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(C) The phenomenon of interference occurs between two waves that have the same frequency and a constant phase difference.Looking at the given waves:$(

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$(i)$ and (iii)

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(C) The phenomenon of interference occurs between two waves that have the same frequency and a constant phase difference.Looking at the given waves:$(i)$ $y_1 = a \sin(\omega t + \phi_1)$ has angular frequency $\omega$.(ii) $y_2 = a \sin(2\omega t)$ has angular frequency $2\omega$.(iii) $y_3 = d' \sin(\omega t + \phi_2)$ has angular frequency $\omega$.(iv) $y_4 = d' \sin(3\omega t + \phi)$ has angular frequency $3\omega$.Since waves $(i)$ and (iii) have the same angular frequency $\omega$,their superposition will result in interference.Therefore,the correct option is $(C)$.

Four light sources produce the following four waves:
$(i)$ $y_1 = a \sin(\omega t + \phi_1)$
(ii) $y_2 = a \sin(2\omega t)$
(iii) $y_3 = d' \sin(\omega t + \phi_2)$
(iv) $y_4 = d' \sin(3\omega t + \phi)$
Superposition of which two waves gives rise to interference?

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Four light sources produce the following four waves:$(i)$ $y_1 = a \sin(\omega t + \phi_1)$(ii) $y_2 = a \sin(2\omega t)$(iii) $y_3 = d' \sin(\omega t + \phi_2)$(iv) $y_4 = d' \sin(3\omega t + \phi)$Superposition of which two waves gives rise to interference?