CD Diffraction

[sweetsis994]

a very nice video indeed! do you have anyvideo on diffraction gration and multiple slits!? you have a lot of videos and its a bit difficult to surf through all of them! TY

[ramrao101]

...hey great video!!! are you a physics highschool teacher...??

[falconphysics]

Yup. I teach at Divine Child High School in Dearborn, MI

[M0reTeaVicar]

Would be useful to point out that this would only work on a commercial, read-only CD, not a CD-R or CD-RW, at least for the part where you remove the backing, and not for incident diffraction if the disc is blank!

[falconphysics]

I do this with CD-R discs regularly both blank and otherwise. Take a look, you can see the same diffraction pattern.

[M0reTeaVicar]

Then this is quite strange. A blank disc should have no tracks, and therefore no groove grating for diffraction to occur. I wonder if you are seeing dispersion instead?

[falconphysics]

Definitely diffraction. The math works out too well to be anything else (it even works with blank DVD-R).

[M0reTeaVicar]

Well, does the math really work out? For a start, in the diffraction formula, d is actually the grating spacing, not the spacing of the diffraction pattern between different bands. You will see a spacing of d between different orders, however there is no higher-order diffraction in what you show. Yes, the angle increases with wavelength - and we see longer wavelengths are on the outside, but this could be coincidence, especially if you see it on a blank CDR!

[M0reTeaVicar]

And because this is a CD - a disc of plastic with refractive index, then dispersion will also occur. The CD interfaces with air are reflective - and like a real rainbow by water droplets, you will see a combination of multiple reflections and dispersion, so that longer wavelengths can exit with a larger angle (in dispersion, longer wavelengths are dispersed a smaller angle, but reflections can invert this relationship). Multiple reflections also also give appearance of orders n>1!

[M0reTeaVicar]

Otherwise Blank CDs are not really blank but have a track laid out... Sadly I have no redundant blank CDRs to butcher and test this out. However, using a used CDR, we can work things out: d, the track spacing, is ~1.5 microns. Taking red 620nm, you'd expect diffraction angle a=24 deg. From observation r=5cm, s=17cm, angle a is then 16 degrees where Tan(a)=r/s. For 450nm, r=2.5, expected a=17deg, measured a = 8 deg. There is thus no agreement with the standard diffraction formula here.

[M0reTeaVicar]

The reason for such discrepancy is because the diffraction formula is derived for a collimated source, e.g. parallel incoming wavefronts. At close range of 1-2m, a lightbulb is effectively a point source - you thus have a different incident angle at each track of the disc, and the standard derived diffraction formula is invalid. You need a beam expander & collimating lens to achieve incident parallel wavefronts.

[falconphysics]

This activity is not of my own invention. It has been vetted by people with much better credentials than me. I first discovered it in a guide for teachers created by General Atomics (first link below), and it was reinforced at a session given by the FOCUS group at the University of Michigan (second link) and by Chris Chiaverina (a past president of the American Association of Physics Teachers) when he wrote about it for Arbor Scientific (third link).

Links in next comment (ran out of space)

[falconphysics]

Well, looking at the Violet (450nm) band I get a spacing of 1580 nm with a blank CD-R that has had the aluminum peeled off. When I do this activity with my students I've had good success. Careful students can typically get within 10% of the expected value.

Additionally, if you bounce a laser off a CD (even blank) you can reflect the diffraction pattern on a wall and arrive at the expected spacing. When done over a great distance you can get very precise results.

[M0reTeaVicar]

I should forget endorsements and concentrate on the physics. A standard HeNe laser should produce a collimated output, and even if it has a small spot size, it will have a good approximation to parallel wavefronts in the illuminated region at the surface - so you should good agreement. Possibly where errors are reduced, you are actually much further away from the source? The further you are, the flatter the wavefronts.

[M0reTeaVicar]

Indeed, the definition for parallel wavefronts is a circle of infinite radius - light from the sun is often used as a good approximation (for e.g. finding focal length of a lens, simply look for the distance that light from the sun is focused to a spot, or distant objects >100m brought into focus). For small distances, laser is best, or bulb+aperture+collimating lens, to achieve parallel wavefronts. Anyway, interesting results with blank CDR - 1.58 micron, so they must still have tracks! Cheers!

[IanPT]

excellent resource for my senior students, THx!!

[RoadRunnerLaser]

Excellent demo !! Thanks.

[Kapacool]

you sound like grissom form csi xD interesting stuff, though.