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

Easy

A

Infinite
B

Five
C

Three
D

Zero

Solution

For maxima $Δ = dsinθ = nλ$
$\Rightarrow 2 λsinθ = nλ \Rightarrow sinθ = \frac{n}{2}$
since value of sin $θ$ can not be greater 1 .
$∴ n = 0, 1, 2$
Therefore only five maximas can be obtained on both side of the screen.

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