First slide

Young's double slit Experiment

Question

The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double slit experiment is

Moderate

A

infinite
B

5
C

3
D

zero

Solution

The condition for possible interference maxima on the screen is
$2 dsin θ = nλ (2 d = slit width) given 2 d = 2 λ 2 λsin θ = nλ or 2 \sin θ = n n = 2 (maximum value of \sin θ = 1) Eq. (1) will be satisfied for integer values - 2, - 1, 0, 1, and 2, i . e ., 5 . So, the maximum number of possible interference maxima is 5$

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