CUET · PHYSICS · PYQ PAPER 2025
In Young's double slit experiment, the width of one of the two slits is four times the width of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern will be
- A \(4:1\)
- B \(3:1\)
- C \(9:1\)
- D \(16:1\)
Answer & Solution
Correct Answer
(C) \(9:1\)
Step-by-step Solution
Detailed explanation
\( \frac{I_1}{I_2} = \frac{w_1}{w_2} = 4 \) \( \frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{4} = 2 \) \( \frac{I_{max}}{I_{min}} = \left(\frac{A_1+A_2}{A_1-A_2}\right)^2 = \left(\frac{2A_2+A_2}{2A_2-A_2}\right)^2 = \left(\frac{3A_2}{A_2}\right)^2 = 9 \)
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