Class 5 - Fresnel equations
1
Fresnel coefficients for silver
The refractive index of silver at λ 625 nm is n 0.13 4i. Calculate the Fresnel coefficients rs and rp for
incidence angle of 80o .
Solution: In this tutorial we will see that the imaginary part of the refractive index has influence on the
reflection amplitudes.
We can use the same expressions for the Fresnel coefficients derived in lecture, just substitue a complex
refractive index. Let’s recall the definition of Fresnel coefficients for both p (TM) and s (TE) polarizations
θi nt cos θt
EErs nni cos
cos θ
n cos θ
(1.1)
cos θi ni cos θt
EErp nnt cos
θ
n cos θ
(1.2)
s
rs
i
i
i
t
t
p
rp
i
i
To find cos θt we can use Snell’s law
ni sin θi
or
sin θt
2
In case of air ni
t
i
nt sin θt
(1.3)
d
2
ni
nt
t
ñ
sin θi
2
1
cos θt
n1
cos θt
rs
1
ni
nt
sin2 θi
b
cos θi (1.5)
a 2
n sin2 θi
a t2
2
(1.6)
sin θi
a
n2 cos θi n2t sin2 θi
rp at
n2t sin2 θi n2t cos θi
cos θi
(1.4)
n2t sin2 θi
t
so rs and rp become
2
nt
(1.7)
Now we can substitute the values from the question
rs
cos p80o q cos p80o q
b
p0.13
p0.13
b
4iq
sin2 p80o q
0.994 0.084i
2
4iq sin2 p80o q
2
1
(1.8)
b
p0.13 4iq
p0.13 4iq2 sin2 p80o q
rp b
0.3676
p0.13 4iq2 sin2 p80o q p0.13 4iq2 cos p80o q
2
cos p80o q 0.898i
(1.9)
It is convenient to show the coefficients in polar form
rs
0.997e3.057i
(1.10)
0.970e1.96i
(1.11)
rp
Please note that the phase for s polarization and p polarization can differ significantly.
Brewster’s angle is also present in an absorptive medium. Although in this case the, reflection coefficient
for p polarization doesn’t vanish completely.
2
Question 2 - Fresnel rhomb
A linearly polarized plane wave in air (n 1)
Ein
<e
"
E0
?
x̂
ŷ
2
eiωtikz
*
(2.1)
is incident perpendicularly on a triangular prism with refractive index n 2 as in figure below:
We will assume for simplicity that there are no reflections of the incident wave upon entering and exiting
the prism. Calculate the outgoing wave. What is the polarization of the outgoing wave?
Solution: From the figure we can see that the outgoing wave propagates in ŷ direction. We will calculate
explicitly and see that there is a total internal reflection. Part of the incident wave having polarization in
x̂ direction is s polarized and is maintained at the outgoing wave. Although the part that is polarized in ŷ
direction, which is p polarized, is now polarized at ẑ direction :
Ein
<e
"
E0
?
x̂
ŷ
2
e iωt ikz
*
ñ
Eout
2
<e
"
E0
rs x̂
? rp ẑ
2
*
e
iωt iky
(2.2)
All is left is to calculate rs and rp . In our case ni
rs
ni cos θi nt
ni cos θi
rp
ni
1
c
ni
nt
sin2 θi
ni
nt
sin θi
2
1
c
2
ni
nt
2
1
nt
nt cos θi ni
c
c
2
1
ni
nt
sin θi
2
nt cos θi
Indeed we see that
a
2 cos p45o q 1 4 sin2 p45o q
a
2 cos p45o q
1 4 sin2 p45o q
cos p45o q 2 1 4 sin2 p45o q
a
2 1 4 sin2 p45o q cos p45o q
2
sin2 θi
2, nt 1 and θi 45o
a
?
?2 i e1.231i
2
i
?12 2i
?12 2i
e2.462i
|rs | |rp | 1
(2.3)
(2.4)
(2.5)
which is the case of total internal reflection. The difference between the polarizations is in phase. In
conclusion, the outgoing wave is given by
Eout
<e
"
E0
1.231i
e
x̂
e2.462i ẑ
2
?
*
eiωt
(2.6)
iky
To see the polarization more clearly we will take out e2.462i as global phase
Eout
<e
"
E0
1.231i
e
x̂
?
2
ẑ
eiωt
*
iky 2.462i
<e
"
E0
0.4πi
e
x̂
?
2
ẑ
eiωt
iky 2.462i
*
(2.7)
which is an elliptically polarized light! In fact, making a unique geometrical shape of a rhomb we can achieve
an affect of quarter wave plate, convert linearly polarized light into circularly polarized. What is the major
difference between Fresnel retarder and half wave plate retarder? Here some actual products that are made
in this principle
Figure 2.1: Fresnel rhombs
3
3
A polarizer based on Brewster’s angle
A slab of glass (nt 1.5) can be used
as a polarizer, when unpolarized light is incident on it from air in the
Brewster’s angle θB arctan nnti :
Figure 3.1: Brewster angle . www.tydexoptics.com/© and the reflection coefficients . wikipedia.com©
Unpolarized light is incident on the glass slab at the Brewster angle ( 56o ). At this angle, there is no
reflection of the p polarized light, so all of the reflected light is s polarized. Note that the light exiting the
slab is also incident at the appropriate Brewster angle ( 34o ). The reflected light is therefore s polarized this is how polarization is achieved.
1. The intensity of the unpolarized incident light is I0 . What is the intensity of the polarized light after
the slab? Ignore multiple-reflections inside the glass.
4
2. Approximate the error in your answer to the previous part, due to ignoring multiple-reflections inside
the glass. Do so by calculating the effect of the 2nd order reflection.
Solution:
1. The reflectance of s polarized part at Brewster’s angle is:
rs
a
cos θB n2t sin2 θB
a 2
2
cos θB
nt
sin
θB
b
p1.5q2 sin2 p56o q
b
0.384
p1.5q2 sin2 p56o q
cos p56o q
cos p56o q (3.1)
Therefore the intensity from the 1st reflection is
Ir
1
2
21 I0 |rs |2 0.074I0
(3.2)
is due to equal distribution of intensity between s and p polarizations.
2. Solution: Note that in this case Rs is the same for inner and outer reflection, meaning Rs,airÑglass Rs,glassÑair . The intensity of the light due to reflectance from the right facet of the slab, and then
transmittance from the left facet is:
Irr2s
0.5I0 TT E,12 RT E,21 TT E,21
1 RT E 0.852
ñ Irr2s 0.5 0.852 0.148 0.852 I0 0.054I0
TT E
5
(3.3)
(3.4)
(3.5)