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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2022
  • Hard
  • Numerical

Two coherent sources of light interfere. The intensity ratio of two sources is \(1: 4\). For this interference pattern if the value of \(\frac{I_{\max }+I_{\min }}{I_{\max _{\max }}-I_{\min }}\) is equal to \(\frac{2 \alpha+1}{\beta+3}\), then \(\frac{\alpha}{\beta}\) will be.

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Correct Answer

(B) \(2\)

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\(\frac{ I _{1}}{ I _{2}}=\frac{1}{4}\)

\(I _{2}=4 I _{1}\)

\(I _{\max }= I _{1}+4 I _{1}+2 \sqrt{ I _{1} 4 I _{1}}=9 I _{1}\)

\(I_{\min }=I_{1}+4 I_{1}-2 \sqrt{I_{1} 4 I_{1}}=I_{1}\)

\(\therefore \frac{9 I _{1}+ I _{1}}{9 I _{1}- I _{1}}=\frac{10}{8}=\frac{5}{4}=\frac{2 \alpha+1}{\beta+1}\)

\(\alpha=2 \beta=1\)

\(\therefore \frac{\alpha}{\beta}=\frac{2}{1}=2\)

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