Newton's ring? setup It can be made by pointing a laser pointer directly to a mirror at low angle and observing the reflection behind the laser pointer. No convex or concave lens needed unlike the Newton's ring experiment we know.

So far, Newton's ring experiments and theories I read only involves a combination of concave and convex lenses, as Wikipedia says: Newton's rings is a phenomenon in which an interference pattern is created by the reflection of light between two surfaces—a spherical surface and an adjacent touching flat surface. The pattern is created by placing a very slightly convex curved glass on an optical flat glass. The two pieces of glass make contact only at the center, at other points there is a slight air gap between the two surfaces, increasing with radial distance from the center.

None of Newton's ring theories I read in Wikipedia or somewhere else involves a laser pointer and a mirror. Is there the same fundamental physics involved in both experiments? Or could it be a completely different thing?

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    $\begingroup$ That's a remarkable picture, a textbook example of Newton's rings! Perhaps the mirror has a tiny curvature? How does it change if you vary where on the mirror you hit? $\endgroup$ – knzhou Mar 17 '19 at 2:03
  • $\begingroup$ As you move the mirror or laser pointer slightly, the pattern moves inward or outward like the waves on water after you throw a rock to it. You can try it yourself using a higher power green laser. 5mw ordinary red laser also work but you need a dark room. Any flat mirrors should work. $\endgroup$ – Yudhi G. Mar 17 '19 at 4:10
  • $\begingroup$ The pattern position is exactly halfway between the laser pointer and the laser reflection. $\endgroup$ – Yudhi G. Mar 17 '19 at 4:24
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    $\begingroup$ ^ If you have the dot separate of the pattern then it certainly isn't the aperture diffraction. $\endgroup$ – dmckee --- ex-moderator kitten Mar 17 '19 at 4:37
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    $\begingroup$ I've already tried with different mirrors no matter how clean or dirty they are. I've also tried it with different laser color. Same result. Those rings keeps appear. Dirty mirror won't make a perfect circular patterns or even imitate the Newton's ring. $\endgroup$ – Yudhi G. Mar 17 '19 at 16:08

The rings are probably not "Newton's rings", though of course they are a related phenomenon. Probably they are due to scatter at the front and back surfaces of the mirror.

In effect, light scattered from the front surface acts as a point source. Light scattered from the back surface will also act as a point source. The interference from two point sources, close together and pretty much in line with each other, results in a pattern like that in your photo.

I'd guess that if you examine the pattern closely you'll see that it is composed of speckles, which would support the idea that it's due to scatter from those surfaces.


My guess is you have Newton’s rings due to intrinsic wavefront curvature in your laser beam. Interference between the curved wavefront reflecting from the front and rear surfaces of the glass gives you concentric rings. If everything were perfectly parallel, the interference would coincide with the reflected laser spot, so my guess is that your first glass surface is slightly non-parallel with the metallized surface.

In the textbook example, there is a collimated beam, and wavefront curvature is introduced by a curved surface. In your case, your laser is not quite collimated, so there is intrinsic wavefront curvature. Thus, you can see circular interference fringes with flat surfaces.

  • $\begingroup$ That's what I'm thinking. The length of the non-perfectly collimated laser beam acts as a radius of the curved wavefront. Thus, the front of the beam acts as if it is the curved surface of the convex lens sitting on a flat glass surface, or in my case, a mirror. $\endgroup$ – Yudhi G. Mar 28 '19 at 4:28

$\let\d=\delta \let\lam=\lambda \let\th=\theta \def\half{{\textstyle {1 \over 2}}}$ I'm leaning towards @S.McGrew's proposal, although there are some points I haven't fully understood. I want to suggest a little computation you could test through some measurements.

Let's define

  • $S$ distance screen-mirror

  • $L$ distance laser-mirror

  • $t$ mirror thickness

  • $n$ index of refraction of mirror glass

  • $\lam$ wavelength.

Neglecting reflection angle (this is a weak point I've still to explore) you have two virtual images of laser source. The first I$_1$ coming from reflection on the front surface of glass, is placed at a distance $S+L$ from screen. The second I$_2$ (reflection on the rear, metal surface of mirror) would be at a distance $S+L+2t$, but as far as optical path is concerned it must be treated as if it were at a distance $S+L+2nt$.

Consider a point P on the screen, at a distance $r$ from centre C of fringe pattern. Let $\th$ be the angle CI$_1$P so that $$r = (S+L)\,\tan\th \simeq (S+L)\,\th.$$ The difference between the optical paths from I$_2$ to P and from I$_1$ to P is $$\d \simeq 2\,n\,t\,\cos\th \simeq 2\,n\,t\,(1 - \half\,\th^2).$$

If for $\th=0$ (P=C) $\d$ is a multiple of $\lam$ $$2\,n\,t = k\,\lam \qquad \hbox{$k\ $ integer}$$ we'll have a maximum in C. The other maxima will be at $$\d = 2\,n\,t\,(1 - \half\,\th^2) = (k-k')\,\lam \qquad \hbox{$k'\ $ integer}$$ $$\th = \sqrt{k' \lam \over n\,t}$$ $$r = (S+L)\,\sqrt{k' \lam \over n\,t}$$

You could easily test the last formula. A similar formula holds for minima.


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