Question

Two waves of equal frequencies, have their amplitudes in the ratio of $4:5$. They are superimposed on each other. Calculate the ratio of maximum and minimum intensity of the resultant wave.

• A. $3:1$
• B. $9:1$
• C. $27:1$
• D. $81:1$

Answer 1

The correct option is D $81:1$

Let the amplitudes of the waves are $A_1$ and $A_2$, then :
$\frac{A_1}{A_2}=\frac{4}{5}$ ....(i)

The relation between intesity $\left(I\right)$ and amplitude $\left(A\right)$ is given by :
$I\propto A^2$
$\Rightarrow \frac{I_1}{I_2}=\frac{A_1^2}{A_2^2}=\frac{16}{25}$

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

$={\left(\frac{\sqrt{\frac{I_1}{I_2}}+1}{\sqrt{\frac{I_1}{I_2}}-1}\right)}^2$

As, $\frac{I_1}{I_2}=\frac{16}{25}$

$\Rightarrow \sqrt{\frac{I_1}{I_2}}=\frac{4}{5}$

$\therefore \frac{I_{max}}{I_{min}}={\left(\frac{\frac{4}{5}+1}{\frac{4}{5}-1}\right)}^2={\left(\frac{9}{-1}\right)}^2$

$\Rightarrow \frac{I_{max}}{I_{min}}=81:1$

Hence, $\left(D\right)$ is the correct answer

$$\begin{array}{c}\text{Why this Question ?}\\ \text{Concept : The intensity of the wave is directly proportional to the square of the amplitude. i.e.}(I\propto A^2)\end{array}$$