Two coherent monochromatic light beams of intensities 16I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are -
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 36I and 4I
For coherent beams the amplitudes add, not the intensities, and intensity is proportional to the square of the amplitude, \( a \propto \sqrt{I} \).
\( \displaystyle \sqrt{I_1} = \sqrt{16I} = 4\sqrt{I} \)
\( \displaystyle \sqrt{I_2} = \sqrt{4I} = 2\sqrt{I} \)
Constructive interference (amplitudes add):
\( \displaystyle I_{\max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^{2} = \left(4 + 2\right)^{2} I = 36I \)
Destructive interference (amplitudes subtract):
\( \displaystyle I_{\min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^{2} = \left(4 - 2\right)^{2} I = 4I \)