JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2024
- Hard
- MCQ
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 _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(9: 1\)
Step-by-step Solution
Detailed explanation
Since, Intensity \(\propto\) width of slit \((\omega)\)
\(\text { so, } I_1=I, I_2=4 \mathrm{I}\)
\(I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2=I\)
\(I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=9 \mathrm{I}\)
\(\frac{I_{\max }}{I_{\min }}=\frac{9 \mathrm{I}}{\mathrm{I}}=\frac{9}{1}\)
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