Abstract

The depolarization of white light upon passage through a birefringent crystal, due to the unequal phase retardation of different wavelengths is considered. Assuming the incident light to be completely linearly polarized with the spectral power distribution of a blackbody radiator, the degree of polarization remaining after passage through various thicknesses of birefringent material is calculated. The problem is simplified by assuming that there is no dispersion in the crystal. Results show that under the assumptions made, the light from a 3000 K source is less than 1% polarized after passage through 2.66 × 10−1 mm of quartz and 1.41 × 10−2 mm of calcite.

© 1975 Optical Society of America

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References

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  1. William A. Shurcliff, Polarized Light, Production and Use (Harvard U.P., Cambridge, Mass., 1962), pp. 11 and 12.
  2. Reference 1, p. 22.
  3. Reference 1, pp. 165 and 169.
  4. I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, Fourth ed. (Translated from the Russian) (Academic, New York, 1965), pp. 325 and 1079.
  5. Reference 4, p. 494.
  6. Reference 4, p. 490.

Gradshteyn, I. S.

I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, Fourth ed. (Translated from the Russian) (Academic, New York, 1965), pp. 325 and 1079.

Ryzhik, I. M.

I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, Fourth ed. (Translated from the Russian) (Academic, New York, 1965), pp. 325 and 1079.

Shurcliff, William A.

William A. Shurcliff, Polarized Light, Production and Use (Harvard U.P., Cambridge, Mass., 1962), pp. 11 and 12.

Other (6)

William A. Shurcliff, Polarized Light, Production and Use (Harvard U.P., Cambridge, Mass., 1962), pp. 11 and 12.

Reference 1, p. 22.

Reference 1, pp. 165 and 169.

I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, Fourth ed. (Translated from the Russian) (Academic, New York, 1965), pp. 325 and 1079.

Reference 4, p. 494.

Reference 4, p. 490.

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Figures (2)

FIG. 1
FIG. 1

Alteration of polarization (Stokes) parameters of light upon passage through a crystal.

FIG. 2
FIG. 2

Degree of polarization (V) remaining (expressed as %) after passage of 45° linearly polarized incident light through crystal thickness (t), corresponding to thickness parameter X(x = 2πkTΔnt/hc). Vc (dashed curve) is the degree of 45° linear polarization. Vs (dotted curve) is the degree of circular polarization. V (solid curve) is the total degree of polarization.

Tables (1)

Tables Icon

TABLE I Degree of polarization (V) remaining after passage of 45° linearly polarized light through crystal thickness (t) corresponding to thickness parameter (X). X = 2πkTΔnt/hc, where Δn = difference between principal refractive indices, T = Kelvin temperature of source, k = Boltzman’s constant, h = Planck’s constant, c = vacuum speed of light.

Equations (34)

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V = ( I d I i ) / ( I d + I i ) .
V = ( M e 2 + C e 2 + S e 2 ) 1 / 2 / I e .
V m = M e / I e ,
V c = C e / I e ,
V s = S e / I e ,
[ V e ] = [ M c ] [ V i ] ,
[ V i ] = [ 1 0 1 0 ]
[ M c ] = [ 1 0 0 0 0 D 2 + G 2 0 0 0 0 D 2 + G 2 2 D G 0 0 2 D G 2 G 2 1 ] ,
D = M sin 1 2 δ , C = cos 1 2 δ , δ = retardance in radians , M = orientation of fast eigenvactor of crystal ,
[ M c ] = [ 1 0 0 0 0 1 0 0 0 0 ( cos 2 1 2 δ sin 2 1 2 δ ) ( 2 sin 1 2 δ cos 1 2 δ ) 0 0 ( 2 sin 1 2 δ cos 1 2 δ ) ( cos 2 1 2 δ sin 2 1 2 δ ) ] .
ψ λ d λ = 8 π h c λ 5 · e h c / λ k T 1 e h c / λ k T .
λ T = 0 ψ λ d λ = 0 8 π h c λ 5 · e h c / λ k T 1 e h c / λ k T d λ ,
[ I i M i C i S i ] = 0 ψ λ d λ [ 1 0 1 0 ]
[ I e M e C e S e ] = [ 1 0 0 0 0 1 0 0 0 0 ( cos 2 1 2 δ sin 2 1 2 δ ) ( 2 sin 1 2 δ cos 1 2 δ ) 0 0 ( 2 sin 1 2 δ cos 1 2 δ ) ( cos 2 1 2 δ sin 2 1 2 δ ) ] · 0 ψ λ d λ [ 1 0 1 0 ] .
I e = 0 ψ λ d λ ,
M e = 0 ,
C e = 0 ψ λ d λ ( cos 2 1 2 δ sin 2 1 2 δ ) ,
S e = 0 ψ λ d λ ( 2 sin 1 2 δ cos 1 2 δ ) .
I e = 0 8 π h c λ 5 e h c / λ k T 1 e h c / λ k T d λ .
I e = C 1 0 1 λ 5 · e C 2 / λ 1 e C 2 / λ d λ ,
I e = C 1 C 2 4 0 x 3 e x 1 d x .
I e = ( C 1 / C 2 4 ) ( π 4 / 15 ) .
C e = C 1 C 2 4 0 x 3 e x 1 cos β x d x ,
C e = C 1 π 4 C 2 4 [ 2 cosh 2 β π + 1 sin h 4 β π 3 β 4 π 4 ] .
S e = C 1 C 2 4 0 x 3 e x 1 sin β x d x
S e = ( C 1 / C 2 4 ) 0 sin β x ( x 3 ) ( e x + e 2 x + e 3 x + ) d x .
G n = ( C 1 / C 2 4 ) 0 sin β x ( x 3 ) e n x d x , n = 1 , 2 , 3 .
G n = C 1 C 2 4 [ ( β 2 n 2 ) 24 β n ( β 2 + n 2 ) 4 ] ,
S e = C 1 C 2 4 0 sin β x ( x 3 e x 1 ) d x = n = 1 C 1 C 2 4 24 β n ( β 2 n 2 ) ( β 2 n 2 ) 4 n = 1 , 2 , 3 , .
V c = C e I e = [ 15 2 cosh 2 β π + 1 sinh 4 β π 3 β 4 π 4 ] .
V c = 15 [ 2 cosh 2 X + 1 sinh 4 X 3 X 4 ] .
V s = 15 n = 1 24 n X π ( X 2 n 2 π 2 ) ( X 2 n 2 π 2 ) .
V = { ( C e / I e ) 2 + ( S e / I e 2 ) } 1 / 2 = { V e 2 + V s 2 } 1 / 2 .
t = X Δ n h c 2 π 2 k T .