Question

Sound of wavelength $\lambda$ passes through a Quincke's tube, which is adjusted to give a maximum intensity $I_o.$ The two interfering waves have equal Intensities. Find the minimum distance the sliding tube should be moved to give an intensity $\frac{I_o}{2}.$

• A. $\frac{\lambda }{8}$
• B. $\frac{\lambda }{2}$
• C. $\frac{\lambda }{4}$
• D. $\frac{3\lambda }{2}$

Answer 1

The correct option is A $\frac{\lambda }{8}$

The maximum intensity occurs during constructive interferece in Quincke's tube.

For two wave interference, we know that,

$I_r=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi ..........\left(1\right)$

But, given that $I_1=I_2=I\left(\text{say}\right)$

So, $I_{max}=4I[\because \varphi =0]$

From the data given in the question,

$I=\frac{I_{max}}{4}=\frac{I_o}{4}.........\left(2\right)$

Given that, required intensity $I_r=\frac{I_o}{2}$

From $\left(1\right)$ we can write that,

$\frac{I_o}{2}=I+I+2\sqrt{I\times I}\cos \varphi$

Using $\left(2\right)$ in the above equation we get,

$\cos \varphi =\frac{\frac{I_o}{2}-\frac{I_o}{2}}{2\sqrt{\frac{I_o}{4}}\sqrt{\frac{I_o}{4}}}=0$

$\Rightarrow \varphi =\frac{(2n-1)\pi }{2}\text{where}n=1,2,3,\mathrm{4.....}$

We know that,
Path difference $\left(\Delta x\right)=\frac{\lambda }{2\pi }\times$ Phase difference $\left(\varphi \right)$
$\therefore \Delta x=\frac{\lambda }{2\pi }\times \frac{(2n-1)\pi }{2}=\frac{(2n-1)\lambda }{4}$

For two successive points of constructive interference, in Quincke's tube, the path difference $\Delta x=2x$, where $x$ is the distance moved by tube.

So, the minimum distance the sliding tube should be moved is,
for $n=1$, $x=\left(\frac{\Delta x}{2}\right)=\frac{\lambda }{8}$