Two waves are superimposed whose ratio of intensities is $9: 1$. The ratio of maximum and minimum intensity is
Two waves are superimposed whose ratio of intensities is $9: 1$. The ratio of maximum and minimum intensity is
9:1
$4: 1$
$3: 1$
$5: 3$
Solution
Given, the ratio of intensities, we can obtain ratio of amplitudes of the waves:
$\begin{aligned} & \frac{I_1}{I_1}=\left(\frac{a_1}{a_2}\right)^2=\frac{9}{1} \\ & \Rightarrow \frac{a_1}{a_2}=\frac{3}{1}\end{aligned}$
$\frac{I_{\max }}{I_{\min }}=\frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\frac{\left(3 a_2+a_2\right)^2}{\left(3 a_2-a_2\right)^2}=\left(\frac{4 a_2}{2 a_2}\right)^2=\frac{4}{1}$
$\frac{I_{\max }}{I_{\min }}=\frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\frac{\left(3 a_2+a_2\right)^2}{\left(3 a_2-a_2\right)^2}=\left(\frac{4 a_2}{2 a_2}\right)^2=\frac{4}{1}$