JEE Mains · Physics · STD 12 - 10. Wave optics
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 _______.
- A \(5\)
- B \(6\)
- C \(4\)
- D \(8\)
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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