Question

Two waves $Y_1=a\sin \omega t$ and $Y_2=a\sin (\omega t+\delta )$ are producing interference, then resultant intensity is proportional to

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

Answer 1

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

$y_1=a\sin wt{y}_2=a\sin (wt+\delta )$

Resultant $=(a^2+a^2+2\times a\times a\cos \delta )=2a^2+2a^2\cos \delta )=2a^2\times 2{\cos }^2\left(\frac{\delta }{2}\right)=4a^2{\cos }^2\left(\frac{\delta }{2}\right)$