Two beams of light having intensities I and $4 \mathrm{I}$ interfere to produce a fringe pattern on a screen…

Two beams of light having intensities I and $4 \mathrm{I}$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\pi / 2$ at point $\mathrm{A}$ and $\pi$ at point $\mathrm{B}$. Then the difference between the resultant intensities at $A$ and $B$ is
  1. 4I
  2. 5I
  3. 2I
  4. 3I

Solution

$\begin{aligned} & \mathrm{I}_{\mathrm{A}}=\mathrm{I}+4 \mathrm{I}+2 \sqrt{\mathrm{I}} \cdot \sqrt{4 \mathrm{I}} \cdot \cos \frac{\pi}{2}=5 \mathrm{I} \\ & \mathrm{I}_{\mathrm{B}}=\mathrm{I}+4 \mathrm{I}+2 \sqrt{\mathrm{I}} \cdot \sqrt{4 \mathrm{I}} \cdot \cos \pi=\mathrm{I} \\ & \therefore \mathrm{I}_{\mathrm{A}}-\mathrm{I}_{\mathrm{B}}=5 \mathrm{I}-\mathrm{I}=4 \mathrm{I}\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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