The following calculations assume you have read the analysis 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)

Concluding, 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°.

Concluding, if the incidence angle is <60°, the spectrum is wider. If >60°, it is shorter. We take this into account later when we measure wavelengths. If we do not get the correct wavelength, it means that our collimator is not at 60° exactly.

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