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:

δ_{1}=arcsin{sin(A)*sqrt(n_{λ}^{2}-sin^{2}(α_{1}))-cos(A)*sin(α_{1})}
(1)

for A=60°, and α_{1}=60°, (1) gives:

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

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

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

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

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

δ_{2}=arcsin{sqrt(3/2)*sqrt(n_{λ}^{2}-sin^{2}(α_{2}))-sin(α_{2}/2)}
(5)

To conclude: First we calculate δ_{1} using (2), then we find
α_{2} using (3) and finally compute δ_{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
Δδ_{2} that way:

for n=1.77578 | for n=1.71681 |

δ°_{1}=65.446902 |
δ°_{1}=58.294643 |

α°_{2}=54.553098 |
α°_{2}=61.705357 |

δ°_{2}=73.569903 |
δ°_{2}=56.730209 |

**=>Δδ _{2}=16.839694**°

Compare this with the ΔE 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
δ(n)=1.77578-1.71681=0.05898 we get ΔE=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 **α _{1}=58**°, then:

for n=1.77578 | for n=1.71681 |

δ°_{2}=79.767447 |
δ°_{2}=58.593937 |

with a **Δδ _{2}=21.17351°**

If **α _{1}=62**°, then:

for n=1.77578 | for n=1.71681 |

δ°_{2}=69.773023 |
δ°_{2}=55.198486 |

with a **Δδ _{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.)