Question

Phase difference ( $\varphi$) and path difference ( $\delta$) are related by
A.)${\displaystyle \frac{2\pi }{\lambda }}\delta$
B.)${\displaystyle \frac{\pi }{2\lambda }}\delta$
C.)${\displaystyle \frac{\lambda }{2\pi }}\delta$
D.)${\displaystyle \frac{2\lambda }{\pi }}\delta$

Answer 1

Hint: The ratio of phase difference $\varphi$ to the total angle ${360}^0(2\pi )$ and that of path difference $\delta$to the wavelength $\lambda$ remains constant. Convert phase angle to radians before substituting.

Formula used:
$\varphi ={\displaystyle \frac{2\pi }{\lambda }}\delta$ Where, $\varphi$denotes the phase angle difference , $\lambda$ denotes the wavelength of the wave $2\pi$ is the total change in angle after travelling a path difference of 1$\lambda$ and $\delta$ shows the path difference.

$y=A\sin (\omega t-kx)$ here $y$ shows the displacement of the progressive wave, $A$ denotes the maximum amplitude ,$\omega$ denotes the angular velocity ,$t$ shows the time ,$k$ denotes the wave constant and $x$ denotes the distance.

Complete step by step answer:
The general equation of a wave can be represented by a sinusoidal equation.
$y=A\sin (\omega t-kx)$
Let us consider two points from a wave. If $x_1$ is the distance of the first point and $x_2$ is the distance of the second point. Their path difference is given by the equation $\delta =x_2-x_1$
Substituting the value of $x_1$ and $x_2$ in the equation of wave phase of one point is $\omega t-k{x}_1$ and of the point is $\omega t-k{x}_2$ Now ,by calculating their difference we get the phase difference
$\varphi =(\omega t-kx{}_1)-(\omega t-k{x}_2)$
$\varphi =k(x_2-x_1)$
By comparing the path difference and phase difference we can see that ${\displaystyle \frac{\varphi }{\delta }}=k$
Wavelength is defined as the length between the points having the same phase angle.
From trigonometry we know that the value of an angle repeats after every $2\pi$ radians or ${360}^0$.
So, $A\sin (\omega t-kx)=A\sin (\omega t-kx+2\pi )$
 $A\sin (\omega t-kx)=A\sin (\omega t-k(x-{\displaystyle \frac{2\pi }{k}}))$
We know that its path difference is $\lambda$.Therefore $x-(x-{\displaystyle \frac{2\pi }{k}})=\lambda$
${\displaystyle \frac{2\pi }{k}}=\lambda$
Therefore $k={\displaystyle \frac{2\pi }{\lambda }}$
Using this in equation ${\displaystyle \frac{\varphi }{\delta }}=k$ we can say ${\displaystyle \frac{\varphi }{\delta }}={\displaystyle \frac{2\pi }{\lambda }}$
The correct option is A.

Note:
This equation can be used in the cases where the waves are travelling in the same media. Because as the medium varies the path difference changes. As a wave travels to a denser medium, it slows down and its wavelength decreases. The frequency remains constant.