Question

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

• A. $\frac{\lambda }{8}$
• B. $\frac{\lambda }{4}$
• C. $\frac{2\lambda }{5}$
• D. $\frac{\lambda }{12}$

Answer 1

The correct option is B $\frac{\lambda }{4}$

The maximum intensity (constructive interference) in Quincke's tube is given by ,

$I_{max}=I_1+I_2+2\sqrt{I_1{I}_2}$ ,

but given $I_1=I_2=I\left(let\right)$ and $I_{max}=I_0$,

hence $I_0=I+I+2\sqrt{I\times I}=4I$ ,

or $I=I_0/4$ ,

now if the intensity is $=I_0/2$ ,

then $I_0/2=I+I+2\sqrt{I\times I}\cos \varphi$ ,

or $I_0/2=I_0/4+I_0/4+2\sqrt{I_0/4\times I_0/4}\cos \varphi$ ,

or

$\cos \varphi =0=\cos \pi /2$ ,

or $\varphi =\pi /2$ (phase difference) ,

therefore path difference ,

$\Delta x=\lambda /2\pi \times \pi /2=\lambda /4$ ,

to produce a path difference of $\lambda /4$ , sliding tube should be

moved by a distance of $\lambda /8$ .