Two light sources do not maintain a constant phase difference. If time-averaged intensity distribution in the interference pattern is obtained, then the average intensity at all points will be
[Here \( I_{0} \) is the intensity of individual sources and \( I \) is resultant intensity]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( I = 2I_{0} \)
Sources that do not keep a constant phase difference are incoherent. Their phase difference changes rapidly, so the interference term averages to zero over time and the intensities simply add.
\( \displaystyle I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\,\overline{\cos\phi}, \quad \overline{\cos\phi} = 0 \)
\( I = I_{0} + I_{0} = 2I_{0} \)
The average intensity is the same at every point: no fringes are seen.
Why the other options are incorrect
Option (1) \( I = I_{0} \) — Incorrect. \( I_{0} \) is the intensity of one source alone; the second source adds its own intensity.
Option (3) \( I = 4I_{0} \) — Incorrect. \( 4I_{0} \) is the intensity at a bright fringe of coherent sources, where the amplitudes add.
Option (4) \( I = I_{0}/2 \) — Incorrect. Two sources together cannot give less than one of them on average.
NCERT (2026-27):
Coherent and Incoherent Addition of Waves – “… However, if the two needles do not maintain a constant phase difference, then the interference pattern will also change with time and, if the phase difference changes very rapidly with time, the positions of maxima and minima will also vary rapidly with time and we will see a “time-averaged” intensity distribution. When this happens, we will observe an average intensity that will be given by
\( \displaystyle I = 2I_0 \qquad (10.12) \)
at all points.”