Let us consider interference between two sources of intensities I and 4I. What will be the intensity at points where the phase difference is \( \pi \)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → I
In interference, the waves add by amplitude, not by intensity. For two coherent sources of intensities \( I_1 \) and \( I_2 \) with phase difference \( \phi \), the resultant intensity is
\( \displaystyle I_{R} = I_1 + I_2 + 2\sqrt{I_1 I_2}\,\cos\phi \)
\( I_1 = I \)
\( I_2 = 4I \)
\( \phi = \pi,\ \cos\pi = -1 \)
\( \displaystyle I_{R} = I + 4I + 2\sqrt{I \times 4I}\,(-1) = 5I - 4I = I \)
A phase difference of \( \pi \) gives destructive interference, so this is the minimum intensity, \( I_{\min} = \left(\sqrt{4I} - \sqrt{I}\right)^{2} = I \).