Question

Four light waves are represented by
$\left(i\right)y=a_1\sin \omega t$
$\left(ii\right)y=a_2\sin (\omega t+\varphi )$
$\left(iii\right)y=a_1\sin 2\omega t$
$\left(iv\right)y=a_2\sin 2(\omega t+\varphi )$

Interference fringes may be observed due to super position of

• A. $\left(iii\right)$ and $\left(iv\right)$
• B. $\left(i\right)$ and $\left(iii\right)$
• C. $\left(ii\right)$ and $\left(iv\right)$
• D. Interference is not possible in any combination of these four given waves.

Answer 1

The correct option is A $\left(iii\right)$ and $\left(iv\right)$

We know interference can be obtained by super position of the waves, which are

$\left(1\right)$ Identical (same wavelength or frequency)

$\left(2\right)$ Coherent (constant phase difference)

Based on these observations, wave $\left(i\right)$ and $\left(ii\right)$ have same frequency $\left(\omega \right)$ and constant phase difference $\left(\varphi \right)$.

Also, wave $\left(iii\right)$ and $\left(iv\right)$ have same frequency $\left(2\omega \right)$ and constant phase difference $\left(2\varphi \right)$.

So, interference can be observed by the combination of waves $\left(i\right)$ and $\left(ii\right)$ and the combination of waves $\left(iii\right)$ and $\left(iv\right)$.

Therefore, option $\left(\text{A}\right)$ is correct.