Two coherent light waves of intensities \( I_1 \) and \( I_2 \) \( (I_1\gt I_2) \) superpose to give an interference pattern. Consider the following statements:
(A) The maximum intensity in the pattern is \( (\sqrt{I_1}+\sqrt{I_2})^{2} \).
(B) The minimum intensity in the pattern is \( I_1-I_2 \).
(C) The difference between the maximum and the minimum intensity is \( 4\sqrt{I_1I_2} \).
(D) The ratio of the maximum to the minimum intensity is \( I_1/I_2 \).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
By the principle of superposition, the displacements of the two waves add. At a maximum the amplitudes add, \( a_1+a_2 \), and at a minimum they subtract, \( a_1-a_2 \). Since the intensity is proportional to the square of the amplitude, each amplitude is proportional to \( \sqrt{I} \).
(A) The maximum intensity in the pattern is \( (\sqrt{I_1}+\sqrt{I_2})^{2} \). — True. At a maximum the resultant amplitude is \( a_1+a_2 \), which is proportional to \( \sqrt{I_1}+\sqrt{I_2} \). Squaring it gives the maximum intensity.
(B) The minimum intensity in the pattern is \( I_1-I_2 \). — False. At a minimum the amplitudes subtract. The intensities do not. The minimum intensity is:
\( \begin{aligned} I_{\mathrm{min}} &= (\sqrt{I_1}-\sqrt{I_2})^{2} \\ &= I_1+I_2-2\sqrt{I_1I_2} \end{aligned} \)
(C) The difference between the maximum and the minimum intensity is \( 4\sqrt{I_1I_2} \). — True. Subtracting the two results:
\( \begin{aligned} I_{\mathrm{max}}-I_{\mathrm{min}} &= (\sqrt{I_1}+\sqrt{I_2})^{2}-(\sqrt{I_1}-\sqrt{I_2})^{2} \\ &= 4\sqrt{I_1I_2} \end{aligned} \)
(D) The ratio of the maximum to the minimum intensity is \( I_1/I_2 \). — False. The ratio is \( \left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^{2} \). For \( I_1=4I \) and \( I_2=I \) it comes to 9, while \( I_1/I_2 \) is only 4.