JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2022
- Medium
- Numerical
Two light beams of intensities \(4\,I\) and \(9\,I\) interfere on a screen. The phase difference between these beams on the screen at point \(A\) is \(zero\) and at point \(B\) is \(\pi\). The difference of resultant intensities, at the point \(A\) and \(B\), will be \(....I\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(24\)
Step-by-step Solution
Detailed explanation
\(I _{\text {met }}=I_{1}+I_{2}+2 \sqrt{I_{1}} \sqrt{I_{2}} \cos \phi\)
\(I_{\max }\) for \(\phi=0 \& I_{\min }\) for \(\phi=\pi\)
\(I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}=(\sqrt{9 I}+\sqrt{4 I})^{2}=25\,I\)
\(I_{\min }=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}=(\sqrt{9 I}-\sqrt{4 I})^{2}=I\)
\(I_{\max }-I_{\min }=25 I - I =24\,I\)
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