Question

Two coherent waves are represented by $y_1=a_1\cos \omega t$ and $y_2=a_2\sin \omega t$. The resultant intensity due to interference will be

• A. $(a_1^2-a_2^2)$
• B. $(a_1^2+a_2^2)$
• C. $(a_1-a_2)$
• D. $(a_1+a_2)$

Answer 1

The correct option is B $(a_1^2+a_2^2)$

If the waves are $y_1=a_1\cos wt$ and $y_2=a_2$ $\sin \omega t$, the resultant wave is $a_1$ $\cos \omega t$ +$a_2$ $\sin \omega t$.
It can be simplified as
$\sqrt{a_1^2}+a_2^2(\frac{a_1}{\sqrt{a_1^2+a_2^2}}\cos \omega t+\frac{a_2}{\sqrt{a_1^2+a_2^2}}\sin \omega t)$
If $\frac{a_1}{\sqrt{a_1^2+a_2^2}}$ is $\cos \delta$ , then $\frac{a_2}{\sqrt{a_1^2+a_2^2}}$ is $\sin \delta$
$\sqrt{a_1^2+a_2^2}$($\cos \omega t\cos \delta +\sin \omega t\sin \delta$)
$\sqrt{a_1^2+a_2^2}\left(\cos \right(\omega t-\delta \left)\right)$
The amplitude is $\sqrt{a_1^2+a_2^2}$
Intensity is square of amplitude.
Intensity $=a_1^2+a_2^2$