Question

Two waves $Y_1=a\sin \omega t$ and $Y_2=\mathrm{asin}(\omega t+\delta )$ are producing interference, then resultent intensity is:

• A. $a^2{\cos }^2\delta /2$
• B. $2a^2{\cos }^2\delta /2$
• C. $3a^2{\cos }^2\delta /2$
• D. $4a^2{\cos }^2\delta /2$

Answer 1

The correct option is D $4a^2{\cos }^2\delta /2$

$Y_1+Y_2=a\sin \omega t+a\sin (\omega t+8)$

$Y=a[\sin \omega t+\sin (\omega t+8\left)\right]$

$Y=a\left[2\sin \left(\frac{\omega t+\omega t+8}{2}\right)\cos \left(\frac{8}{2}\right)\right]$

$Y=2a\sin (\omega t+\frac{8}{2})\cos \left(\frac{8}{2}\right)$

$Y=\left[2a\cos \left(\frac{8}{2}\right)\right]\sin (\omega t+8/2)$

As Intensity $\alpha A^2$

Here $I\alpha {\left[2a\cos \left(\frac{8}{2}\right)\right]}^2$

$I\alpha 4a^2{\cos }^2\left(\frac{8}{2}\right)$

Option $D$ is correct.