Q102 / 120 · VITEEE 2009 · Physics
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) $y_{3}=a \sin \left(\omega t+\phi_{2}\right)$
(iv) $y_{4}=a \sin (3 \omega t+\phi)$
Superposition of which two waves give rise to interference?
- A(i) and (ii)
- B(ii) and (iii)
- Correct answerC(i) and (iii)
- D(iii) and (iv)
Solution
Inteference takes place between two waves having equal frequency and propagate in same direction.
Hence, $y_{1}=a \sin \left(\omega t+\phi_{1}\right)$
$$
y_{3}=a^{\prime} \sin \left(\omega t+\phi_{2}\right)
$$ same frequency $\omega$.
Hence, the correct answer is option: 'C'.