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:
δ₁=arcsin{sin(A)*sqrt(n_λ²-sin²(α₁))-cos(A)*sin(α₁)} (1)
for A=60°, and α₁=60°, (1) gives:
δ₁=arcsin{sqrt(3/2)*sqrt(n_λ²-3/4)-sqrt(3/4)}
=>δ₁=arcsin{sqrt(3/2)*(sqrt(n_λ²-3/4)-1/2)} (2)
Now: 60°+δ₁+α₂=180°=>α₂=120°-δ₁ (3)
=>α₂=120°-arcsin{sqrt(3/2)*sqrt(n_λ²-3/4)-sqrt(3/4)} (4)

Then with incidence angle α₂ at C, the emergence angle at D will be:

δ₂=arcsin{sqrt(3/2)*sqrt(n_λ²-sin²(α₂))-sin(α₂/2)} (5)

To conclude: First we calculate δ₁ using (2), then we find α₂ using (3) and finally compute δ₂ 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 Δδ₂ that way:

=>Δδ₂=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 α₁=58°, then:

with a Δδ₂=21.17351°

If α₁=62°, then:

with a Δδ₂=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.)