MHT CET202519 Apr 2025Evening ShiftPhysicsWave OpticsActual
In Young's double slit experiment, the intensity on screen at a point where path difference is 4 is K 2 . The intensity at a point when path difference is ' ' will be
Options
- A4 K
- B2 K
- CK
- DK 4
Correct answer
C. K
Step-by-step solution
The intensity I in Young's double-slit experiment varies with phase as I = I₀ ^2 ( 2 ) , where I₀ is the maximum intensity, and = 2 x relates the phase difference to the path difference x . Given that the intensity is K 2 for a path difference of 4 , the corresponding phase difference is ₁ = 2 . Since ( 4 ) = 1 2 , the intensity is I₁ = I₀ ^2 ( 4 ) = I₀ 2 . Equating I₀ 2 = K 2 yields I₀ = K . For a path difference of , the phase difference becomes ₂ = 2 . The intensity is then I₂ = I₀ ^2 ( ) = I₀ . Substituting I₀