How do I estimate the diffraction angles of laser light which illuminates a CD? Does the storage capacity have an impact?

Theories I have considered are:

  • refraction
  • diffraction
  • $\sin(\theta)=\frac \lambda d$

However I'm finding it difficult to approach the problem using the theorems stated.

  • $\begingroup$ You may want to check this page: sciencebuddies.org/science-fair-projects/project_ideas/… You can find there details of how to calculate the diffraction pattterns and how experiments for this can be set up. And yes, the storage capacity does have an impact. $\endgroup$ – Merlin1896 Feb 22 '16 at 15:45
  • $\begingroup$ so how does the storage capacity have an effect ? I cant see this is the link $\endgroup$ – darren Feb 22 '16 at 15:51
  • $\begingroup$ If you want to store more information on an object with the same size (CD vs DVD), the areas which store "one information" have to be smaller. In the case of CDs/DVDs these areas are the pits on the backside. The pits on DVDs are smaller and have a smaller spacing. This can be seen in the diffraction experiments. $\endgroup$ – Merlin1896 Feb 22 '16 at 16:00
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    $\begingroup$ You may find my earlier answer to a related question interesting. $\endgroup$ – Floris Feb 22 '16 at 18:08

A cd is basically a diffraction grating with line spacing of $\approx 1.6\; \mu$m. The true value depends on the exact storage capacity of the CD e.g. 650, 700 or 750 Mb for example. The actual data that is stored on a cd has no impact on the diffraction angles. A DVD will produce different angles due to its different line spacing of $0.74\; \mu$m. The CD/DVD can either be used in reflection as it is or in transmission by removing the foil on the backside. I prefered the transmission geometry but its more or less only a geometrical consideration.

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  • $\begingroup$ I am pretty sure that the line spacing (track pitch) differs in dependence on the total capacity of the medium. That is a 700 MB CD-R simply has slighty smaller tracks than a 650 MB disk. Wikipedia confirms this and gives $1.5\,\mathrm{\mu m}$ for a 700 MB disk. (Note: The capacity refer to the conventional data capacity and the actual capacity depends on the disk mode, the physical capacity is even higher due to error correction, sector metadata, ...). $\endgroup$ – Sebastian Riese Feb 22 '16 at 16:52

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