I am trying to figure out if Snell's Law for refraction can be derived from Scalar Diffraction Theory.

The setup is this: light (plane wave, with wave vector $\vec k_i = (k_x, k_y, k_z)$ ) falls on a flat interface which is taken to be x-y plane; the incident side has refractive index $n_i$ . I want to figure out the refracted ray wave vector $\vec k$, on the otherside of the interface, which has refractive index $n$. Of course, I assume that the magnitutes of wave vectors are proportional to the refractive indices of respective media; i.e. $ |k|/|k_i| = n/n_i$.

I start with the Fresnel Diffraction Integral, with plane wave light $$U(\xi,\eta,z=0) = \exp( i \left[k_x \xi + k_y \eta + 0 \right] )$$ and got to the point of showing that $$U(x,y,z) = \exp( i \left[ k_x x + k_y y + k_z^{\prime} z \right] ),$$ where $$ k_z^{\prime} = k - \frac{k_x^2 + k_y^2}{2k}. $$

I was happy to see that this field has the same $k_x$ and $k_y$ values, but no matter how I play with $k_z^{\prime}$, I am unable to show that it is consistent with Snell's Law.

Any pointers?

  • $\begingroup$ @George: Thank you, I have clarified the setup in which the calculations are made, where in the I have indicated where the refractive indices are used. I am aware of the elegant derivation using Fermat's "least time" principle; I am exploring/learning Scalar Diffraction Theory and wanted to see if I can show this! :) $\endgroup$ Oct 9, 2013 at 4:15
  • $\begingroup$ Prabhakar, forgive me if my comment sounded dismissive; it wasn't meant to be. My experience all comes from actual commercial industry; which usually focusses on the simplest way to a solution (less costly too). But if your aim was to understand use of the diffraction theory, that's a different issue. I would also second the mention of Born and Wolfe. One might almost say, if it isn't in there, you don't need to know it ! $\endgroup$
    – user26165
    Oct 9, 2013 at 20:00

3 Answers 3


In scalar wave theory you're simply going to impose the continuity of the scalar field across the boundary.

The reasoning is the same whether or not you consider reflexions, whose magnitude and form can be gotten from scalar theory by imposing a further boundary condition of continuous normal derivatives across the interface.

With or without reflexions, the scalar variation along the interface on the incidence side is $\exp(i\,k\,n_i\,\sin\theta_i\,x)$, where $n_i$ is the refractive index of the medium on the incoming side of the interface, $k$ the freespace wavenumber and $x$ is measured along the interface plane in the plane of incidence (plane containing wavevector and interface normal). The $\sin\theta_i$ factor arises simply because the wavevector is skewed relative to the interface normal: it is the component of the wavevector in the interface plane.

By the same reasoning, the scalar variation along the interface on the output (transmission) side is simply $\exp(i\,k\,n_t\,\sin\theta_t\,x)$, where $n_t$ is the refractive index on that side.

So, for continuity of the scalar field, we must have:

$$\exp(i\,k\,n_i\,\sin\theta_i\,x) = \exp(i\,k\,n_t\,\sin\theta_t\,x),\;\forall x\in\mathbb{R}$$

which can only be true if

$$n_i\,\sin\theta_i = n_t\,\sin\theta_t$$

which is of course Snell's Law.

The full, vector field picture is to be found in the derivation of the Fresnel equations as given, for example, in section 1.5 of the sixth edition of:

Born and Wolf, "Principles of Optics"

however the simpler derivation above is more general, since it works for scalar waves such as sound, too.

  • $\begingroup$ Thank you, much appreciated. However, I am unable to make out the wave you have used in $exp(i k n_i,\sin θ_i x)$; should I take this to mean $\exp ( i k \hat{n}_i + \hat{x} \sin \theta_i )$ ? $\endgroup$ Oct 10, 2013 at 7:35
  • $\begingroup$ Scratch that; I should I take: $\exp(i\,k\,n_i,\,\sin\theta_i\,x)$ means $\exp ( i k n_i (\hat{z} \cos \theta_i + \hat{x} \sin \theta_i ))$ , where $\hat z$ and $\hat x$ are normal and tangential unit vectors. If yes, everything makes sense. $\endgroup$ Oct 10, 2013 at 7:44
  • $\begingroup$ @PrabhakarBhimalapuram Exactly, although we are only looking at variation within the plane, so $z = 0$. I have corrected a couple of rogue commas in my equations - they may read better now. $\endgroup$ Oct 10, 2013 at 7:54

Well I'm not sure where in your math the refractive indices of the media are hidden, or even the propagation velocities, so it's not clear to me that Snell's law should fall out of your diffraction theory.

Snell's law can commonly be obtained from Fermat's principle, by simply solving for the minimum path length across a media boundary.


Answering my own question: George's comment and Rod's answer, while not the exact answer, directly lead to me figure out the missing was the "math" steps! (which is below)

The Fresnel Diffraction Formula basically (somewhere in its derivation) Taylor expands square root i.e. $r=\sqrt(x^2 + y^2 + z^2) = z (1 + \frac{x^2 + y^2}{2z^2})$, with an approximation $x^2 + y^2 << z^2$. So what I had for normal component of the wave-vector after refraction was $ k - \frac{k_x^2 + k_y^2}{2k}$ basically the same approximation for $\sqrt{k^2 - (k_x^2 + k_y^2)}$, showing that the the wave-vector after refraction is $(k_x, k_y, \sqrt{k^2 - (k_x^2+k_y^2})$, where the wave vector before refraction was $(k_x, k_y, k_z)$. Given that $\sqrt{k_x^2+k_y^2+k_z^2}/k = n_i/n$, the wave vectors satisfy Snell Law for refraction!


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