The path difference is the difference between the distances travelled by two waves meeting at a point. Given the path difference, how does one calculate the phase difference?
[Physics] the difference between phase difference and path difference
terminologywaves
Related Solutions
Actually, one wave and its 180 flip phase image can be seen as two waves with the same phase but with opposite amplitude (when one is positive the other is negative).
The reason you cannot hear the difference between the two is that your ear (or microphone) is not sensitive to the amplitude but rather to the intensity of the wave, i.e. the square of the absolute value of the amplitude. As you can see, the difference in amplitude sign (or the phase) does not matter and both the original and its 180 flip phase image will sound the same.
What you would like to measure to see the difference between the two is the phase of the waves but as mentioned earlier, you cannot detect the phase directly. However you can measure the interference between the waves : when waves are emitted simulatenously they sum up. Now if the two waves have the same phase, then the amplitude add up (constructive interference). If the phases are 180 degree apart, then the amplitudes will cancel out (destructive interference, mentioned above). If the phase difference is somewhere inbetween, then you will get something ... inbetween (an interference pattern).
My textbook says that $\frac{\delta}{2\pi}=\frac{\Delta L}{\lambda}$, where $\delta$ is the phase difference and $\Delta L$ is so called path difference. But that's a cheat in my opinion. Even if the geometrical paths of two waves are equal it doesn't imply that their phases will be equal too. This leads me to a conclusion that $\Delta L$ is rather the optical path difference (OPD) than the actual difference in distance the waves have traveled.
Your textbook is describing a special case.
Consider a wave described by the equation
$$\xi = A\sin 2\pi\left(ft-\frac x \lambda\right) = A\sin (\omega t - \varphi)$$
where $x$ is the distance from the source of the wave.
From that we apparently have
$$\varphi = 2\pi\frac x \lambda$$
Now consider two points lying on a line, one with distance $x_1$ from the source and the other with distance $x_2$. Now the path difference is
$$\Delta L\equiv x_2 - x_1$$
and the phase difference
$$\delta\equiv\varphi_2-\varphi_1 = 2\pi\frac{\Delta L}\lambda$$
In vacuum, we obviously don't need to take index of refraction into account.
In case of the entire space being filled with environment with index of refraction $n\neq1$, it doesn't matter either, because we can either take the wavelength of the wave in the environment to be $\lambda$ (in which case the equation stays the same), or we'll take the wavelength in vacuum to be $\lambda$, then the wavelength in the environment becomes
$$\lambda_e = \frac\lambda n$$
and we have
$$\delta = 2\pi\frac{n\Delta L}{\lambda}=2\pi\frac{\text{OPD}}{\lambda}$$
where $\text{OPD}$ is optical path difference.
[L]et's assume we are given two waves $f(x,t)=A\cos{(kx-\omega t)}$ and $f'(x,t)=A'\cos{(k'x-\omega t)}$. Their phase difference at a given point $x$ would obviously be $$ \delta=(k'-k)x=\frac{2\pi}{\lambda}(n'-n)x $$
That's true if you have two waves, each of them traveling through distance $x$ in different environments, the first wave traveling through an environment with index of refraction $n$, and the second one through an environment with $n'$.
(If instead they traveled through the same path in the same environment, and they had the same wavelength in vacuum ($\lambda$), the index of refraction for them would be the same as well, so for two such waves described by your equations, $\delta = 0$.)
So the question is: how on earth can I show that
$$\sum\limits_{i=1}^k(n'_i-n_i)d_i=(n'-n)x$$
Since $x$ is the distance the waves traveled, you can rewrite the right side as
$$\sum\limits_{i=1}^k(n'_i-n_i)d_i=(n'-n)d$$
Now it's apparent that both sides are identical, except the left side takes into account the possibility of each wave travelling through $k$ environments.
Best Answer
Let's assume that, two stones are thrown at two points which are very near, then you will see the following pattern as shown in the figure below:
let's mark the first point of disturbance as $S_1$ and the other as $S_2$, then waves will be emanated as shown above. By having a cross-sectional view, you will see the same waves as shown in the figure below (in the below explanation wavelengths of waves emanated from two different disturbances is assumed to be the same).
The waves emanating from $S_1$ has arrived exactly one cycle earlier than the waves from $S_2$. Thus, we say that, there is a path difference between the two waves of about $\lambda$ (wavelength). If the distance traveled by the waves from two disturbance is same, then path difference will be zero. Once you know the path difference, you can find the phase difference using the formula given below:
$$\Delta{X}=\frac{\lambda\cdot\Delta{\phi}}{2\pi}$$
Here, $\Delta{X}$ is path difference, $\Delta{\phi}$ is phase difference.