Homework on Optical Storage, 2002 version (solutions)

Feel free to work on the homework in groups. The work you hand in, however, should reflect your understanding of the material and be in your own words.  Students who turn in identical (or close to identical) homework assignments will be asked to explain their answers orally to the TA or prof.  A student who cannot explain how he or she arrived at a given answer will be charged with academic dishonesty.

You should show all of your calculations (neatly) and justify all of your answers for full credit.

 
Diffraction in CD-ROMs
One way to look at the smallest amount bit of data that can be stored on an optical disk is to calculate the smallest spot size that can be generated by the laser being used to detect the bits of data. When a laser is focused down by a lens, the smallest spot size achievable is limited by diffraction.
1.) Currently, a CD-ROM uses an infrared laser with a wavelength of 780 nm (780 x 10-7 cm). Calculate the radius of the central diffraction maximum (distance from the center to the first minimum) for a circular lens of 1 mm diameter after traveling 1 mm beyond the lens. Your answer will be the minimum spot size that a perfect lens (with a 1 mm focal length) can produce.

sin q = 1.22  / d; with  = 780 x 10-9 m and d = 1 x 10-3 m, sin q = 9.516 x 10-4. This is small enough for the small angle approximation (sin q = q = tan q) to apply. tanq = r / D. D = 1 x 10-3 m, so the radius of the spot is 951.6 nm.

2.) Explain why your answer to question (1) limits the density of data that can be put on a CD-ROM.

It is impossible to resolve objects that are smaller than the spot of light created by diffraction, and so the spots on the disc that represent our data (one bit per spot) must be larger than this limit.

3.) Suppose you use a 2 mm diameter lens (keeping the 1 mm focal length) instead of a 1 mm diameter lens. What is the diffraction limited spot size now? (This is the same calculation as problem 1 except that now you have a 2 mm diameter circular aperture.) Please be aware that this larger lens (and larger laser beam required to take advantage of it) will make the optical system at least 10 times more expensive. Would it be worth the added cost?

d is increased by a factor of 2, and everything else remains constant, so the spot radius would be half as big – 475.8 nm. This will increase our data density by only a factor of four (area ~ radius2), and is probably not worth the cost.

4.) Suppose that cost is not an issue. What is the diameter of the very smallest spot size that you can make with the 780 nm laser? Use the theory for circular diffraction and recall that sin q can be no greater than 1.

Light of a particular wavelength cannot be used to resolve objects smaller than that wavelength. (Full credit was awarded for this question for any response.)

5.) Blue solid-state lasers are currently being produced (although they remain expensive) that have a wavelength of 420 nm. Repeat your calculation for question (1) with this new wavelength. What spot size is achievable now?

Method is same as for (1).   Plugging in l= 420 nm gives a spot radius of 512.4 nm.

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