Two waves of equal frequencies have their amplitudes in the ratio of 3:5. They are superimposed on each other. Calculate the ratio of maximum and minimum intensities of the resultant wave is ${\alpha }^2$ then  $\alpha$ is  

Given,  $\frac{A_1}{A_2}=\frac{3}{5}$
 $\therefore \sqrt{\frac{I_1}{I_2}}=\frac{3}{5}$
Maximum intensity is obtained, where 
   $\cos \varphi =1$
And $I_{\max }={(\sqrt{I_1}+\sqrt{I_2})}^2$
Minimum intensity is found, where 
$\cos \varphi =-1$ 
And  $I_{\min }={(\sqrt{I_1}-\sqrt{I_2})}^2$  
Hence,  $\frac{I_{\max }}{I_{\min }}={\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)}^2={\left(\frac{\sqrt{\frac{I_1}{I_2}}+1}{\sqrt{\frac{I_1}{I_2}}-1}\right)}^2$
$={\left(\frac{3/5+1}{3/5-1}\right)}^2=\frac{64}{4}=\frac{16}{1}$