Question
Physics Question on Wave optics
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
9 :1
Solution
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 / Imin = 9I / I = 9 : 1.
Answer: 9 : 1