Appendix: The Fresnel Equations
87
r || = r y =
E 0r
E 0i
y
=
n 2 cos θ 1 − n 1 cos θ 2
n 1 cos θ 2 + n 2 cos θ 1
(2.45)
t ⊥ = t x =
E 0t
E 0i
x
=
2n 1 cos θ 1
n 1 cos θ 1 + n 2 cos θ 2
(2.46)
t || = t y =
E 0t
E 0i
y
=
2n 1 cos θ 1
n 1 cos θ 2 + n 2 cos θ 1
(2.47)
If light is incident perpendicularly on the material interface, then the angles are θ
1 = θ 2 = 0. From Eqs. (2.44) and (2.45) it follows that the reflection coefficients are
r x (θ 1 = 0) = −r y (θ 2 = 0) =
n 1 − n 2
n 1 + n 2
(2.48)
Similarly, for θ 1 = θ 2 = 0, the transmission coefficients are
t x (θ 1 = 0) = t y (θ 2 = 0) =
2n 1
n 1 + n 2
(2.49)
Example 2A.1 Consider the case when light traveling in air (n air = 1.00) is incident
perpendicularly on a smooth glass surface that has a refractive index n tissue = 1.48.
What are the reflection and transmission coefficients?
Solution From Eq. (2.48) with n 1 = n air and n 2 = n glass it follows that the reflection
coefficient is
r x = −r x = (1.48 − 1.00)/(1.48 + 1.00) = 0.194
and from Eq. (2.49) the transmission coefficient is
t x = t y = 2(1.00)/(1.48 + 1.00) = 0.806
The change in sign of the reflection coefficient r x means that the field of the
perpendicular component shifts by 180° upon reflection.
The field amplitude ratios can be used to calculate the reflectance R (the ratio of
the reflected to the incident flux or power) and the transmittance T (the ratio of the
transmitted to the incident flux or power). For linearly polarized light in which the
vibrational plane of the incident light is perpendicular to the interface plane, the total
reflectance and transmittance are
R ⊥ =
E 0r
E 0i
2
x
= R x = r
2
x
(2.50)
R || =
E 0r
E 0i
2
y
= R y = r
2
y
(2.51)
87
r || = r y =
E 0r
E 0i
y
=
n 2 cos θ 1 − n 1 cos θ 2
n 1 cos θ 2 + n 2 cos θ 1
(2.45)
t ⊥ = t x =
E 0t
E 0i
x
=
2n 1 cos θ 1
n 1 cos θ 1 + n 2 cos θ 2
(2.46)
t || = t y =
E 0t
E 0i
y
=
2n 1 cos θ 1
n 1 cos θ 2 + n 2 cos θ 1
(2.47)
If light is incident perpendicularly on the material interface, then the angles are θ
1 = θ 2 = 0. From Eqs. (2.44) and (2.45) it follows that the reflection coefficients are
r x (θ 1 = 0) = −r y (θ 2 = 0) =
n 1 − n 2
n 1 + n 2
(2.48)
Similarly, for θ 1 = θ 2 = 0, the transmission coefficients are
t x (θ 1 = 0) = t y (θ 2 = 0) =
2n 1
n 1 + n 2
(2.49)
Example 2A.1 Consider the case when light traveling in air (n air = 1.00) is incident
perpendicularly on a smooth glass surface that has a refractive index n tissue = 1.48.
What are the reflection and transmission coefficients?
Solution From Eq. (2.48) with n 1 = n air and n 2 = n glass it follows that the reflection
coefficient is
r x = −r x = (1.48 − 1.00)/(1.48 + 1.00) = 0.194
and from Eq. (2.49) the transmission coefficient is
t x = t y = 2(1.00)/(1.48 + 1.00) = 0.806
The change in sign of the reflection coefficient r x means that the field of the
perpendicular component shifts by 180° upon reflection.
The field amplitude ratios can be used to calculate the reflectance R (the ratio of
the reflected to the incident flux or power) and the transmittance T (the ratio of the
transmitted to the incident flux or power). For linearly polarized light in which the
vibrational plane of the incident light is perpendicular to the interface plane, the total
reflectance and transmittance are
R ⊥ =
E 0r
E 0i
2
x
= R x = r
2
x
(2.50)
R || =
E 0r
E 0i
2
y
= R y = r
2
y
(2.51)