## Spectrum Angular Width in The Phasmatron Spectroscope

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

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:
ä1=arcsin{sin(A)*sqrt(në2-sin21))-cos(A)*sin(á1)} (1)
for A=60°, and á1=60°, (1) gives:
ä1=arcsin{sqrt(3/2)*sqrt(në2-3/4)-sqrt(3/4)}
=>ä1=arcsin{sqrt(3/2)*(sqrt(në2-3/4)-1/2)} (2)
Now: 60°+ä12=180°=>á2=120°-ä1 (3)
=>á2=120°-arcsin{sqrt(3/2)*sqrt(në2-3/4)-sqrt(3/4)} (4)

Then with incidence angle á2 at C, the emergence angle at D will be:

ä2=arcsin{sqrt(3/2)*sqrt(në2-sin22))-sin(á2/2)} (5)

To conclude: First we calculate ä1 using (2), then we find á2 using (3) and finally compute ä2 from (5):

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 Ää2 that way:

 for n=1.77578 for n=1.71681 ä1=65.446902° ä1=58.294643° á2=54.553098° á2=61.705357° ä2=73.569903° ä2=56.730209°

=>Ää2=16.839694°.

Compare this with the ÄE 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 ä(n)=1.77578-1.71681=0.05898 we get ÄE=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 á1=58°, then:

 for n=1.77578 for n=1.71681 ä2=79.767447° ä2=58.593937°

with a Ää2=21.17351°

If á1=62°, then:

 for n=1.77578 for n=1.71681 ä2=69.773023° ä2=55.198486°

with a Ää2=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.)