Question

The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum of the minimum intensity in the interference pattern is :

• A. 9 :1
• B. 16 : 1
• C. 1 : 1
• D. 4 : 1

Answer 1

The Correct Option is A

Since the intensity is proportional to the width of the slit (ω):
I₁ = I, I₂ = 4I.

1. **Minimum Intensity (I_min):**
The minimum intensity in an interference pattern is given by:
I_min = (√I₁ - √I₂)².
Substituting I₁ = I and I₂ = 4I:
I_min = (√I - √4I)² = (√I - 2√I)² = I.

2. **Maximum Intensity (I_max):**
The maximum intensity in an interference pattern is given by:
I_max = (√I₁ + √I₂)².
Substituting I₁ = I and I₂ = 4I:
I_max = (√I + 2√I)² = 9I.

3. **Ratio of Maximum to Minimum Intensity:**
I_max / I_min = 9I / I = 9 : 1.

Answer: 9 : 1