First slide

Young's double slit Experiment

Question

In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D, the slits are separated by d and are illuminated by light of wavelength $λ$ the distance from the central point where the intensity falls to half the maximum is

Easy

A

$\frac{λD}{3 d}$
B

$\frac{λD}{2 d}$
C

$\frac{λD}{d}$
D

$\frac{λD}{4 d}$

Solution

the intensity at any point on the screen due to two identical slits is given by I = 4 I₁ cos² $(δ / 2)$ where I₁intensity due to either slit. At the center of a bright fringe $δ = 2 nπ$ At center I = 4I₁ ( 8 = 0) this is maximum intensity Let at a distance x from the center, the intensity is half of maximum intensity, i.e.2 I₁ Hence $2 I_{1} = 4 I_{1} \cos^{2} \frac{δ}{2}$
or $or \cos^{2} \frac{δ}{2} = \frac{1}{2} or \cos \frac{δ}{2} = \frac{1}{\sqrt{2}} or \frac{δ}{2} = \cos^{- 1} \frac{1}{\sqrt{2}} = \frac{π}{4} or δ = \frac{π}{2} Phase difference π / 2 corresponds to path difference λ / 4 \frac{xd}{D} = \frac{λ}{4} or x = \frac{λD}{4 d}$

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