First slide

Young's double slit Experiment

Question

Two identical radiators have a separation of $d = λ / 4$ where $λ$ is the wavelength of the waves emitted by either source. The initial phase difference between the sources is $λ / 4$ . Then the intensity on the screen at a distant point situated at an angle $θ = 30^{\circ}$ from the radiators is (here I₀ is intensity at that point due to one radiator alone)

Moderate

A

I₀
B

2I₀
C

3I₀
D

4I₀

Solution

The intensity at a point on screen is given by
$I = 4 I_{0} \cos^{2} (ϕ / 2)$
where $ϕ$ is the phase difference. In this problem $ϕ$ arises (i) due to initial phase difference of $π / 4$ and (ii) due to path difference for the observation point situated at $θ = 30^{\circ} .$ Thus
$ϕ = \frac{π}{4} + \frac{2 π}{λ} (dsinθ) = \frac{π}{4} + \frac{2 π}{λ} \cdot \frac{λ}{4} (\sin 30^{\circ})$
$= \frac{π}{4} + \frac{π}{4} = \frac{π}{2}$
Thus $\frac{ϕ}{2} = \frac{π}{4} and I = 4 I_{0} \cos^{2} (π / 4) = 2 I_{0}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App