JEE MainPhysicsWave Optics
In a Young's double slit experiment, the interference pattern formed by two coherent sources has a ratio of maximum to minimum intensity of 25:9 . The ratio of the widths of the two slits is
Options
- A4:1
- B5:3
- C16:1
- D25:9
Correct answer
C. 16:1
Step-by-step solution
Let the amplitudes of the waves from the two slits be A₁ and A₂ . The ratio of maximum to minimum intensity is given by: I_ I_ = (A₁ + A₂)^2 (A₁ - A₂)^2 = 25 9 Taking the square root on both sides: A₁ + A₂ A₁ - A₂ = 5 3 Cross-multiplying yields: 3A₁ + 3A₂ = 5A₁ - 5A₂ 2A₁ = 8A₂ A₁ A₂ = 4 1 The intensity of light from a slit is directly proportional to its width ( w I A^2 ). Therefore, the ratio of the slit widths is: w₁ w₂ = ( A₁ A₂ )^2 = ( 4 1 )^2 = 16 1 Answer: 16:1