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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 _______.

  1. A \(5\)
  2. B \(6\)
  3. C \(4\)
  4. D \(8\)
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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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