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Question

Physics Question on Youngs double slit experiment

Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are

A

5I and I

B

5I and 3 I

C

9I and I

D

9I and 3 I

Answer

9I and I

Explanation

Solution

Imax=(I1+I2)2=(4I+I)2=9II_{max}=(\sqrt{I_1}+\sqrt{I_2})^2=(\sqrt{4I}+\sqrt I)^2=9I Imin=(I1I2)2=(4II)2=II_{min}=(\sqrt{I_1}-\sqrt{I_2})^2=(\sqrt{4I}-\sqrt I)^2=I \therefore Correct option is (c ).