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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?

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  • $\begingroup$ It is a homework type - atleast not thinking booster question. $\endgroup$
    – Dvij D.C.
    Feb 9 '14 at 13:07
  • $\begingroup$ Pun not intended $\endgroup$ Sep 8 '17 at 14:41
  • $\begingroup$ You need to be alert to the role of refractive index when considering this, because for any given wave frequency, the wavelength depends on the refractive index. $\endgroup$ Dec 27 '20 at 20:02
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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:
enter image description here

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).

enter image description here

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.

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path difference is the difference in path traversed by the two waves , measured in terms of wavelength of the associated wave. It has a direct relation with phase difference. Phase difference decides the nature of interference pattern but phase difference is found out by path difference. Phase difference is related to quantum mechanics. If path difference b/w 2 waves is integral multiple of wavelength, which satisfies condition for constructive interference. Whereas, if path difference b/w 2 waves is odd multiple of half wavelength, it satisfies condition for destructive interference.

whereas phase difference is the difference between some reference point in 2 waves. Its how much one wave is shifted from the other. For eg. at the origin, if one wave's displacement is zero and the other one has some displacement, then there is a shift, which is their phase difference. For eg. sine wave is zero at the origin, but cos wave is not zero at the origin, its zero at pi/2. So the phase difference is pi/2. Infact, cos wave is just the sine wave phase-shifted.

If two waves have zero phase difference, then their crests occur at the same time and so do their troughs. Its like moving together. They will add up (constructive interference).

For only one wave, phase difference means how much the wave is shifted from the origin. Like in your example, Acos(wt + Φ), at time t=0, cos has some value other than 1. So, the cos wave is slightly shifted by Φ. Physically, what this means is when u started measuring the time, the system oscillates, not from either origin or amplitude, but somewhere in between.

Ofcourse, the concept of phase difference is not useful if there is only one wave. You could always choose time to start from when the system is at an extreme position. Phase difference concept is useful when there are atleast 2 waves, then when u start the time when one wave starts from zero, the other wave could already be in some other position.

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  • $\begingroup$ Phase difference and path difference are the same thing but for different scale of same wave $\endgroup$
    – Jdeep
    Dec 27 '20 at 18:18
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Consider two waves coming from different places and arriving at the same point. Also, let those two waves initially be in phase (i.e when one wave is at a maximum so is the other one). Furthermore, let the wavelength of each wave be the same (i.e. the distance between consecutive maximums).

Now let the first wave travel a distance equal to ten times its wavelength. Let the second wave travel a distance equal to 10.5 times its wavelength. That means that upon arriving at the same point, if the first wave is at a maximum then the second wave is at a minimum. We call this 180 degrees out of phase.

From this simple case, we can see that if both waves travel an integer number of wavelengths before interfering with each other, then the two waves will be "in phase" (assuming that both waves started out in phase). However, if one wave travel's some fraction of a wavelength more than the other than there will be a phase difference $\phi=2\pi * \text{(path difference)}/\text{wavelength}$. This formula satisfies the simple example above and is easier to see if you draw out some cosine or sine waves (note that a cosine wave is just a sine wave phase shifted by $\pi/2$.

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  • $\begingroup$ This seems to be more of a discussion about phase difference than path difference...The important thing is that you know the difference in numbers of wavelengths, assuming the waves are coherent, or in meters. $\endgroup$
    – sqrtbottle
    Apr 16 '15 at 20:00
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Path difference and Phase difference are very similar things. Let me show this in a bit more intuitive way!

In our world, we have "good" waves and "bad" waves (atleast for calculations)

For example :

  • A good wave is y=sinx plotted on a graph having crests and troughs at $\frac{n\pi}{2}$ alternatively.

sinewave


  • A bad wave is basically what I like to call a "light wave". I mean light waves do look like sine waves (amplitude=1, wavelength=$2\pi$) but they have different amplitudes and wavelengths.

badwave


What happens to a bad wave?

Well, we really love to work with $\pi$ rather than with plain numbers like $632nm$. You will get to know why.Just read till the end.

Suppose a physicist is given two coherent sources of light of wavelength $600nm$ (lets say, for performing Y.D.S.E).

Now, what physicists secretly do is that they do a mathematical transformation of the light wave (to make a new wave) in such a way that the wavelength of the light wave is made equal to $2\pi$ units


pux


Now the physicist does all his calculations on the new transformed wave. Suppose he transformed the light waves of the two sources in the similar way and finds that the difference between two nearest crests of the two waves is $\pi$ (a.k.a phase difference)


final pic


What the physicist found is called as phase difference of two waves. But as you know, the physicist is doing all his calculations on a transformed wave. Meaning, that the real life phase difference of the wave will be very different than the calculated value $\pi$.

So our task is to get the "real life phase difference". We use pretty basic unitary methods to find the relation between the phase difference of our transformed wave and the real wave.

Since $600nm$ (wavelength of real light) in real light corresponds to $2\pi$ in transformed wave, therefore ,

$\Delta x$ phase difference in the real waves is equal to $\frac{2\pi\phi}{\lambda}$ where $\phi$ is the phase difference of transformed waves.

Now, the moment of truth. This $\Delta x$ that you calculated is the phase difference of real waves. This is also known by the term "path difference"

Therefore path difference is nothing but phase difference of the transformed waves. Phase difference is just an aid for ease of calculations. The real thing that matters to get correct results is the path difference.

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The optical path difference is the length difference $d$ (dimensions of length: $[L]$) in the paths travelled by two different rays from one plane, at which they are typically assumed to have the same phase, to a second location (typically another plane, surface or point).

Light is a wave with a wavelength of $\lambda$, or corresponding angular spatial frequency of $k = 2\pi/\lambda$ (dimensions of inverse length: $[L^{-1}]$). As a wave propagates through space it accumulates phase at a rate of $k$.

Thus to convert optical path difference to optical phase difference we multiply by $k$:

$$ \phi = kd. $$

Note that when $d=\lambda$ the phase difference is $\phi=2\pi$ as expected.

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Suppose two waves, from S1 and S2 respectively, are meeting at a point P. The path difference will be equal to S2P-S1P. Note that the difference is always a whole number of wavelengths i.e. Nx(wavelength)

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    $\begingroup$ What are S2P and S1P? Note that the site has MathJax enabled, so you can write equations as Feynman did in his answer above. $\endgroup$
    – Kyle Kanos
    Oct 20 '15 at 11:40
  • $\begingroup$ @KyleKanos he assumed basic and standard notations in the general ydse setup. S1 and S2 are sources or the slits from which waves emanate and P is a point on the screen. $\endgroup$
    – OhMyGauss
    May 20 '20 at 2:14

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