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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.

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Correct Answer

(D) \(2\)

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\(\phi_{ A }=\frac{\pi}{2}\)

\(\phi_{ B }=\frac{\pi}{3}\)

\(I _{ A }= I +4 I +2 \sqrt{ I } \sqrt{4 I } \cos \left(\frac{\pi}{2}\right)\)

\(=5 I +4 I (0)=5 I\)

\(I _{ B }= I +4 I +2 \sqrt{ I } \sqrt{4 I } \cos \left(60^{\circ}\right)\)

\(=5 I +4 I \times \frac{1}{2}=7 I\)

\(I _{ B }- I _{ A }=7 I -5 I =2 I ,( x =2)\)

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