Questions about measuring data track spacing with laser

[BAJA241]

Is the incident beam the original laser beam shining onto the CD in the first place? If we are shining the laser directly from normal (90 degrees) then the incident angle is 0? Looking at the picture on the procedure page it shows the incident beam at 20 degrees.Is that because they are shining it from that angle and not from 90 as it explains to do in the procedure?

Also on the formula, for calculating d, does the / mean to divide M x wavelength by the other part (sin 0m -sin 0i)? I am getting a weird solution for m = -1 every time like between 8.33 micrometers and 10.22. All the others are 1.39 to 1.6 or so which seems like the right range for CD. But this might be because I am confused about the incident angle from looking at the picture on the procedure. I've been shining the laser right at 90 degrees and then counting the second beam over to the right as the incident beam (which is always about 25 to 27 degrees from normal). So confused~!

[deleted-90600]

Hello,
The incident angle is indeed the angle at which the laser hits the CD in the first place. It is measured in relation to the normal. Your incident angle should be some degrees off from the normal, like 20 degrees, as shown in the picture. The procedure probably says to line up the laser at 90 degrees only as a starting point when looking for a diffraction pattern, not where the incident angle should be.

And yes, to use the formula you multiply m by the wavelength and then divide by (sin θm − sin θi ). The calculations may be wrong because the angles are not correct. The beam coming straight from the laser is always the incident beam, the other beams are diffracted.

I hope everything is cleared up, I wish you luck with your project!