Two coherent light beams of intensity 2I and 18I are superposed with each other. The maximum and minimum possible intensities in the resulting beam are
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 32I, 8I
For coherent beams the amplitudes add, and the intensity is proportional to the square of the resultant amplitude. The amplitudes are proportional to \( \sqrt{2I} \) and \( \sqrt{18I} = 3\sqrt{2I} \). The maximum occurs when the waves are in phase and the minimum when they are in opposite phase.
\( I_{\max} = \left(\sqrt{2I} + \sqrt{18I}\right)^{2} = \left(\sqrt{2} + 3\sqrt{2}\right)^{2}I = \left(4\sqrt{2}\right)^{2}I = 32I \)
\( I_{\min} = \left(\sqrt{18I} - \sqrt{2I}\right)^{2} = \left(3\sqrt{2} - \sqrt{2}\right)^{2}I = \left(2\sqrt{2}\right)^{2}I = 8I \)
Shortcut: The amplitudes are in the ratio \( 1 : 3 \). So \( I_{\max}\propto(3 + 1)^{2} = 16 \) and \( I_{\min}\propto(3 - 1)^{2} = 4 \), where one unit corresponds to the intensity 2I. This gives \( 32I \) and \( 8I \) at once.