Optical Characteristics of SF10 Crystal Prisms

Version 1.2 of 1/2/2011-10:52 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[1]

graph of sellmeier's equation for 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.

dn/dλ:

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.

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[1].

The Resolution is then defined to be R=λ/Δλ.

Resolution: R=λ/Δλ.

For λ=5892.938A and Δλ=0.3866148A, R=15242.4.

Consider the formula of minimum deviation angle for one prism[2]:

sin{(D+A)/2}/sin{A/2}=nλ. (1)

The Phasmatron spectroscope has 60 degree prisms so A=60°. 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)

Here A=60°, (3) => dD/dn=1/cos{sin-1{nλ/2}}.

two-prism dispersion

Note E=2*D, so dE/dn=2*dD/dn, so using (3) we get for nD=1.72803, dD/dn=1.9869763 rad.

For the Phasmatron spectroscope which has two prisms,

dE/dn=2*dD/dn=2/cos{sin-1{nλ/2}}.

For nλ=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).

f is the camera's lens focal length, using a 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:

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°[3].

The width of the entire spectrum on film can be calculated.

Width of Spectrum on Film: Δs=f*ΔE.

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[4].

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[5].

(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.

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=arctan(nλ).

For nλ=1.72803, => θ=59.945468°.

Follows the internal transmittance table for the SF10 crystal:

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

internal transmittance for 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 : Ti(Kmm)=Ti(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.

Notes

  1. 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.
  2. Where D is the minimum deviation angle and A is the apical prism angle.
  3. 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. The true width will be around 16.84° owing to correct measuring of the internal angles. Consult article on Spectrum Angular Width.
  4. Also consult article on Phasmatron Photography.
  5. F=4861.327A, C=6562.8A, D=5892.938A.

References

  1. Linos "Optics Catalog" (formerly Spindler & Hoyer), 1999. (online)
  2. K.D. Alexopoulos, "General Physics: Optics", 1st Edition, Athens 1966 (in Greek).
  3. G. Bruhat, "Cours De Physique Generale: Optique", Sixieme Edition, Paris 1965 (in French).

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