# Difference between phase difference and path difference

1. Apr 12, 2007

### logearav

respected members,
could anyone explain the difference between path difference and phase difference ? what does the term phase exactly mean? thanks in advance

2. Apr 12, 2007

### lpfr

Path difference is the difference (in meters) between the lengths of two paths.

In periodic phenomena, the phase is the relative position in the cycle. That is if you divide all the cycle in 360 degrees (or better in 2 pi radians), and decide to start counting at a given position. Each time that the process attains the same position its phase will have advanced 360 degrees.
Take as periodic process a walking person, and place the start of phase when the right foot touches the earth. When the phase advances 180°, the left foot touches the earth. See why?
No take two persons walking beside. They start touching the earth with their right foot simultaneously: they are in phase, their phase difference is zero.
If they are parading soldiers they will stay in phase and their right foots will continue to touch the earth simultaneously.
But if they are not parading, the pace period of each one will be probably different. If the period of person B is shorter, his right foot will touch earth a little before person A. We say that person B is in advance on phase relative to A. Or that phase of A lags B. If the periods of each one stay so, the difference of phase will increase. Eventually B will be 180° in advance (or B 180° in retard).

When two waves have the same periodicity, that is same frequency, and travel at the same speed, if they travel the same distance they will keep the same phase difference. But if one of the waves has to travel a longer path it will be in retard of phase.

3. Apr 12, 2007

### Edgardo

Consider a lightsource that emits a monochromatic wave A(x,t),

Code (Text):

_         _
/   \     /   \
/     \   /     \
x/       \_/       \_...       Wave A(x,t)

<-----d--------->
_        _         _
/   \     /  \     /   \
/     \   /    \   /     \
x/       \_/      \_/       \....       Wave B(x,t)

and another lightsource emitting a wave B(x,t) below the first one that is shifted to the left by
the distance d. Both waves shall have the same frequency. This distance d is called path difference.
We say: These two waves have the path difference d.

What is the phase difference?
To answer this question we first have to look at the definition
of the phase. The phase occurs with the mathematical form of a wave:

$$A(x,t) = A_{0} \cdot sin(kx-\omega t)$$

Definition: The phase is the term in the brackets of the sin-function, so

$$\mbox{phase}=kx-\omega t$$

The phase is often denoted by the greek letter $$\phi$$,
so let us also write:

$$\phi = kx- \omega t$$

Thus, our wave can be written as:

$$A(x,t) = A_{0} \cdot sin(\phi) = A_{0} \cdot sin(kx-\omega t)$$

Now have a look at the second wave B(x,t). It is shifted to the left.
What does this shift mean?
Imagine that both waves travel to the right.
If you compare how far both waves travel, then you will notice
that the left-shifted wave travels farther by the path d.
This can be expressed mathematically:

$$B(x,t) = B_{0} \cdot sin(\phi_2) = B_{0} \cdot sin(k(x+d)-\omega t) = B_{0} \cdot sin(kx- \omega t +kd)$$

What is the phase difference? It's the difference between $$\phi_2$$ and $$\phi$$,
thus:

$$\mbox{phase difference} = \mbox{Phase of wave } B(x,t) - \mbox{Phase of wave } A(x,t)}$$
$$= \phi_2-\phi = (k(x+d)-\omega t)-(kx-\omega t) = kx+kd- \omega t-kx+ \omega t = kd$$

Result: For a path difference d, we get a phase difference
$$\phi_2-\phi=kd$$

We say The phase difference between the two waves is $\phi_2-\phi=kd$

But remember: The general definition of the phase difference is

$$\phi_2 - \phi$$

Questions for you:
(a) What happens with B(t) if $$kd = 2 \pi$$? Call the result B_new(t) (Hint: sin(x+2Pi) = sin(x))
(b) What is the path difference between B_new(t) and A(t)?

(c) Let $$k= \frac{2 \pi}{500 \mbox{nm}}$$. What is the phase difference between the two waves B(x,t) and A(x,t),
if I have a path difference of d=300 nm?
How do I have to choose the shift d such that the phase difference
is zero?

Last edited: Apr 12, 2007
4. Apr 16, 2007

### logearav

thanks a lot, edgardo and lpfr for your informative clarifications. i ll solve the problem and get back to you.

5. Nov 9, 2009

### Ref3t_Bekhit

thanx first for this valuable explanation :)