JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2021
- Hard
- MCQ
In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(4: 1\)
Step-by-step Solution
Detailed explanation
Amplitude \(\propto\) Width of slit \(\Rightarrow A _{2}=3 A _{1}\)
\(\frac{ I _{\max }}{ I _{\min }}=\left(\frac{\sqrt{ I _{1}}+\sqrt{ I _{2}}}{\left|\sqrt{ I _{1}}-\sqrt{ I _{2}}\right|}\right)^{2}\)
\(\because\) Intensity \(I \propto A ^{2}\)
\(\Rightarrow \frac{I_{\max }}{I_{\min }}=\left(\frac{A_{1}+A_{2}}{\left|A_{1}-A_{2}\right|}\right)^{2}\)
\(=\left(\frac{ A _{1}+3 A _{1}}{\left| A _{1}-3 A _{1}\right|}\right)^{2}\)
\(=\left(\frac{4 A _{1}}{2 A _{1}}\right)^{2}=4: 1\)
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