Question

Sound of wavelength $\lambda$ passes through a Quincke’s tube, which is adjusted to give a maximum intensity $I_0$. Through what distance should the sliding tube be moved to give an intensity $\frac{I_0}{2}$?

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

Answer 1

The correct option is D

$\frac{\lambda }{8}$

$(\sqrt{I}+\sqrt{I}{)}^2=I_0\Rightarrow I=\frac{I_0}{4}$

Now,

$\frac{I_0}{2}=\frac{I_0}{4}+\frac{I_0}{4}+2\frac{I_0}{4}cos\varphi$

$\Rightarrow cos\varphi =0\&\varphi =\frac{\pi }{2}$

Phase difference

$\varphi =\frac{2x}{\lambda }.2\pi$

$\frac{\pi }{2}=\frac{2x}{\lambda }.2\pi$

$x=\frac{\lambda }{8}$