3
$\begingroup$

I'm coming to you with a strange physical/optical phenomenon I noticed today for which I'm looking for an explanation. I can't seem to find references to this, and it was pretty surprising to see for the first time.

Background: I'm a photographer, and I am trying to experiment more with monochromatic light sources. I hacked together a fixture for a relatively bright low pressure sodium (SOX 90 Watt) lamp, and it worked beautifully, producing a bright near-monochromatic yellow-orange light. The first thing I did once I had it working was to take it into the bathroom. There, I experienced the phenomenon: when I looked in the mirror, there was an interference pattern around my eyes in the reflection. It looked a lot like Newton's Rings. When I opened only my left eye, only my left eye's reflection had the rings around it. When I tried to replicate this with a camera, I did not get the same effect, which leads me to believe it's something specific to the optics of the eye.

What's more, I have a pretty monochromatically red CFL in my bathroom, which I use successfully as a safe light. I do not see this effect with the red light source.

Is this a well-known phenomenon that I just don't know the right words to search for? Is this a function of wavelength? I have also read that technically, low pressure sodium bulbs have two emission peaks, at 589.0nm and 589.6nm. Does the gap between these cause some sort of constructive/destructive interference?

Thank you for your time. As spooky as this was to discover, I am genuinely interested in how and why this was happening.

UPDATE: I got my housemate to come in the bathroom with me, and he confirmed my observation. What's more, we could see the rings in our own eyes, but not each other's eyes.

$\endgroup$
6
  • $\begingroup$ Just checking....you don't wear glasses do you? $\endgroup$
    – DKNguyen
    Jul 19 at 5:32
  • $\begingroup$ Hello, no glasses or contacts. My vision is generally pretty good. I also see this effect in my reflection on panes of glass and several different mirrors. $\endgroup$ Jul 19 at 5:33
  • $\begingroup$ Is it possible to take a photo of the reflection? If the same is in the image then it's not your eyes. $\endgroup$
    – joseph h
    Jul 19 at 6:55
  • 1
    $\begingroup$ @josephh , As stated, taking a photo was my first reaction, but so far I have not been able to replicate the ringed effect. Next time I have someone over, I will also see if they experience the same phenomenon. $\endgroup$ Jul 19 at 7:02
  • 1
    $\begingroup$ OK, I didn't notice that part. It seems strange. Yeah, get a friend to confirm and then let us know. Good luck. $\endgroup$
    – joseph h
    Jul 19 at 7:06
1
$\begingroup$

Sorry for my poor english. My native language is French.

It is possible that what you saw were Quetelet fringes, or Newton diffusion rings. They are discribed in this link : Quetelet rings with another interesting link.

In general, in physical optics, the fringes only appear with well-worked glass slides, under certain lighting conditions. While these fringes can be observed with thick slides of any kind of glass. They are well described in old optical books and dew on the glass helps to see them, hence the interest of a bathroom !

They are also described in the book by Craig F. Bohren: "What Light Through Yonder Window Breaks" Chapter 2 : interference patterns on garage door windows".

I can't do better than quote it :

A particle on a window illuminated by a beam scatters toward the back surface of the window, and part of this scattered light is reflected to the observer. But light from the incident beam also is reflected by the back-surface and illuminates the particle, which scatters some of this reflected light to the observer. Interference between these two beams with different histories -scattered by the particle, then reflected by the glass; or reflected by the glass, then scattered by the particles- is the origin of the beautiful colored fringes I saw.

$\endgroup$
1
  • $\begingroup$ It's an interesting hypothesis, and it was my first guess. I'm somewhat familiar with Newton rings, as they can be an annoying issue with scanning film. This bathroom did not have a shower in it (half bath) and the humidity in my house is very low, so no dew. In my experience Newton rings generally are created at the boundary of two pieces of glass, though that doesn't seem to be the same situation as your quote. But why would this only be the case under a monochromatic light source? $\endgroup$ Jul 19 at 16:49
0
$\begingroup$

You may be seeing an Airy disk. However i would expect rings about the light source rather than about the mirror image of your eyes. Could it be that the direct light does not reach your eyes but is covered?

Sodium light comes from two very close, narrow transitions of the sodium dimer, Na$_2$. It is strongly monochromatic, much more so than the red light you mention. I assume that you would have stated so if that is from a laser.

Your eye's pupil has a finite aperture of about 5 mm. This 1/8500 times the wavelength so the first interference minimum is expected at 0.04 degree. This is about a tenth of the angular diameter of the moon so it should be clearly perceptible.

It is possible that you use a much larger diaphragm when you take the photo so that the rings are much closer.

This analysis assumes a small pointlike light source.

$\endgroup$
2
  • $\begingroup$ This is interesting. This effect worked both when I was backlit and side lit. The light source is relatively large, about 50cm long and in a diffusing enclosure, which seems in many ways like the opposite of a pointlike source. Yes, I suppose it is probably much more monochromatic than the red light. The red only has to stay below a certain wavelength, and usually CFL phosphors have a pretty non-monochrome emission spectrum. $\endgroup$ Jul 19 at 16:56
  • $\begingroup$ @ my2cts The pupil itself may act as the point-like source, so maybe this proposal is ok $\endgroup$ Jul 20 at 2:26
0
$\begingroup$

It would be interesting to know the radius of the ring.

Here is an idea.

Presumably you were looking straight towards the mirror let's say from a distance $d$.

If there was constructive interference between the two wavelengths that you mentioned, if $n$ wavelengths of the longer wavelength light fitted in the distance $d$ and $n+1$ for the other wavelength, then

$$\frac{d}{\lambda_1} = n$$

$$\frac{d}{\lambda_2} = n+1$$

so

$$\frac{d}{\lambda_2} - \frac{d}{\lambda_1} = 1$$

$$d = \frac{\lambda_1\lambda_2}{\lambda_1 - \lambda_2}$$

about 0.58mm

This distance is the smallest distance where the different wavelengths constructively interfere.

However it'll happen for any distance that is a multiple of the 0.58mm

If it happens for the straight ahead distance, from your eye to the mirror - doing a Pythagoras theorem for a right angled triangle from eye to mirror ($EM$) and a distance $r$ along the mirror at right angles to $EM$, we can find another distance where the constructive interference happens.

Taking $EM$ as 1 metre

$$1^2+ r^2 = 1.00058^2$$

A radius of 3.4cm

It's possible that the rings occur with a radius of about 3.4cm, the dark rings would be between the bright rings where the path difference was half a wavelength.


Update to include the radii for the first 10 maxima:

Using $d^2+ r^2 = (d + 0.00058n)^2$, where $n$ is the maxima number and $d$ is distance $EM$, now 0.5 metres

$r$ is approximately $$\sqrt{2dn \times 0.00058}$$ or $$\sqrt{5.8n}$$ in cm.

The radii, to the middle of each bright fringe, are

\begin{array}{r|r} n & \text{ radius (mm)}\\ \hline 1 & 24 \\ 2 & 34\\ 3 & 42\\ 4 & 48\\ 5 & 54\\ 6 & 59\\ 7 & 64\\ 8 & 68\\ 9 & 72\\ 10 & 76 \end{array}

$\endgroup$
13
  • $\begingroup$ The wavelength is 589 nm so 1000 times smaller than what this answer uses. $\endgroup$
    – my2cts
    Jul 19 at 12:26
  • $\begingroup$ @my2cts the answer uses 589 nm the 0.58mm is from the formula $d =\dots$ $\endgroup$ Jul 19 at 14:24
  • $\begingroup$ @JohnHunter , this is a good idea. When I get a chance, I will attempt to measure this distance with a marker and ruler. $\endgroup$ Jul 19 at 16:59
  • $\begingroup$ Ok I now see what you mean. However the two spectral lines are mutually incoherent and would not interfere. $\endgroup$
    – my2cts
    Jul 19 at 21:07
  • $\begingroup$ @my2cts more data would be good. Upon 'reflection' the Airy Disk idea might be the right one, with the pupil itself acting as the point-like source - (the circular aperture), that gives a radius of order mm. Data about the ring size at different distances, maybe a sketch of the image and setup, and if it happens the same on a polished or different mirror would all help. $\endgroup$ Jul 20 at 2:24

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.