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Dispersion of a Prism:
A
p
G
a1
D
a1’
d
a2
a2’
F
E
B
C
From the geometry:
Since DE and EF are normals to AB and AC:
ADE = EFA = 900
(eq 1)
From polygon ADEF:
90 + 90 + p + DEF = 360  p + DEF = 180
(eq 2)
From triangle DEF:
a1’ + a2 + DEF = 180
(eq 3)
So, from eq 1 and eq 2:
a1’ + a2 = p
Note that:
So, from DFG:
GDF = a1 – a1’ and GFD = a2’ – a2
DGF + GDF + GFD = 180
and, d = 180 - DGF

d = a1 + a2’ - p
One can use the two boxed equations and Snell’s law to calculate the deviation of a
prism.
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Thick Prisms:
It is possible to combine the previous equations and Snell’s law to get an expression for
the deviation of a general prism of index n, in terms of the incident and prism angles:
1


d  a1  p  sin 1  n 2  sin 2 a1  2 sin  p   cos p sin a1 


although it is generally more convenient to simply do the ray trace using the previous
page’s equations.


Plotting the above equation for a prism with prism angle d = 600 over a range of
incidence angles gives (assuming n = 1.5):
Evidently, the deviation angle is a strong function of the angle of incidence. Angles
below ~290 also result in TIR inside the prism, rather than a deviated exit ray.
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Thin Prisms:
Plotting the deviation of a prism with a small (p = 50) apex angle:
We see that the deviation angle is nearly a constant over a wide range of incident angles.
Apparently we need a “thin-prism approximation” to this deviation angle:
Assuming that the angles are small enough to substitute the angle for the sin, and
referring to the prism drawing, we get:
a
a1  1
n
a
a2  p  a1  p  1
n

a2  na2  np  a1
d  a1  a2  p  np  p
Hence:
d  p(n  1)
which, for p=5 and n=1.5 gives 2.5 degrees.
By using the Taylor expansion for sin, we can show (but won’t here) that the third order
expression for the deviation is the paraxial displacement plus a correction factor:
 a 2 n  1

d  pn  11  1
 
2n


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If we consider dispersion (variation of index with wavelength), we can calculate:
d d  pnd  1 ; d F  pnF  1 ; dC  pnC  1 at the usual wavelengths of
d = 588nm, F = 486nm, and C = 656nm.
d
, we get:
d
d d F  dC pnF  1  pnC  1 1



d
dd
pnd  1
Vd
Hence, the relative deviation of a thin prism is just the inverse of the V-number.
Defining the relative deviation as
Achromatic Prism:
The analogy with an achromat is obvious. If we wish to construct a prism with non-zero
deviation, d, but zero change in deviation, d, we can add two thin prisms of different
glasses such that:
d1 d 2
d  d1  d 2 and

0
V1 V2
(where the d’s are understood to be at the d = 588nm line)
Substituting the paraxial equation for the deviation, d=p(n-1), and solving these two
equations simultaneously, we get:
dV1
p1 
n1  1V1  V2 
(where d is the total deviation of the pair)
dV2
p2 
n2  1V2  V1 
It’s apparent that one prism will have a negative apex angle. What does this mean, given
our figure?
Rotation from 1st face to 2nd face:
CCW > 0
CW < 0
n
n
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Diagram of an achromatic prism
p1
d
red ray
blue ray
p2
The Direct Vision Prism:
Prisms are often used to separate wavelengths. The deviation from the initial ray paths
causes bends in the optical system and is inconvenient. Hence, it is sometimes required
that a prism have a large wavelength dependent deviation, but zero average deviation.
We can do this by solving the equations: d1  d 2  0, and
d 
d1 d 2

V1 V2
Making the appropriate substitutions, we end up with:
d  V1V2
p1 
n1  1V2  V1 
d  V1V2
n2  1V1  V2 
Which gives you the apex angles, given the V-numbers and the desired deviation.
p2 
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Diagram of a Direct Vision Prism
F
d
C
An instrument called a hand spectroscope is often constructed by stacking multiple Direct
Vision dispersing prisms together:
Hand Spectroscope
Eye
Slit
Collimating
lens
Scanning:
Two thin prisms that can be rotated independently, can form a beam scanner. Depending
on the prism angles, very fine control of beam direction is possible.
Consider coupling two optical fibers using a free-space connection:
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Typical values:
 core dia = 10m,
 Acceptable misalignment of focused spot on core = 1m
 Lens focal length = 5mm
  max angular misalignment = 1/5000 = 40 arc sec
This is beyond economically achievable manufacturing accuracy – the coupling must be
adjusted after manufacture. This is expensive.
Optical system for automating fiber coupling:
US Pat # 6,597,829 “1xN optical fiber switch”
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Appendix: Zemax’s “Glass Map”:
The index of refraction of glasses is traditionally taken at three wavelengths (for use in
the visible). These are the Fraunhofer lines for hydrogen and helium ( a nearby line from
mercury was used before helium became widely available). These were convenient,
because they spanned the visible spectrum and could be duplicated with great precision in
any well-equipped laboratory.
F (hydrogen)  486nm
d (helium)  588 nm
C (hydrogen)  656 nm
In the “glass map”, the index of refraction is the index at the d line (nd), and the ‘Abbe
Number’ (often called the “V-number” by those who aren’t sure how to pronounce
“Abbe”) is:
n 1
Vd  d
nF  nC
pg. 8