Question

Two sound waves having a phase difference of \[{60^0}\] have path difference of
A. $2\lambda $.
B. $\dfrac{\lambda }{2}$.
C. $\dfrac{\lambda }{6}$.
D. $\dfrac{\lambda }{3}$.

Answer 1

Hint: In this question, when a phase difference is given, we will use the method of finding the path difference of two waves. We're going to use the formula; $\vartriangle x = \dfrac{\lambda }{{2\pi }}\vartriangle \phi $ where $\vartriangle x$ is the difference in path and where $\vartriangle \phi $ is the difference in phase.

Formula used: $\vartriangle x = \dfrac{\lambda }{{2\pi }}\vartriangle \phi $.

Complete step-by-step solution -

Given that, the difference in phase, $\vartriangle \phi $ = \[{60^0}\].

We need to find out the path difference.
We know that the difference in phase angle of two waves is the phase difference and the difference in direction is the direction that the waves follow.
The relation between the difference in path and the difference in phase is given by,
$ \Rightarrow \vartriangle x = \dfrac{\lambda }{{2\pi }}\vartriangle \phi $, …….(i)

Where $\vartriangle x$ is the path difference and $\vartriangle \phi $ is the phase difference. In order to figure out the path difference, we should note that the units must be the same in the relation. We have phase differences in degrees here. So we will convert $2\pi $ in degrees (which is in radian).
$ \Rightarrow 2\pi = 2\pi \times \dfrac{{{{180}^0}}}{\pi } = {360^0}$

Following this, equation (i) will become
$ \Rightarrow \vartriangle x = \dfrac{\lambda }{{360}}\vartriangle \phi $
Then, now
$
   \Rightarrow \vartriangle x = \dfrac{\lambda }{{360}} \times 60 \\
   \Rightarrow \vartriangle x = \dfrac{\lambda }{6} \\
$
Thus, we can see the path difference is $\dfrac{\lambda }{6}$ metre.
So option (C) is the right one.

Note: We will use the relation between the phase difference and the path difference in this type of question. When one of them is provided we can easily identify any one of them. We should remember that the units must be the same as in the question. The difference in step will be either in radian or in degree only. Accordingly we can modify the formula. We can get the required answer by putting the values in.