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:

ä_{1}=arcsin{sin(A)*sqrt(n_{ë}^{2}-sin^{2}(á_{1}))-cos(A)*sin(á_{1})}
(1)

for A=60°, and á_{1}=60°, (1) gives:

ä_{1}=arcsin{sqrt(3/2)*sqrt(n_{ë}^{2}-3/4)-sqrt(3/4)}

=>ä_{1}=arcsin{sqrt(3/2)*(sqrt(n_{ë}^{2}-3/4)-1/2)}
(2)

Now: 60°+ä_{1}+á_{2}=180°=>á_{2}=120°-ä_{1}
(3)

=>á_{2}=120°-arcsin{sqrt(3/2)*sqrt(n_{ë}^{2}-3/4)-sqrt(3/4)}
(4)

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

ä_{2}=arcsin{sqrt(3/2)*sqrt(n_{ë}^{2}-sin^{2}(á_{2}))-sin(á_{2}/2)}
(5)

To conclude: First we calculate ä_{1} using (2), then we
find á_{2} using (3) and finally compute ä_{2} 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
Ää_{2} that way:

for n=1.77578 |
for n=1.71681 |

ä°_{1}=65.446902 |
ä°_{1}=58.294643 |

á°_{2}=54.553098 |
á°_{2}=61.705357 |

ä°_{2}=73.569903 |
ä°_{2}=56.730209 |

**=>Ää _{2}=16.839694**°

Compare this with the ÄE we get from dE/dn=2/cos(arcsin(n_{D}/2)).
This formula gives for n_{D}=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 n_{D}=1.72803.

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

For example: If **á _{1}=58**°, then:

for n=1.77578 |
for n=1.71681 |

ä°_{2}=79.767447 |
ä°_{2}=58.593937 |

with a **Ää _{2}=21.17351°**

If **á _{1}=62**°, then:

for n=1.77578 |
for n=1.71681 |

ä°_{2}=69.773023 |
ä°_{2}=55.198486 |

with a **Ää _{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.)