Facts about CD's (Pg. 307, Wolfson-Pasachoff "Phyiscs", 3rd. edition)
Proceedure: Hold the CD roughly
perpendicular to a distant bright white light source.
Look into the CD and find the first order.
Move it towards your eye until the spectrum, which is red to violet (from
outer to inner tracks) fills the entire CD recorded region width along
a radial strip.
Record the distance between your eye and the CD.
Discussion: Make sure that you understand why the colors are in the order that they are in. Assume that all the white light is coming in nearly perpendicular to the CD. Which color is being scattered at the largest angle from the incoming white light beam?
Calculation: The angular width, deltaphi, of the spectrum (which can be found from the width of the CD recording area and the distance that the CD is held from the eye) is related wavelength range in the spectrum. One can find (see above reference, pg. 895) that the wavelength range of visible light is from about 730nm (red light) to 380nm (violet). The diffraction satisfies
d sin(phi) = n lambda
And since we are looking in the first order, phi ~ 0. Then the angular dispersion of the spectrum is given by differentiating the diffraction equation above, namely,
d cos(phi) deltaphi = n deltalambda
So, approximately for the first order n=1 and phi ~ 0 so, the track distance is d ~ deltalambda/deltaphi.
Example: I found that the deltaphi
is nearly 1/4 radian. Thus, d ~ 1.4 microns. (compare with 1.6 microns
from the reference above).
EXPERIMENT 2: Determination of the wavelength of laser pen light
Proceedure: Now we use the intertrack distance that we computed (or were given) to determine the wavelength of laser pen light. We shine the light nearly perpendicular to the CD and record the angular distance to the n=1 and n=2 diffraction peaks.
Calculation: Use then the relation
d sin(phi) = n lambda
to where d = 1.6 microns and phi is the angle that you measure. Do this for both diffraction peaks. Find the average lambda.
Example: We find that the angles
are (for n=1, the first diffraction peak) phi = 24.2 degrees
and for (n=2) phi = 52.1 degrees. Plugging in we find that the two
orders give the lambda estimates of .657 microns and .631 microns
respectively. Averaging we have lambda = .644 microns. The laser pen itself
says that the wavelength range is from .63 to .68 microns.
ADDITIONAL SUPPORTING CALCULATIONS:
1) Run time and speed to determine d independently: Since we know that the CD contains 74 minutes of music, and is read out at at a uniform speed of 1.3 m/s, the total track length is 5770 meters. Since the outer radius of the tracks are .06m and the inner track is about .025m, the average track radius is .0425m. Thus the average track circumfrence is 2 pi times that average radius, or .27 m. Thus, there are 5770/.25 = tracks on a CD, spread over .06-.025 = .035 m. THUS, the intertrack spacing is .035/5770 = 1.5 E -6 m , or 1.5 microns.
2) How big is a bit? We just determined that the total track length on a CD is about 5770 m. But in that tracklength one can record 750 MB of data. At 8 bits each, this is 750E6 x 8 = 6E9 bits, or about a bit for each person on the earth. Thus the typical track feature is 5770m/6E9 = .96 microns.