JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2024
- Hard
- Numerical
Two coherent monochromatic light beams of intensities \(I\) and \(4I\) are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is \(x\)I. The value of \(x\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
\(I_{\max }=(\sqrt{\mathrm{I}}+\sqrt{4 \mathrm{I}})^2=9 \mathrm{I}\)
\(\mathrm{I}_{\min }=(\sqrt{4 \mathrm{I}}-\sqrt{\mathrm{I}})^2=\mathrm{I}\)
\(\therefore \mathrm{I}_{\max }-\mathrm{I}_{\min }=8 \mathrm{I}\)
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