# Phase difference and optical path difference (OPD)

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 ($$\textrm{OPD}$$) than the actual difference in distance the waves have traveled. Hence we arrive to a following formula: $$\delta=2\pi \frac{\textrm{OPD}}{\lambda}.$$ The problem is (assuming that my reasoning is valid) I have no idea how to prove it. Here's my attempt: let'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 seems totally cool, except for one thing. The $$\textrm{OPD}$$ is defined as $$\sum_{i=1}^k(n'_i-n_i)d_i$$ where $$d_i$$ is the distance a wave $$f$$ travels in environment with index of refraction equal to $$n_i$$. So the question is: how on earth can I show that $$\sum_{i=1}^k(n'_i-n_i)d_i=(n'-n)x$$ ?

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.