JEE Mains · Physics · STD 12 - 10. Wave optics
Two beams of light having intensities \(I\) and \(4\,I\) interfere to produce a fringe pattern on a screen. The phase difference between the two beams are \(\pi / 2\) and \(\pi / 3\) at points \(A\) and \(B\) respectively. The difference between the resultant intensities at the two points is \(x I\). The value of \(x\) will be.
- A \(1\)
- B \(3\)
- C \(4\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(\phi_{ A }=\frac{\pi}{2}\) \(\phi_{ B }=\frac{\pi}{3}\) \(I _{ A }= I +4 I +2 \sqrt{ I } \sqrt{4 I } \cos \left(\frac{\pi}{2}\right)\) \(=5 I +4 I (0)=5 I\) \(I _{ B }= I +4 I +2 \sqrt{ I } \sqrt{4 I } \cos \left(60^{\circ}\right)\) \(=5 I +4 I \times \frac{1}{2}=7 I\)…
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