Question

Sound of wavelength $ \lambda $ passes through a Quincke’s tube, which is adjusted to give a minimum intensity of $ {I_0} $ .The two interfering waves have equal intensity. Find the minimum distance through the sliding tube should be moved to give an intensity $ \dfrac{{{I_0}}}{2} $
(A) $ \dfrac{\lambda }{8} $
(B) $ \dfrac{\lambda }{4} $
(C) $ \dfrac{{2\lambda }}{5} $
(D) $ \dfrac{\lambda }{{12}} $

Answer 1

Hint: For a Quincke’s tube the maximum intensity of sound occurs only during constructive interference. The intensity of two sound waves in interference is given by $ I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $ . Also, the path difference is related to the phase difference by $ \Delta x = \dfrac{\lambda }{{2\pi }}\Delta \phi $ .

Formulas Used: We will be using $ I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $ where $ I $ is the intensity of interference, $ {I_1} $ and $ {I_2} $ are the intensity of the waves that are involved in the interference and $ \phi $ is the phase angle at which the interference occurs. We will also be using $ \Delta x = \dfrac{\lambda }{{2\pi }}\Delta \phi $ where $ \Delta x $ is the path difference, $ \lambda $ is the wavelength of the wave and $ \Delta \phi $ is the phase difference.

Complete Step by Step solution
A Quincke’s tube is an experimental setup that is used to demonstrate sound interference. A single sound source is passed on through one slit to divide and interfere before being detected at the other slit in the tube. Thus, interference is demonstrated by keeping one of the U tubes used in the experiment fixed and the other movable.
However, we know that the intensity of interference is given by $ I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $ . Also the maximum intensity of interference occurs at phase angle $ \phi = 0 + 2\pi n $ for any integer $ n $ . Thus maximum intensity of interference is given by $ {I_{\max }} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $ [since $ \cos (0 + 2\pi n) = 1 $ ].
Given the maximum intensity $ {I_{\max }} = {I_0} $ and both the interfering waves have equal intensity $ {I_1} = {I_2} = I $ , now rewriting the equation for maximum intensity of interference, $ {I_0} = I + I + 2\sqrt {I \times I} \cos (0 + 2\pi n) $
 $ \Rightarrow {I_0} = 2I + 2\sqrt {{I^2}} (1) $
Since $ \cos (2\pi n) = 1 $ and $ I \times I = {I^2} $ . Simplifying the equation further we get,
 $ {I_0} = 4I $
 $ \Rightarrow \dfrac{{{I_0}}}{4} = I $
Now we must find the intensity of interference for $ \dfrac{{{I_0}}}{2} $ . Rewriting the intensity equation with $ I = \dfrac{{{I_0}}}{2},{I_1} = \dfrac{{{I_0}}}{4} $ and $ {I_2} = \dfrac{{{I_0}}}{4} $ we get,
 $ \dfrac{{{I_0}}}{2} = \dfrac{{{I_0}}}{4} + \dfrac{{{I_0}}}{4} + 2\sqrt {\dfrac{{{I_0}}}{4}\dfrac{{{I_0}}}{4}} \cos \phi $
Simplifying the equation, we get,
 $ \Rightarrow \dfrac{{{I_0}}}{2} = \dfrac{{{I_0} + {I_0}}}{4} + 2\sqrt {\dfrac{{{I_0}}}{4}\dfrac{{{I_0}}}{4}} \cos \phi $
 $ \Rightarrow \dfrac{{{I_0}}}{2} = \dfrac{{2{I_0}}}{4} + 2\sqrt {\dfrac{{{I_0}}}{4}\dfrac{{{I_0}}}{4}} \cos \phi $
Taking $ \dfrac{{2{I_0}}}{4} $ to the L.H.S,
 $ \Rightarrow \dfrac{{{I_0}}}{2} - \dfrac{{2{I_0}}}{4} = \dfrac{{2{I_0} - 2{I_0}}}{4} = 2\sqrt {\dfrac{{{I_0}}}{4}\dfrac{{{I_0}}}{4}} \cos \phi $ [ taking L.C.M to the L.H.S]
 $ \Rightarrow 0 = 2\dfrac{{{I_0}}}{4}\cos \phi $ [Since $ \sqrt {\dfrac{{{I_0}}}{4}\dfrac{{{I_0}}}{4}} = \dfrac{{{I_0}}}{4} $ ]
So, now there can be two possibilities such that the R.H.S equates to 0, either $ {I_0} = 0 $ or $ \cos \phi = 0 $ . However, $ {I_0} = 0 $ is not possible because it is the maximum intensity of interference. So, $ \cos \phi = 0 $ .
 $ \cos \phi = 0 $ only when $ \phi = \dfrac{\pi }{2} $ .
 $ \Rightarrow \Delta \phi = \dfrac{\pi }{2} - 0 = \dfrac{\pi }{2} $
Using the phase difference $ \Delta \phi $ we can find path difference by $ \Delta x = \dfrac{\lambda }{{2\pi }} \times \dfrac{\pi }{2} $
 $ \Rightarrow \Delta x = \dfrac{\lambda }{4} $
Thus the minimum distance that the wave should travel to give intensity $ \dfrac{{{I_0}}}{2} $ is $ \Delta x = \dfrac{\lambda }{4} $ .
So the correct answer is option B.

Note
The interference of the waves and its intensity are affected by a large factor when there is a change in the phase angle between the waves. Also here $ {I_1} = {I_2} = I $ , hence the intensity is affected only by the phase angle $ \phi $ .