JEE MainPhysicsWave Optics
In a Young's double-slit experiment, the interference pattern on the screen has a maximum to minimum intensity ratio of 25 : 9 . The ratio of the widths of the two slits used in the experiment is :
Options
- A4 : 1
- B5 : 3
- C256 : 1
- D16 : 1
Correct answer
D. 16 : 1
Step-by-step solution
The ratio of maximum to minimum intensity in the interference pattern is given by: I_ I_ = ( I₁ + I₂ I₁ - I₂ )^2 Given that I_ I_ = 25 9 , taking the square root on both sides yields: I₁ + I₂ I₁ - I₂ = 5 3 Cross-multiplying to solve for the ratio of the amplitudes: 3 I₁ + 3 I₂ = 5 I₁ - 5 I₂ 8 I₂ = 2 I₁ I₁ I₂ = 4 1 Squaring this ratio gives the ratio of the intensities of the two slits: I₁ I₂ = 16 1 Since the intensity of light passing through a slit is directly proportional to its width ( I w ), the ratio of the sl