Question

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

Answer 1

Hint: The ratio of phase difference $\phi $ 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:
$\phi = \dfrac{{2\pi }}{\lambda }\delta $ Where, $\phi $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
$\phi = (\omega t - kx{}_1) - (\omega t - k{x_2})$
$\phi = k({x_2} - {x_1})$
By comparing the path difference and phase difference we can see that $\dfrac{\phi }{\delta } = k$
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, $A\sin (\omega t - kx) = A\sin (\omega t - kx + 2\pi )$
 $A\sin (\omega t - kx) = A\sin (\omega t - k(x - \dfrac{{2\pi }}{k}))$
We know that its path difference is $\lambda $.Therefore $x - (x - \dfrac{{2\pi }}{k}) = \lambda $
$\dfrac{{2\pi }}{k} = \lambda $
Therefore $k = \dfrac{{2\pi }}{\lambda }$
Using this in equation $\dfrac{\phi }{\delta } = k$ we can say $\dfrac{\phi }{\delta } = \dfrac{{2\pi }}{\lambda }$
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.