Graph of Sellmeier's Equation for the SF10 Crystal

All the calculations below are valid only for the area of the sodium D lines.

dn/dλ:

First from the table of the SF10 prism we get:

dn/dλ~Δn/Δλ=(1.72825-1.72803)/ |5875.618-5892.938|A=1.2702*10^-5/A.

dn/dλ~1.2702*10^-5/A (in the area of the sodium D lines).

Then, the resolving power (or separation limit) Δλ is defined by the Lord Rayleigh's formula as Δλ=λ/{B(dn/dλ)}, where B is the length of the prism Base (in Angstroms).

Resolving power: Δλ=λ/{B(dn/dλ)}.

For the sodium D area we get:

λ=5892.938A, B=2*6*10⁸A, (1cm=10⁸A, 2 prisms x 6cm Base each), dn/dλ =1.2702*10^-5/A, => Δλ=0.3866148A (in the area of the sodium lines).

If you work in the sodium D area observe that the sodium D line is actually two lines with wavelengths D₁=5895.923A, D₂=5889.953A. The Phasmatron spectroscope will easily resolve them, since their corresponding difference of 5.97A is greater than Δλ=0.3866148A.

The resolution is then defined to be R=λ/Δλ, so

Resolution: R=λ/Δλ.

For λ=5892.938A, and Δλ=0.3866148A, R=15242.4 (in the area of the sodium lines).

Then consider the formula of minimum deviation angle for one prism:

sin{(D+A)/2}/sin{A/2}=n_λ.     (1)

(Where D is the minimum deviation angle, and A is the apical prism angle). The Phasmatron spectroscope has 60 degree prisms so A=60°. But we cannot apply the formula above directly, because the two prism geometry is different from the one prism case. What we can do however, is calculate the dispersion of the system, which is defined as dE/dλ. For the one prism case it is easy to see that in the position of minimum deviation, we can differentiate (1) to get:

cos{(D+A)/2}(dD/dn)(1/2)=sin{A/2} =>

dD/dn=2sin{A/2}/cos{(D+A)/2} (2)

Given that (1) => (D+A)/2=sin^-1{n_λ*sin{A/2}}

(2) => dD/dn=2sin{A/2}/cos{sin^-1{n_λ*sin{A/2}}} (3)

since here A=60°, (3) => dD/dn=1/cos{sin^-1{n_λ/2}}.

However note that E=2D so dE/dn=2*dD/dn, so using (3) we get for n_D=1.72803, dD/dn=1.9869763 rad.

For the Phasmatron spectroscope which has two prisms, we thus get:

dE/dn=2dD/dn=2/cos{sin^-1{n_λ/2}}.

For n_λ=1.72803, =>

dE/dn=3.9724624 rad.

Accordingly we now define dispersion:

Dispersion: dE/dλ=(dE/dn)*(dn/dλ).

For dn/dλ=1.2702*10^-5/A, for dE/dn=3.9724624 rad, => dE/dλ=5.04582*10^-5 rad/A.

On the photographic plate we also define the dispersive power useful in creating scales. We need to know how many mm's of film fit in an Angstrom under the current circumstances. This is expressed as:

Dispersive Power: ds/dλ=f*(dE/dλ).

Where f is the camera's lens focal length, using a camera lens in front of prism. For example:

Next we calculate the angle of the spectrum:

Spectrum Angular Width: ΔE=(dE/dn)*Δn.

We take Δn=1.77578-1.71682, the limits of the visible spectrum.

dE/dn=3.9724624 rad,  =>

ΔE=13.42° (as we shall see this is not entirely accurate, since dE/dn is valid only in the area of the sodium D lines, and not on the entire spectrum. In fact, the true width will be around 16.84° owing to correct measuring of the internal angles. See section on Spectrum Angular Width).

Next, the width of the entire spectrum on the film can be calculated.

Width of Spectrum on Film: Δs=f*ΔE,

where f is the focal length of the lens used on the camera. Take f=50mm, ΔE=16.84°=0.2939134 rad,  => Δs=14.69567mm on film.

If we took instead f=700mm then Δs=205.73938mm on film (see also section on Phasmatron Photography).

Next we indicate some common constants of the SF10 crystal:

n_F, n_C, n_D, are the refraction indexes for the Hydrogen F, C lines and the sodium D line (4861.327A=F, 6562.8A=C, 5892.938A=D).

(n_F- n_C) is called the mean dispersivity for n_F=1.74648,  n_C=1.72085 and  n_D=1.72803 we get:

SF10 Mean Dispersivity: (n_F- n_C)=0.02563.

Let ω=(n_F- n_C)/(n_D- 1)=0.0352045. ω is the inverse of the Abbe number. So:

SF10 Abbe # =1/ω=28.405384.

The Brewster angle is the angle for which the reflected beam is totally linearly polarized. SF10  Brewster angle: θ=tan^-1(n_λ). For n_λ=1.72803, => θ=59.945468° (for sodium light).

Next follows the internal transmittance table for the SF10 crystal:

Internal Transmittance graph for the SF10 crystal

To find the internal transmittance for a block of thickness K other than 5mm we can use the following formula:

SF10 Internal Transmittance : T_i(Kmm)=T_i(5mm)^{K/5}

For example the internal transmittance for 25mm for the 420nm line will be:

T₄₂₀(25mm)=0.981^{25/5}=0.981⁵=0.908542.

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