Question

Two light waves are represented by $y_1=3\sin \omega t$ and $y_2=4\sin (\omega t+\frac{\pi }{3})$. The resultant amplitude due to interference will be

• A. $\sqrt{21}$
• B. $\sqrt{26}$
• C. $\sqrt{37}$
• D. $\sqrt{41}$

Answer 1

Answer: (C)

$\sqrt{37}$

The resultant wave $y=y_1+y_2$
$y=3\sin \omega t+4\sin (\omega t+\pi /3)$
$=3\sin \omega t+2\sin \omega t+2\sqrt{3}\cos \omega t$
$=5\sin \omega t+2\sqrt{3}\cos \omega t$

Amplitude of the resultant wave is given by $=\sqrt{(5{)}^2+(2\sqrt{3}{)}^2}$
$=\sqrt{25+12}$
$=\sqrt{37}$