Four light sources produce the following four waves : (i) $y_1=a \sin \left(\omega t+\phi_1\right)$ (ii)…

Four light sources produce the following four waves : (i) $y_1=a \sin \left(\omega t+\phi_1\right)$ (ii) $y_2=a \sin 2 \omega t$ (iii) $y_3=d^{\prime} \sin \left(\omega t+\phi_2\right)$ (iv) $y_4=d^{\prime} \sin (3 \omega t+\phi)$ Superposition of which two waves give rise to interference?
  1. (i) and (ii)
  2. (ii) and (iii)
  3. (i) and (iii)
  4. (iii) and (iv)

Solution

Interference phenomenon takes place between two waves which have equal frequency and propagate in same direction. Hence, $\begin{aligned} & y_1=a \sin \left(\omega t+\phi_1\right) \\ & y_3=d^{\prime} \sin \left(\omega t+\phi_2\right) \end{aligned}$ will give rise to interference as the two waves have same frequency $\omega$.

Asked in: AP EAMCET 2009

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