Question

In Quincke's experiment, the sliding tube is pulled out by such a distance that sound intensity is changed from minimum to maximum. Find the ratio of the amplitudes of the two waves arriving at the detector, if maximum intensity is 16 times that of the minimum intensity.

• A. 1 : 2
• B. 2 : 1
• C. 4 : 5
• D. 5 : 3

Answer 1

The correct option is D 5 : 3

Resultant intensity of two waves of intensities $I_1$ and $I_2$ is:
$I_{res}=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$

For $I_{max},\cos \varphi =1$ and for $I_{min},\cos \varphi =-1$
So, $I_{max}=I_1+I_2+2\sqrt{I_1{I}_2}$
$=(\sqrt{I_1}+\sqrt{I_2}{)}^2$
and $I_{min}=(\sqrt{I_1}-\sqrt{I_2}{)}^2$
From question,
$\frac{I_{max}}{I_{min}}=16$
$\Rightarrow \frac{(\sqrt{I_1}+\sqrt{I_2}{)}^2}{(\sqrt{I_1}-\sqrt{I_2}{)}^2}=16\Rightarrow \frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}=4$
$\Rightarrow \sqrt{I_1}+\sqrt{I_2}=4\sqrt{I_1}-4\sqrt{I_2}$
$\Rightarrow 3\sqrt{I_1}=5\sqrt{I_2}$
$\Rightarrow \frac{\sqrt{I_1}}{\sqrt{I_2}}=\frac{5}{3}$
and we know, $I\propto A^2$
$\Rightarrow \sqrt{I}\propto A$
That means
$\frac{\sqrt{I_1}}{\sqrt{I_2}}=\frac{A_1}{A_2}=\frac{5}{3}$
So, $A_1:A_2=5:3$