Birefringence and wave plates
In anisotropic crystals, the refraction index depends on the propagation direction and on the polarization of the EM wave. We consider a particular material for which the index of refraction has the
values ns and np for S-polarized and P -polarized waves (i.e. for polarizations along z or in the xy
plane), respectively. Both ns and np are assumed to be real and positive.
a) Assume that a light wave is incident on the surface of the material with wavevector k in the xy plane and a linear polarization directed at an angle θ with respect
α
y
to the incidence plane (i.e. the polarization is a mixing of S- and P -polarizations).
z x
In this configuration, the incident ray splits up into two refracted rays at different
θi
angles θr± = θr ± α/2. Show how from the measurement of θr and α the values of
ns and np may be obtained (you can assume np,s = n̄ ± δn/2 with δn/n̄ ≪ 1 and
keep only first order therms in δn/n̄).
b) Now assume normal incidence (θi = 0) and the polarization to be oriented at an angle π/4 with respect to the ŷ and
ẑ axes, i.e. along the bietrix of the yz plane. The crystal
has a thickness equal to d(≫ λ). Fin the values of d such y
that the light exiting the surface opposite to the incidence
E
one is either circularly polarized or linearly polarized but in
x
a direction rotated by π/2 with respect to the incident one.
z
d
(Neglect the difference in the reflection coefficient between
S- and P -polarizations).
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Solution
a) The incident wave can be written as the overlap of two waves having P - and S-polarization (i.e.
lying in the xy plane or directed along z), respectively. The observed birefringence is due to the P
and S components satysfying Snell’s refraction law with different values of the refraction index. The
corresponding angles are
sin θr,p =
sin θi
,
np
sin θr,s =
sin θi
.
ns
By posing θr,(s,p) = θr ± α/2 and np,s = n̄ ± δn/2 we obtain (up to first order in δn/n̄ ≪ 1)
α
sin θi
δn
sin θr ± cos θr ≃
,
1±
2
2n̄
n̄
(1)
(2)
from which sin θr = sin θi /n̄ and
α=
sin2 θr
sin θi δn
.
=
δn
cos θr n̄2
cos θr sin θi
(3)
Thus the values of np and ns are obtained from the measured angles as
np,s =
sin θi
sin2 θr
±
.
sin θr 2 cos θr sin θi
(4)
b) In order to turn the incident linearly polarized light into circularly polarized light it is necessary
for the two components to get out of phase by π/2 thanks to the difference in the optical path. This
brings the condition
δφ = kδnd =
2πδnd . π
= ,
λ
2
(5)
i.e. d = λ/(4δn). This is called a quarter-wave plate. If δφ = π instead, i.e. d = λ/(2δn), there is a
relative change of sign between the two components, which leads to a polarization rotation of π/2;
this is an half-wave plate.
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