.

This means that at though the central maxima of the two coincide, but first maximum for λ1=900nm will be further away from the first maxima for λ2=750nm, and so on.

A stage may come when this mismatch equals b2, then again maxima of λ1=900nm, will coincide with a maxima of λ2=750nm,

let this correspond to nth order fringe for l1. Then it will correspond to (n+1)th order fringe for l2.

Therefore nλ1Dd=(n+1)λ2Dd

n×900×109=(n+1)750×109n=5

Minimum distance from Central maxima =nλ1Dd=5×900×109×22×103

=45×104m=4.5mm



NCERT Chapter Video Solution

Dual Pane