First slide

Young's double slit Experiment

Question

Two coherent monochromatic light beams of intensities I and 4 l are superposed, The maximum and minimum possible intensities in the resulting beam
are

Moderate

A

5 I and I
B

5I and 3 I
C

9I and I
D

9 I and 3 I

Solution

$I_{max} = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} = I + 4 I + 2 \sqrt{4 I^{2}} = 9 I and I_{min} = a_{1}^{2} + a_{2}^{2} - 2 a_{1} a_{2} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}} = I + 4 I - 4 I = I$

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