MHT CET Medical202624 April 2026Morning ShiftPhysicsWave OpticsActual
In Young's double slit experiment using a monochromatic light of wavelength . The path difference (in terms of an integer n) corresponding to any point having one fourth of its peak intensity is ( /3)=1/2, (4 /3)=-1/2
Options
- A(n 1 2 ) , n=0,1,2
- B(n 1 3 ) , n=0,1,2
- C(n 1 4 ) , n=0,1,2
- D(n 1 5 ) , n=0,1,2
Correct answer
B. (n 1 3 ) , n=0,1,2
Step-by-step solution
The intensity at any point in Young's double slit experiment is given by I = I₀ ^2 ( 2 ) Given that I = I₀ 4 I₀ 4 = I₀ ^2 ( 2 ) ( 2 ) = 1 2 2 = n 3 = 2n 2 3 The phase difference is related to the path difference x by = 2 x 2 x = 2 (n 1 3 ) x = (n 1 3 ) Answer: (n 1 3 ) , n=0,1,2