Question

Two waves $y_1=2\sin \omega t$ and $y_2=4\sin (\omega t+\delta )$ superimpose. The ratio of the maximum to the minimum intensity of the resultant wave is :

• A. $9$
• B. $3$
• C. Infinity
• D. Zero

Answer 1

The correct option is A $9$

The resultant wave
$y=a_1\sin \omega t+a_2\sin (\omega t+\varphi )$
$y=2\sin \omega t+4\sin (\omega t+\delta )$
From equation of wave
$a_1=2$ and $a_2=4$
$\frac{I_{max}}{I_{min}}={\left(\frac{a_1+a_2}{a_1-a_2}\right)}^2$
$={\left(\frac{2+4}{2-4}\right)}^2$
$\frac{I_{max}}{I_{min}}=\frac{36}{4}=9$.