Question

Two waves of equal frequencies have their amplitudes in the ratio of $3:5$. They are superimposed on each other. The ratio of maximum and minimum intensities of the resultant wave is

• A. $16:1$
• B. $1:16$
• C. $8:1$
• D. $1:8$

Answer 1

The correct option is A $16:1$

Given,
$\frac{A_1}{A_2}=\frac{3}{5}$

As we know, $I\propto A^2$

$\therefore \sqrt{\frac{I_1}{I_2}}=\frac{3}{5}.......\left(1\right)$

The formula for the resultant intensity is given by,

$I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$

Maximum intensity is obtained, when
$\cos \varphi =1$

$\Rightarrow I_{max}={(\sqrt{I_1}+\sqrt{I_2})}^2$

Minimum intensity is obtained, when
$\cos \varphi =-1$

$\Rightarrow I_{min}={(\sqrt{I_1}-\sqrt{I_2})}^2$

Hence,
$\frac{I_{max}}{I_{min}}={\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)}^2={\left(\frac{\sqrt{\frac{I_1}{I_2}}+1}{\sqrt{\frac{I_1}{I_2}}-1}\right)}^2$

Putting the value fron equation $\left(1\right)$,

$\frac{I_{max}}{I_{min}}=={\left(\frac{\frac{3}{5}+1}{\frac{3}{5}-1}\right)}^2=\frac{16}{1}=16:1$
​​​​​​​
Hence, option $\left(\text{A}\right)$ is correct.