MHT CET202120 Sep 2021Morning ShiftPhysicsWave OpticsActual
In Young's double slit experiment, the intensity at a point where the path difference is 4 [ is wavelength of light used] is ' I '. If ' I ₀ ' is the maximum intensity, then I I ₀ is equal to
Options
- A3: 2
- B2: 3
- C3: 4
- D1: 2
Correct answer
D. 1: 2
Step-by-step solution
If I' is the intensity of each wave, then resultant intensity is given by I =4 I ^2 2 I will have maximum value when ^2 2 =1 maximum intensity, I ₀=4 I When path difference is 4 , the phase difference, = 2 The resultant intensity, I=4 I^ ^2 4 =4 I^ 1 2 =2 I^ I I ₀ = 1 2