MHT CET202617 April 2026Morning ShiftPhysicsWave OpticsActual
In Young's double slit experiment, width of the second slit is double the width of first slit, consequently the amplitude of the light from two slits. ' I_m ' is the maximum intensity. The resultant intensity ' I ' when they interfere with the phase difference of is given by
Options
- AI_m 9 (1 + 8 ^2 2 )
- BI_m 7 (3 + 5 ^2 2 )
- CI_m 5 (1 + 2 ^2 2 )
- DI_m 3 (1 + 6 ^2 2 )
Correct answer
A. I_m 9 (1 + 8 ^2 2 )
Step-by-step solution
Based on the given conditions, the amplitude of light from the second slit is twice that of the first slit. Let A₁ = A , then A₂ = 2A . The corresponding intensities are proportional to the square of the amplitudes, so I₁ = I₀ and I₂ = 4I₀ . The maximum intensity I_m occurs when the phase difference is zero: I_m = ( I₁ + I₂ )^2 I_m = ( I₀ + 4I₀ )^2 = (3 I₀ )^2 = 9I₀ I₀ = I_m 9 The resultant intensity I at a phase difference is given by: I = I₁ + I₂ + 2 I₁ I₂ I = I₀ + 4I₀ + 2 4I₀^2 I = 5I₀ + 4I₀ Using the trigonomet