Question

Two waves of equal frequencies moving in the same direction have their amplitudes in the ratio $7:5$. They are superimposed on each other. Find the ratio of maximum to minimum intensity of the resultant wave.

• A. $16:25$
• B. $25:16$
• C. $36:1$
• D. $1:36$

Answer 1

The correct option is C $36:1$

Given that,
$\frac{A_1}{A_2}=\frac{7}{5}$
where, $A_1\to$ Amplitude of wave $1$
$A_2\to$ Amplitude of wave $2$

We know that,
Intensity of wave, $I\propto A^2$
Let $I_1$ and $I_2$ be the intensities of wave $-1$ and wave $-2$ respectively.
Then, $\sqrt{\frac{I_1}{I_2}}=\frac{7}{5}......\left(1\right)$

After superimposition of two waves,
Maximum intensity is given by
$I_{max}\propto (\sqrt{I_1}+\sqrt{I_2}{)}^2....(2)$
Minimum intensity is given by
$I_{min}\propto (\sqrt{I_1}-\sqrt{I_2}{)}^2....(3)$

From $\left(2\right)$ and $\left(3\right)$, we get
$\frac{I_{max}}{I_{min}}={\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)}^2$

Using $\left(1\right)$ in the above equation,
$\frac{I_{max}}{I_{min}}={\left(\frac{\sqrt{\frac{I_1}{I_2}}+1}{\sqrt{\frac{I_1}{I_2}}-1}\right)}^2$
$\Rightarrow \frac{I_{max}}{I_{min}}={\left(\frac{\frac{7}{5}+1}{\frac{7}{5}-1}\right)}^2={\left(\frac{\left(\frac{12}{5}\right)}{\left(\frac{2}{5}\right)}\right)}^2$
$\Rightarrow \frac{I_{max}}{I_{min}}=\frac{36}{1}$

Thus, option (c) is the correct answer.