Interference fringes may be observed due to the superposition of two…

The correct answer is Option (3) → \( y_1 = 2a\sin 2\omega t \) and \( y_2 = 2a\sin 2(\omega t + \delta) \)

Steady interference fringes need coherent sources: waves of the same frequency with a constant phase difference.

Option (1) \( y_1 = a\sin\omega t \) and \( y_2 = a\sin 2\omega t \) — Incorrect. The frequencies are \( \omega \) and \( 2\omega \). They differ, so the phase difference changes with time and no steady fringes form.

Option (2) \( y_1 = a\sin\omega t \) and \( y_2 = 2a\sin(2\omega t + \delta) \) — Incorrect. The frequencies are \( \omega \) and \( 2\omega \); they differ.

Option (3) \( y_1 = 2a\sin 2\omega t \) and \( y_2 = 2a\sin 2(\omega t + \delta) \) — Correct. Writing \( y_2 = 2a\sin(2\omega t + 2\delta) \), both waves have frequency \( 2\omega \) and a constant phase difference \( 2\delta \). They are coherent.

Option (4) \( y_1 = 2a\sin\omega t \) and \( y_2 = 2a\sin(2\omega t + \delta) \) — Incorrect. The frequencies are \( \omega \) and \( 2\omega \); they differ.

NCERT (2026-27):
Coherent and Incoherent Addition of Waves – “… Now if the two sources are coherent (i.e., if the two needles are going up and down regularly) then the phase difference φ at any point will not change with time and we will have a stable interference pattern; i.e., the positions of maxima and minima will not change with time.”