MHT CET · Physics · Waves and Sound
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
- A 4I
- B 5I
- C 2I
- D 3I
Answer & Solution
Correct Answer
(A) 4I
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{I}_{\mathrm{A}}=\mathrm{I}+4 \mathrm{I}+2 \sqrt{\mathrm{I}} \cdot \sqrt{4 \mathrm{I}} \cdot \cos \frac{\pi}{2}=5 \mathrm{I} \\ & \mathrm{I}_{\mathrm{B}}=\mathrm{I}+4 \mathrm{I}+2 \sqrt{\mathrm{I}} \cdot \sqrt{4 \mathrm{I}} \cdot \cos \pi=\mathrm{I} \\ & \therefore \mathrm{I}_{\mathrm{A}}-\mathrm{I}_{\mathrm{B}}=5 \mathrm{I}-\mathrm{I}=4 \mathrm{I}\end{aligned}\)
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