## Optical Characteristics of SF10 Crystal Prisms

### Version 1.2 of 17/3/2021-4:00 a.m.

 λ (A) Element nλ 23254 Hg arc 1.68218 19701 Hg arc 1.68750 15296 - 1.69378 10600 Nd laser 1.70227 10140 Hg arc 1.70345 8521.1 Cs arc 1.70889 7067.217 Ar arc 1.71682 6562.8 H arc 1.72085 6438.4696 Cd arc 1.72200 6328 HeNe laser 1.72309 5892.938 Na arc 1.72803 5875.618 He arc 1.72825 5460.74 Hg arc 1.73430 4861.327 H arc 1.74648 4799.9107 Cd arc 1.74805 4358.35 Hg arc 1.76197 4046.561 Hg arc 1.77578

Refractive index table for the SF10 crystal

Graph of Sellmeier's Equation for the SF10 Crystal

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

PHASMATRON 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).

PHASMATRON 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=0.3866148Angstroms. (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
PHASMATRON 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}=nλ.     (1)
(where D is the minimum deviation angle, and A is the apical prism angle). The phasmatron 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{nλ*sin{A/2}}
(2) => dD/dn=2sin{A/2}/cos{sin-1{nλ*sin{A/2}}} (3)
since here A=60°, (3) => dD/dn=1/cos{sin-1{nλ/2}}

However note that E=2D so dE/dn=2*dD/dn
Using (3) we get for nD=1.72803, dD/dn=1.9869763 rad.
For the PHASMATRON which has two prisms, we thus get:

PHASMATRON dE/dn=2dD/dn=2/cos{sin-1{nλ/2}}:

for nλ=1.72803, =>

Accordingly we now define dispersion:
PHASMATRON dispersion: dE/dλ=(dE/dn)(dn/dλ):

for dn/dλ=1.2702*10-5/A

On the photographic plate we also define the dispersive power usuful 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:
PHASMATRON dispersive power: ds/dλ=f*(dE/dλ).
where f is the camera"s lens focal length, using camera lens in front of prism. For example:

 f: ds/dλ: dλ/ds: 50mm 2.52291*10-3mm/A 396.36A/mm 135mm 6.81185*10-3mm/A 146.80A/mm 300mm 0.0151374mm/A 66.06A/mm 700mm 0.0353207mm/A 28.31A/mm

Next we calculate the angle of the spectrum:
PHASMATRON spectrum angular width: ΔE=(dE/dn)Δn.
We take Δn=1.77578-1.71682, the limits of the visible spectrum.
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.

PHASMATRON width of spectrum on film: Δs=f*ΔE
where f is the focal length of the lens used on the camera.
take f=50mm,
Δs=14.69567mm on film.
If we took instead f=700mm then

Next we indicate some common constants of the SF10 crystal:

nF, nC, nD, are the refraction indexes for the Hydrogen F, C lines and the sodium D line. (4861.327A=F, 6562.8A=C, 5892.938=D)
(nF- nC) is called the mean dispersivity
for nF=1.74648,  nC=1.72085 and  nD=1.72803 we get:
SF10 mean dispersivity: (nF- nC)=0.02563
Let ω=(nF- nC)/(nD- 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(nλ).
for nλ=1.72803, =>
θ=59.945468° (for sodium light)

Next, the internal transmittance table for the SF10 crystal follows:

 λ (nm) Ti (5mm) 2325.4 0.959 1970.1 0.990 1529.6 0.999 1060.0 0.999 700 0.999 660 0.999 620 0.999 580 0.999 546.1 0.999 500 0.998 460 0.995 435.8 0.990 420 0.981 404.7 0.952 400 0.93 390 0.83 380 0.59 370 0.21 365.0 0.06

Internal transmittance for the SF10 crystal

Transmittance graph for the SF10 crystal[1]

To find the internal transmittance for a block of thickness K other than 5mm use the following formula[2]:
SF10 internal transmittance : Ti(Kmm)=Ti(5mm){K/5}
For example the internal transmittance for 25mm for the 420nm line will be[3]:
T420(25mm)=0.981{25/5}=0.9815=0.908542.

Notes/References

1. SCHOTT AG Technical Information (Transmittance of SF10 Optical Glass).
2. SCHOTT AG Technical Information, Optics for Devices (Transmittance of Optical Glass).
3. To compare between different available optical glass materials for UV/vis/IR, consult this Quora answer.