Spectrum Angular Width in The Phasmatron Spectroscope

Version 1.0 of 6/3/2004-8:10 a.m.

fig6.gif

The following calculations assume you have read the analysis on Calculating the deviation angle for a prism.

Note that the first emergence angle at B is given by:
1=arcsin{sin(A)*sqrt(n2-sin2(1))-cos(A)*sin(1)} (1)
for A=60, and 1=60, (1) gives:
1=arcsin{sqrt(3/2)*sqrt(n2-3/4)-sqrt(3/4)}
=>1=arcsin{sqrt(3/2)*(sqrt(n2-3/4)-1/2)} (2)
Now: 60+1+2=180=>2=120-1 (3)
=>2=120-arcsin{sqrt(3/2)*sqrt(n2-3/4)-sqrt(3/4)} (4)

Then with incidence angle 2 at C, the emergence angle at D will be:

2=arcsin{sqrt(3/2)*sqrt(n2-sin2(2))-sin(2/2)} (5)

To conclude: First we calculate 1 using (2), then we find 2 using (3) and finally compute 2 from (5):

twoprisms.gif

The method above generalizes to n prisms easily, provided we know the angles between them.

For example, we can calculate the spectrum's angular width 2 that way:

for n=1.77578 for n=1.71681
1=65.446902 1=58.294643
2=54.553098 2=61.705357
2=73.569903 2=56.730209

=>2=16.839694.

Compare this with the E we get from dE/dn=2/cos(arcsin(nD/2)). This formula gives for nD=1.72803, dE/dn=3.9724624 rad, so with a (n)=1.77578-1.71681=0.05898 we get E=0.2342958 rad=13.424162. The difference is understandable, since the formula for dE/dn is valid essentially only for the area of the sodium line D, from where we picked nD=1.72803.

CAUTION:THE WIDTH OF THE SPECTRUM CHANGES AS WE CHANGE THE INCIDENCE ANGLE!

For example: If 1=58, then:

for n=1.77578 for n=1.71681
2=79.767447 2=58.593937

with a 2=21.17351

If 1=62, then:

for n=1.77578 for n=1.71681
2=69.773023 2=55.198486

with a 2=14.574537

Conclusion: If the incidence angle is <60, the spectrum is wider. If >60,it is shorter. (Take this into account later when you measure wavelengths. If you do not get the correct wavelength, it means that your collimator is not at 60 exactly.)

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