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 :
- 9 :1
- 16 : 1
- 1 : 1
- 4 : 1
The Correct Option is A
Solution and Explanation
Since the intensity is proportional to the width of the slit (ω):
I₁ = I, I₂ = 4I.
1. **Minimum Intensity (Imin):**
The minimum intensity in an interference pattern is given by:
Imin = (√I₁ - √I₂)².
Substituting I₁ = I and I₂ = 4I:
Imin = (√I - √4I)² = (√I - 2√I)² = I.
2. **Maximum Intensity (Imax):**
The maximum intensity in an interference pattern is given by:
Imax = (√I₁ + √I₂)².
Substituting I₁ = I and I₂ = 4I:
Imax = (√I + 2√I)² = 9I.
3. **Ratio of Maximum to Minimum Intensity:**
Imax / Imin = 9I / I = 9 : 1.
Answer: 9 : 1
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