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Homework on Optical Storage, 2001 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 of the CD-ROMs in my office holds 600 Mbytes of data.  We will use our knowledge of optics to determine the lens size needed to read this data.  We will make some approximations along the way, but our end result will be a good estimate.
 
1. How many bits are stored on this CD-ROM?   Remember, a byte is 8 bits, a kilobye is 1024 bytes, and a megabyte is 1024 kilobytes.  (We made an approximation in my lecture slides, but you should try to be more exact.
600 MB * 1024 kB/MB * 1024 bytes/kB * 8 bits/byte = 5.03 x 109 bits
2. This CD-ROM has data stored on a ring with outer radius a=5.9 cm and inner radius b=2.9 cm.  The surface area of a ring is found from
A = p(a2 - b2 ),
(This is just the area of the bigger circle minus the area of the smaller circle).  What is the area used by my CD-ROM for storage?
3.14159 * ((5.9 cm)2 - (2.9 cm)2) = 82.9 cm2
3. What is the average area of one bit?
82.9 cm2 / (5.03 x 109 bits) = 1.6 x 10-8 cm2/bit
4. If the bits are circular (an approximation), what is the radius of one bit?
A = pr2; sqrt (1.6 x 10-8 cm2 / 3.14159) = 724 nm
5. CD-ROMs use infrared lasers with a wavelength of 780 nm (780 x 10-7 cm).  If the lens is 0.1 cm away from the disc, how large does the lens need to be for the central diffraction maximum to be smaller than one bit?  (Hint:  use the expression for diffraction through a circular aperture, with the lens diameter as the aperture diameter d.  You may use the small angle approximation in this situation.)
sin q = 1.22 l/d; small angle approximation says sin q @ q @ tan q = y/D;
(724 x 10-9 m) / (0.1 x 10-2 m) = 0.000724; (1.22 * 780 nm) / 0.000724 = 0.131 cm
6. The lens will focus the light at a point one focal length away from the lens.  So in our application, the distance between the lens and the disc is the focal length.  As mentioned in your reading, lenses with a diameter larger than the focal length (F-stop less than 1) are very difficult to make and are certainly not going to appear in mass-produced electronics.  Compare your result from part (e) with this practical limit.
0.131 cm > 0.1 cm (0.001 m), so this is a difficult lens to make
7. As we mentioned in class, the laser light must travel through a plastic coating (n=1.55) before striking the lands and pits.  What is the wavelength of light when it hits the lands and pits?
ln = l0 / n; 780 nm / 1.55 = 503 nm
8. Using this new wavelength, re-calculate the size of lens required to produce a central maximum smaller than one bit (question 5).
(1.22 * 503 nm) / 0.000724 = 0.085 cm
9. What F-stop (D/d) does this require?  Is this practical?
0.085 cm < 0.1 cm (0.001 mm), so this lens is easy to make.

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