Earlier I asked whether diffraction could bend light more than 90 degrees.

earlier question can light be diffracted backward?

Lambda gave me an answer. He stuck a razor blade into a piece of wood, and he shone a green laser at it, and he photographed the green light shining from the razor blade at an acute angle. It had to have bent more than ninety degrees.

Lambda's picture

This is the picture he posted. There is a small glowing green dot on the back side of the razorblade, seen from behind and above. This light has somehow looped around the edge of the blade.

I tried it myself. I got a cheap laser from Petco, and used a c-clamp to keep it turned on and to provide a steady stand for it. I stuck a single-edge razor blade into a piece of wood, turned it so it was 45 degrees, and aimed the laser at it so that about half of the laser light was blocked by the razor blade. When I looked at the razor blade from an angle more than 90 degrees, it shone red at the spot the laser hit it.

My laser or my camera (one or the other or both) weren't good enough to photograph it, but I could see it clearly. I could see the red dots on the razor blade when my head was at extreme angles and also up and down out from various skew angles. I could see them from at least 270 degrees, the light I saw was clearly going rather much backward.

When I put the razor blade at a distance, it looked like there were two bright spots at the edges of the laser beam, right where it seemed to go from bright to dark. When the distance from the laser was shorter, it looked more like four bright spots. That might be because my cheap laser has a dim second beam that makes a circle of light offcenter. Maybe it shows up when it's close, and gets either dim or off at some angle from a bigger distance.But when I rotated the laser 90 degrees (which would change its polarization, assuming it's polarized), then it looked like there were a lot of bright spots. I'm not sure. It might be something about how the razor is honed.

I tried putting two razor blades into the wood, at right angles with a small distance between the blades at the bottom and with them touching at the top. When I aimed the laser just right near the very top edge I could see a diffraction pattern. So the laser is probably pretty much OK. I didn't play with that much because everything had to be lined up just right, and I wasn't sure whether the two shiny razors might do weird reflections and/or they might not be lined up just right which might cause a weird final reflection.

What Lambda showed is definitely real. I'm not sure how to explain it. Is it really diffraction? Could it be something about how razor blades are ground, that allows a special reflection? Or something else?

Here is the diffraction pattern expected if the razor blade is at 90 degrees to the light.

fresnel edge

We get some light bleeding around the edge, and there could be some visible at 45 degrees, and maybe tiny amounts of light extending almost 90 degrees to the edge of the screen. Not very much.

With the razor blade itself at 45 degrees it's more like this:

weird diffraction

There is light that would not hit the screen at all. The simple descriptions of Fresnel diffraction do not discuss this possibility.

Is this diffraction or something else?

If it is diffraction, are there known formulas that say how much light to expect at extreme angles?

I'm very excited, but it might turn out to be something trivial.

  • $\begingroup$ A photo and a diagram would be a very big help. $\endgroup$
    – S. McGrew
    Dec 10, 2018 at 4:12
  • $\begingroup$ the image at the top should be lower, I think.as it describes your experiment? $\endgroup$
    – anna v
    Dec 10, 2018 at 5:22
  • $\begingroup$ What is your question? (Other than the one already answered.) $\endgroup$
    – safesphere
    Dec 10, 2018 at 5:52
  • $\begingroup$ My question is how to explain this. Is it diffraction or something else? Is it an example of some fundamental phenomenon, or is it just a trick with razor blades? $\endgroup$
    – J Thomas
    Dec 10, 2018 at 14:43

3 Answers 3


From the diagram, I think you are making a point source geometry

I tried putting two razor blades into the wood, at right angles with a small distance between the blades at the bottom and with them touching at the top.

You are generating a coherent wavefront spherically symmetric, so it is not surprising it will go over 90 degrees.

  • $\begingroup$ I saw diffraction when I did that, but I didn't look for high-angle diffraction that way, it took too much precision. The observed high-angle diffraction was with a single edge. When I look at intro texts about diffraction, I see formulas to predict the intensity at angles less than 90 degrees. Pick a point on a flat wall that the diffraction pattern arrives on, and the formulas predict the light intensity. Fraunhoffer formulas do it with simplifying assumptions, Fresnel do it more exactly. I don't see formulas to predict light diffracting backward, toward 180 degrees. $\endgroup$
    – J Thomas
    Dec 10, 2018 at 14:49
  • $\begingroup$ the solution has to be a spherical wavefront,imo. in the extra diagrams you introduced, the tip becomes a point source., maybe because it is metal the irregularities in the lattice do that? try painting the edge black? $\endgroup$
    – anna v
    Dec 10, 2018 at 17:06
  • $\begingroup$ or the edge a cylindrical wavefront $\endgroup$
    – anna v
    Dec 10, 2018 at 17:15
  • $\begingroup$ I'm surprised. I didn't see anything about this from the Frauhofer formula or from the Fresnel formula. Did I miss it? How would I find out more about a cyllindrical wavefront diffracting backward? $\endgroup$
    – J Thomas
    Oct 30, 2019 at 15:29

I was thinking of the razor blade as a very thin straight edge. But real razor blades are not straight.

Here is a photo of one edge. It has peaks around 1 micrometer apart. A bit larger than one wavelength.

enter image description here

Some pictures of double-edge razors showed the striations at sub-micron distances. This is not necessarily quite how my razor blade works.

But suppose it is. If light that hits those peaks scatters at all angles rather than reflecting at the angle of incidence, that would get the result we see.

Possibly light that travels THROUGH those valleys is getting the single-slit treatment, or I guess there are thousands of slits. That doesn't explain it diffracting more than 90 degrees, but it could do things I wasn't expecting.

These are possible explanations, but there's no certainty that they are important. I should try to get a straight edge. Possibly a piece of glass or rock with a conchoidal fracture. But I'd need it to be opaque. Maybe something else with a conchoidal fracture. Those can be very smooth but might not be.

enter image description here


TL;DR: Yes, it is indeed diffraction, and it's perfectly normal that it reaches all the way backwards.

Your general setup is known as the Sommerfeld diffraction problem. It has an analytical solution in terms of error function $\operatorname{erf}$ (on a complex line of its domain) or Fresnel integrals (with real-valued parameters).

Consider a monochromatic plane with a wave vector $\begin{pmatrix}k\cos\theta_0\\ k\sin\theta_0\end{pmatrix}$ incident on a semi-infinite perfect reflector (a knife) at $y=0,$ $x>0$. The space at $x<0$ is free. Introducing polar coordinates $r\ge0$ and $\theta\ge0,$ and choosing $\theta_0$ so that it's also non-negative (i.e. adding $2\pi$ to it if necessary), we get the following solution of the wave equation (assuming homogeneous Dirichlet conditions at the reflecting surface):

$$ U_{\text{s}}(r,\theta)=\exp\left(\frac{i\pi}4\right)\frac{1-i}{2\sqrt2}\times\\ \left( \exp(ikr\cos(\theta-\theta_0)) \left[1+\operatorname{erf}\left((i+1)\sqrt{kr}\cos\frac{\theta-\theta_0}2\right)\right]-\\ \exp(ikr\cos(\theta+\theta_0)) \left[1+\operatorname{erf}\left((i+1)\sqrt{kr}\cos\frac{\theta+\theta_0}2\right)\right] \right). $$

The $\operatorname{erf}$ function of the complex argument can be rewritten in terms of Fresnel integrals as

$$\operatorname{erf}\left(\frac{i+1}{\sqrt2}x\right)= (i+1)\left[\operatorname{C}\left(\sqrt{\frac2\pi} x\right)-i\operatorname{S}\left(\sqrt{\frac2\pi} x\right)\right].$$

Here's an example for $\theta_0=2\pi-\pi/4$:

If we plot how the intensity changes with as the probe angle goes off axis towards the shadow area at $r=10^8\lambda,$ we'll get the following result.

As we can see, even at 20° from the surface (relative to the edge) we still observe about $10^{-6}$ of the incident intensity. Given that laser light is quite concentrated, and human eyes are very sensitive, it's what you see in the experiment.


  1. A. D. Rawlins, Plane-wave diffraction by a rational wedge, Proc. R. Soc. A 411, 265-283 (1987).

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.