JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2022
- Medium
- Numerical
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.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(\phi_{ A }=\frac{\pi}{2}\)
\(\phi_{ B }=\frac{\pi}{3}\)
\(I _{ A }= I +4 I +2 \sqrt{ I } \sqrt{4 I } \cos \left(\frac{\pi}{2}\right)\)
\(=5 I +4 I (0)=5 I\)
\(I _{ B }= I +4 I +2 \sqrt{ I } \sqrt{4 I } \cos \left(60^{\circ}\right)\)
\(=5 I +4 I \times \frac{1}{2}=7 I\)
\(I _{ B }- I _{ A }=7 I -5 I =2 I ,( x =2)\)
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