S and P polarizations

The plane of incidence is defined by the incoming radiation's propagation vector and the normal vector of the surface.

There are two sets of Fresnel coefficients for two different linear polarization components of the incident wave. Since any polarization state can be resolved into a combination of two orthogonal linear polarizations, this is sufficient for any problem. Likewise, unpolarized (or "randomly polarized") light has an equal amount of power in each of two linear polarizations.

The s polarization refers to polarization of a wave's electric field normal to the plane of incidence (the z direction in the derivation below); then the magnetic field is in the plane of incidence. The p polarization refers to polarization of the electric field in the plane of incidence (the xy plane in the derivation below); then the magnetic field is normal to the plane of incidence.

Although the reflection and transmission are dependent on polarization, at normal incidence (θ = 0) there is no distinction between them so all polarization states are governed by a single set of Fresnel coefficients (and another special case is mentioned below in which that is true).

Special cases

Normal incidence

For the case of normal incidence, ${\displaystyle \theta _{\mathrm {i} }=\theta _{\mathrm {t} }=0}$, and there is no distinction between s and p polarization. Thus, the reflectance simplifies to

$${\displaystyle R_{0}=\left|{\frac {n_{1}-n_{2}}{n_{1}+n_{2}}}\right|^{2}\,.}$$

For common glass (n₂ ≈ 1.5) surrounded by air (n₁ = 1), the power reflectance at normal incidence can be seen to be about 4%, or 8% accounting for both sides of a glass pane.

45° Incidence

Reflection at 45° incidence is very commonly used for making 90° turns. For the case of light traversing from a less dense medium into a denser one at 45° incidence (θ = 45°), it follows algebraically from the above equations that R_p equals the square of R_s:

$${\displaystyle R_{\text{p}}=R_{\text{s}}^{2}}$$

This can be used to either verify the consistency of the measurements of R_s and R_p, or to derive one of them when the other is known. This relationship is only valid for the simple case of a single plane interface between two homogeneous materials, not for films on substrates, where a more complex analysis is required.

Measurements of R_s and R_p at 45° can be used to estimate the reflectivity at normal incidence. The "average of averages" obtained by calculating first the arithmetic as well as the geometric average of R_s and R_p, and then averaging these two averages again arithmetically, gives a value for R₀ with an error of less than about 3% for most common optical materials. This is useful because measurements at normal incidence can be difficult to achieve in an experimental setup since the incoming beam and the detector will obstruct each other. However, since the dependence of R_s and R_p on the angle of incidence for angles below 10° is very small, a measurement at about 5° will usually be a good approximation for normal incidence, while allowing for a separation of the incoming and reflected beam.

Alternative forms

In the above formula for r_s‍, if we put ${\displaystyle n_{2}=n_{1}\sin \theta _{\text{i}}/\sin \theta _{\text{t}}}$ (Snell's law) and multiply the numerator and denominator by 1/n₁ sin θ_t‍, we obtain ^[10][11]

$${\displaystyle r_{\text{s}}=-{\frac {\sin(\theta _{\text{i}}-\theta _{\text{t}})}{\sin(\theta _{\text{i}}+\theta _{\text{t}})}}.}$$

If we do likewise with the formula for r_p‍, the result is easily shown to be equivalent to ^[12][13]

$${\displaystyle r_{\text{p}}={\frac {\tan(\theta _{\text{i}}-\theta _{\text{t}})}{\tan(\theta _{\text{i}}+\theta _{\text{t}})}}.}$$

These formulas ^[14][15][16] are known respectively as Fresnel's sine law and Fresnel's tangent law.^[17] Although at normal incidence these expressions reduce to 0/0, one can see that they yield the correct results in the limit as‍ θ_i → 0.

Material parameters

In order to compute meaningful Fresnel coefficients, we must assume that the medium is (approximately) linear and homogeneous. If the medium is also isotropic, the four field vectors‍ E, B, D, H  are related by

$${\displaystyle {\begin{aligned}\mathbf {D} &=\epsilon \mathbf {E} \\\mathbf {B} &=\mu \mathbf {H} \,,\end{aligned}}}$$

where ϵ and μ are scalars, known respectively as the (electric) permittivity and the (magnetic) permeability of the medium. For a vacuum, these have the values ϵ₀ and μ₀, respectively. Hence we define the relative permittivity (or dielectric constant) ϵ_rel = ϵ/ϵ₀ , and the relative permeability μ_rel = μ/μ₀.

In optics it is common to assume that the medium is non-magnetic, so that μ_rel = 1. For ferromagnetic materials at radio/microwave frequencies, larger values of μ_rel must be taken into account. But, for optically transparent media, and for all other materials at optical frequencies (except possible metamaterials), μ_rel is indeed very close to 1; that is, μ ≈ μ₀.

In optics, one usually knows the refractive index n of the medium, which is the ratio of the speed of light in a vacuum (c) to the speed of light in the medium. In the analysis of partial reflection and transmission, one is also interested in the electromagnetic wave impedance Z, which is the ratio of the amplitude of E to the amplitude of H. It is therefore desirable to express n and Z in terms of ϵ and μ, and thence to relate Z to n. The last-mentioned relation, however, will make it convenient to derive the reflection coefficients in terms of the wave admittance Y, which is the reciprocal of the wave impedance Z.

In the case of uniform plane sinusoidal waves, the wave impedance or admittance is known as the intrinsic impedance or admittance of the medium. This case is the one for which the Fresnel coefficients are to be derived.

Electromagnetic plane waves

In a uniform plane sinusoidal electromagnetic wave, the electric field E has the form

${\displaystyle \mathbf {E_{k}} e^{i(\mathbf {k\cdot r} -\omega t)},}$

(1)

where E_k is the (constant) complex amplitude vector,  i is the imaginary unit,  k is the wave vector (whose magnitude k is the angular wavenumber),  r is the position vector,  ω is the angular frequency,  t is time, and it is understood that the real part of the expression is the physical field.^{[Note 1]}  The value of the expression is unchanged if the position r varies in a direction normal to k; hence k is normal to the wavefronts.

To advance the phase by the angle ϕ, we replace ωt by ωt+ϕ  (that is, we replace −ωt by −ωt−ϕ),  with the result that the (complex) field is multiplied by e^−iϕ. So a phase advance is equivalent to multiplication by a complex constant with a negative argument. This becomes more obvious when the field (1) is factored as‍ E_k e^ik⋅re^−iωt,  where the last factor contains the time-dependence. That factor also implies that differentiation w.r.t. time corresponds to multiplication by −iω. ^{[Note 2]}

If ℓ is the component of r in the direction of k , the field (1) can be written‍ E_k e^{i(kℓ−ωt)}.  If the argument of e^i(⋯) is to be constant,  ℓ must increase at the velocity‍ ${\displaystyle \omega /k\,,\,}$ known as the phase velocity  (v_p). This in turn is equal to‍ ${\displaystyle c/n}$. Solving for k gives

${\displaystyle k=n\omega /c\,.}$

(2)

As usual, we drop the time-dependent factor e^−iωt which is understood to multiply every complex field quantity. The electric field for a uniform plane sine wave will then be represented by the location-dependent phasor

${\displaystyle \mathbf {E_{k}} e^{i\mathbf {k\cdot r} }.}$

(3)

For fields of that form, Faraday's law and the Maxwell-Ampère law respectively reduce to ^[33]

$${\displaystyle {\begin{aligned}\omega \mathbf {B} &=\mathbf {k} \times \mathbf {E} \\\omega \mathbf {D} &=-\mathbf {k} \times \mathbf {H} \,.\end{aligned}}}$$

Putting  B = μH  and  D = ϵE‍,‍ as above, we can eliminate B and D to obtain equations in only E and H:

$${\displaystyle {\begin{aligned}\omega \mu \mathbf {H} &=\mathbf {k} \times \mathbf {E} \\\omega \epsilon \mathbf {E} &=-\mathbf {k} \times \mathbf {H} \,.\end{aligned}}}$$

If the material parameters ϵ and μ are real (as in a lossless dielectric), these equations show that k , E , H  form a right-handed orthogonal triad, so that the same equations apply to the magnitudes of the respective vectors. Taking the magnitude equations and substituting from (2), we obtain

$${\displaystyle {\begin{aligned}\mu cH&=nE\\\epsilon cE&=nH\,,\end{aligned}}}$$

where H and E are the magnitudes of H and E.  Multiplying the last two equations gives

${\displaystyle n=c\,{\sqrt {\mu \epsilon }}\,.}$

(4)

Dividing (or cross-multiplying) the same two equations gives  H = YE‍,‍ where

${\displaystyle Y={\sqrt {\epsilon /\mu }}\,.}$

(5)

This is the intrinsic admittance.

From (4) we obtain the phase velocity‍ ${\displaystyle c/n=1{\big /}\!{\sqrt {\mu \epsilon \,}}}$.  For a vacuum this reduces to‍ ${\displaystyle c=1{\big /}\!{\sqrt {\mu _{0}\epsilon _{0}}}}$.  Dividing the second result by the first gives

$${\displaystyle n={\sqrt {\mu _{\text{rel}}\epsilon _{\text{rel}}}}\,.}$$

For a non-magnetic medium (the usual case), this becomes ${\displaystyle n={\sqrt {\epsilon _{\text{rel}}}}}$.

(Taking the reciprocal of (5), we find that the intrinsic impedance is ${\textstyle Z={\sqrt {\mu /\epsilon }}}$. In a vacuum this takes the value ${\textstyle Z_{0}={\sqrt {\mu _{0}/\epsilon _{0}}}\,\approx 377\,\Omega \,,}$‍ known as the impedance of free space. By division, ${\textstyle Z/Z_{0}={\sqrt {\mu _{\text{rel}}/\epsilon _{\text{rel}}}}}$.  For a non-magnetic medium, this becomes ${\displaystyle Z=Z_{0}{\big /}\!{\sqrt {\epsilon _{\text{rel}}}}=Z_{0}/n.}$)

The wave vectors

Incident, reflected, and transmitted wave vectors (k_i‍, k_r‍, and k_t ), for incidence from a medium with refractive index n₁ to a medium with refractive index n₂. The red arrows are perpendicular to the wave vectors.

In Cartesian coordinates (x, y,‍z), let the region‍ y < 0‍ have refractive index n₁ , intrinsic admittance Y₁ , etc., and let the region‍ y > 0‍ have refractive index n₂ , intrinsic admittance Y₂ , etc. Then the xz plane is the interface, and the y axis is normal to the interface (see diagram). Let i and j (in bold roman type) be the unit vectors in the x and y directions, respectively. Let the plane of incidence be the xy plane (the plane of the page), with the angle of incidence θ_i measured from j towards i. Let the angle of refraction, measured in the same sense, be θ_t , where the subscript t stands for transmitted (reserving r for reflected).

In the absence of Doppler shifts, ω does not change on reflection or refraction. Hence, by (2), the magnitude of the wave vector is proportional to the refractive index.

So, for a given ω, if we redefine k as the magnitude of the wave vector in the reference medium (for which n =‍1), then the wave vector has magnitude n₁k in the first medium (region‍ y < 0‍ in the diagram) and magnitude n₂k in the second medium. From the magnitudes and the geometry, we find that the wave vectors are

$${\displaystyle {\begin{aligned}\mathbf {k} _{\text{i}}&=n_{1}k(\mathbf {i} \sin \theta _{\text{i}}+\mathbf {j} \cos \theta _{\text{i}})\\[.5ex]\mathbf {k} _{\text{r}}&=n_{1}k(\mathbf {i} \sin \theta _{\text{i}}-\mathbf {j} \cos \theta _{\text{i}})\\[.5ex]\mathbf {k} _{\text{t}}&=n_{2}k(\mathbf {i} \sin \theta _{\text{t}}+\mathbf {j} \cos \theta _{\text{t}})\\&=k(\mathbf {i} \,n_{1}\sin \theta _{\text{i}}+\mathbf {j} \,n_{2}\cos \theta _{\text{t}})\,,\end{aligned}}}$$

where the last step uses Snell's law. The corresponding dot products in the phasor form (3) are

${\displaystyle {\begin{aligned}\mathbf {k} _{\text{i}}\mathbf {\cdot r} &=n_{1}k(x\sin \theta _{\text{i}}+y\cos \theta _{\text{i}})\\\mathbf {k} _{\text{r}}\mathbf {\cdot r} &=n_{1}k(x\sin \theta _{\text{i}}-y\cos \theta _{\text{i}})\\\mathbf {k} _{\text{t}}\mathbf {\cdot r} &=k(n_{1}x\sin \theta _{\text{i}}+n_{2}y\cos \theta _{\text{t}})\,.\end{aligned}}}$

(6)

Hence:

At  ${\displaystyle y=0\,,~~~\mathbf {k} _{\text{i}}\mathbf {\cdot r} =\mathbf {k} _{\text{r}}\mathbf {\cdot r} =\mathbf {k} _{\text{t}}\mathbf {\cdot r} =n_{1}kx\sin \theta _{\text{i}}\,.}$

(7)

The s components

For the s polarization, the E field is parallel to the z axis and may therefore be described by its component in the z direction. Let the reflection and transmission coefficients be r_s and t_s , respectively. Then, if the incident E field is taken to have unit amplitude, the phasor form (3) of its z component is

${\displaystyle E_{\text{i}}=e^{i\mathbf {k} _{\text{i}}\mathbf {\cdot r} },}$

(8)

and the reflected and transmitted fields, in the same form, are

${\displaystyle {\begin{aligned}E_{\text{r}}&=r_{s\,}e^{i\mathbf {k} _{\text{r}}\mathbf {\cdot r} }\\E_{\text{t}}&=t_{s\,}e^{i\mathbf {k} _{\text{t}}\mathbf {\cdot r} }.\end{aligned}}}$

(9)

Under the sign convention used in this article, a positive reflection or transmission coefficient is one that preserves the direction of the transverse field, meaning (in this context) the field normal to the plane of incidence. For the s polarization, that means the E field. If the incident, reflected, and transmitted E fields (in the above equations) are in the z direction ("out of the page"), then the respective H fields are in the directions of the red arrows, since  k , E , H  form a right-handed orthogonal triad. The H fields may therefore be described by their components in the directions of those arrows, denoted by  H_i , H_r‍, H_t .  Then, since  H = YE‍,

${\displaystyle {\begin{aligned}H_{\text{i}}&=\,Y_{1}e^{i\mathbf {k} _{\text{i}}\mathbf {\cdot r} }\\H_{\text{r}}&=\,Y_{1}r_{s\,}e^{i\mathbf {k} _{\text{r}}\mathbf {\cdot r} }\\H_{\text{t}}&=\,Y_{2}t_{s\,}e^{i\mathbf {k} _{\text{t}}\mathbf {\cdot r} }.\end{aligned}}}$

(10)

At the interface, by the usual interface conditions for electromagnetic fields, the tangential components of the E and H fields must be continuous; that is,

${\displaystyle \left.{\begin{aligned}E_{\text{i}}+E_{\text{r}}&=E_{\text{t}}\\H_{\text{i}}\cos \theta _{\text{i}}-H_{\text{r}}\cos \theta _{\text{i}}&=H_{\text{t}}\cos \theta _{\text{t}}\end{aligned}}~~\right\}~~~{\text{at}}~~y=0\,.}$

(11)

When we substitute from equations (8) to (10) and then from (7), the exponential factors cancel out, so that the interface conditions reduce to the simultaneous equations

${\displaystyle {\begin{aligned}1+r_{\text{s}}&=\,t_{\text{s}}\\Y_{1}\cos \theta _{\text{i}}-Y_{1}r_{\text{s}}\cos \theta _{\text{i}}&=\,Y_{2}t_{\text{s}}\cos \theta _{\text{t}}\,,\end{aligned}}}$

(12)

which are easily solved for r_s and t_s‍, yielding

${\displaystyle r_{\text{s}}={\frac {Y_{1}\cos \theta _{\text{i}}-Y_{2}\cos \theta _{\text{t}}}{Y_{1}\cos \theta _{\text{i}}+Y_{2}\cos \theta _{\text{t}}}}}$

(13)

and

${\displaystyle t_{\text{s}}={\frac {2Y_{1}\cos \theta _{\text{i}}}{Y_{1}\cos \theta _{\text{i}}+Y_{2}\cos \theta _{\text{t}}}}\,.}$

(14)

At normal incidence‍ (θ_i = θ_t = 0), indicated by an additional subscript 0, these results become

${\displaystyle r_{\text{s0}}={\frac {Y_{1}-Y_{2}}{Y_{1}+Y_{2}}}}$

(15)

and

${\displaystyle t_{\text{s0}}={\frac {2Y_{1}}{Y_{1}+Y_{2}}}\,.}$

(16)

At grazing incidence‍ (θ_i → 90°), we have  cos θ_i → 0‍, hence  r_s → −1  and  t_s → 0.

The p components

For the p polarization, the incident, reflected, and transmitted E fields are parallel to the red arrows and may therefore be described by their components in the directions of those arrows. Let those components be  E_i , E_r‍, E_t  (redefining the symbols for the new context). Let the reflection and transmission coefficients be r_p and t_p. Then, if the incident E field is taken to have unit amplitude, we have

${\displaystyle {\begin{aligned}E_{\text{i}}&=e^{i\mathbf {k} _{\text{i}}\mathbf {\cdot r} }\\E_{\text{r}}&=r_{p\,}e^{i\mathbf {k} _{\text{r}}\mathbf {\cdot r} }\\E_{\text{t}}&=t_{p\,}e^{i\mathbf {k} _{\text{t}}\mathbf {\cdot r} }.\end{aligned}}}$

(17)

If the E fields are in the directions of the red arrows, then, in order for  k , E , H  to form a right-handed orthogonal triad, the respective H fields must be in the −z direction ("into the page") and may therefore be described by their components in that direction. This is consistent with the adopted sign convention, namely that a positive reflection or transmission coefficient is one that preserves the direction of the transverse field (the H field in the case of the p polarization). The agreement of the other field with the red arrows reveals an alternative definition of the sign convention: that a positive reflection or transmission coefficient is one for which the field vector in the plane of incidence points towards the same medium before and after reflection or transmission.^[34]

So, for the incident, reflected, and transmitted H fields, let the respective components in the −z direction be  H_i , H_r‍, H_t .  Then, since  H = YE‍,

${\displaystyle {\begin{aligned}H_{\text{i}}&=\,Y_{1}e^{i\mathbf {k} _{\text{i}}\mathbf {\cdot r} }\\H_{\text{r}}&=\,Y_{1}r_{p\,}e^{i\mathbf {k} _{\text{r}}\mathbf {\cdot r} }\\H_{\text{t}}&=\,Y_{2}t_{p\,}e^{i\mathbf {k} _{\text{t}}\mathbf {\cdot r} }.\end{aligned}}}$

(18)

At the interface, the tangential components of the E and H fields must be continuous; that is,

${\displaystyle \left.{\begin{aligned}E_{\text{i}}\cos \theta _{\text{i}}-E_{\text{r}}\cos \theta _{\text{i}}&=E_{\text{t}}\cos \theta _{\text{t}}\\H_{\text{i}}+H_{\text{r}}&=H_{\text{t}}\end{aligned}}~~\right\}~~~{\text{at}}~~y=0\,.}$

(19)

When we substitute from equations (17) and (18) and then from (7), the exponential factors again cancel out, so that the interface conditions reduce to

${\displaystyle {\begin{aligned}\cos \theta _{\text{i}}-r_{\text{p}}\cos \theta _{\text{i}}&=\,t_{\text{p}}\cos \theta _{\text{t}}\\Y_{1}+Y_{1}r_{\text{p}}&=\,Y_{2}t_{\text{p}}\,.\end{aligned}}}$

(20)

Solving for r_p and t_p‍, we find

${\displaystyle r_{\text{p}}={\frac {Y_{2}\cos \theta _{\text{i}}-Y_{1}\cos \theta _{\text{t}}}{Y_{2}\cos \theta _{\text{i}}+Y_{1}\cos \theta _{\text{t}}}}}$

(21)

and

${\displaystyle t_{\text{p}}={\frac {2Y_{1}\cos \theta _{\text{i}}}{Y_{2}\cos \theta _{\text{i}}+Y_{1}\cos \theta _{\text{t}}}}\,.}$

(22)

At normal incidence‍ (θ_i = θ_t = 0), indicated by an additional subscript 0, these results become

${\displaystyle r_{\text{p0}}={\frac {Y_{2}-Y_{1}}{Y_{2}+Y_{1}}}}$

(23)

and

${\displaystyle t_{\text{p0}}={\frac {2Y_{1}}{Y_{2}+Y_{1}}}\,.}$

(24)

At grazing incidence‍ (θ_i → 90°), we again have  cos θ_i → 0‍, hence  r_p → −1  and  t_p → 0.

Comparing (23) and (24) with (15) and (16), we see that at normal incidence, under the adopted sign convention, the transmission coefficients for the two polarizations are equal, whereas the reflection coefficients have equal magnitudes but opposite signs. While this clash of signs is a disadvantage of the convention, the attendant advantage is that the signs agree at grazing incidence.

Power ratios (reflectivity and transmissivity)

The Poynting vector for a wave is a vector whose component in any direction is the irradiance (power per unit area) of that wave on a surface perpendicular to that direction. For a plane sinusoidal wave the Poynting vector is‍ 1/2‍Re{E × H^∗}, where E and H are due only to the wave in question, and the asterisk denotes complex conjugation. Inside a lossless dielectric (the usual case), E and H are in phase, and at right angles to each other and to the wave vector k ; so, for s polarization, using the z and xy components of E and H respectively (or for p polarization, using the xy and -z components of E and H), the irradiance in the direction of k is given simply by EH/2 ,‍ which is‍ E²/2Z  in a medium of intrinsic impedance Z = 1/Y. To compute the irradiance in the direction normal to the interface, as we shall require in the definition of the power transmission coefficient, we could use only the x component (rather than the full xy component) of H or E or, equivalently, simply multiply EH/2 by the proper geometric factor, obtaining (E²/2Z ) cos θ.

From equations (13) and (21), taking squared magnitudes, we find that the reflectivity (ratio of reflected power to incident power) is

${\displaystyle R_{\text{s}}=\left|{\frac {Y_{1}\cos \theta _{\text{i}}-Y_{2}\cos \theta _{\text{t}}}{Y_{1}\cos \theta _{\text{i}}+Y_{2}\cos \theta _{\text{t}}}}\right|^{2}}$

(25)

for the s polarization, and

${\displaystyle R_{\text{p}}=\left|{\frac {Y_{2}\cos \theta _{\text{i}}-Y_{1}\cos \theta _{\text{t}}}{Y_{2}\cos \theta _{\text{i}}+Y_{1}\cos \theta _{\text{t}}}}\right|^{2}}$

(26)

for the p polarization. Note that when comparing the powers of two such waves in the same medium and with the same  cos θ, the impedance and geometric factors mentioned above are identical and cancel out. But in computing the power transmission (below), these factors must be taken into account.

The simplest way to obtain the power transmission coefficient (transmissivity, the ratio of transmitted power to incident power in the direction normal to the interface, i.e. the y direction) is to use R + T = 1‍ (conservation of energy). In this way we find

${\displaystyle T_{\text{s}}=1-R_{\text{s}}=\,{\frac {4\,{\text{Re}}\{Y_{1}Y_{2}\cos \theta _{\text{i}}\cos \theta _{\text{t}}\}}{\left|Y_{1}\cos \theta _{\text{i}}+Y_{2}\cos \theta _{\text{t}}\right|^{2}}}}$

(25T)

for the s polarization, and

${\displaystyle T_{\text{p}}=1-R_{\text{p}}=\,{\frac {4\,{\text{Re}}\{Y_{1}Y_{2}\cos \theta _{\text{i}}\cos \theta _{\text{t}}\}}{\left|Y_{2}\cos \theta _{\text{i}}+Y_{1}\cos \theta _{\text{t}}\right|^{2}}}}$

(26T)

for the p polarization.

In the case of an interface between two lossless media (for which ϵ and μ are real and positive), one can obtain these results directly using the squared magnitudes of the amplitude transmission coefficients that we found earlier in equations (14) and (22). But, for given amplitude (as noted above), the component of the Poynting vector in the y direction is proportional to the geometric factor cos θ‍ and inversely proportional to the wave impedance Z. Applying these corrections to each wave, we obtain two ratios multiplying the square of the amplitude transmission coefficient:

${\displaystyle T_{\text{s}}=\left({\frac {2Y_{1}\cos \theta _{\text{i}}}{Y_{1}\cos \theta _{\text{i}}+Y_{2}\cos \theta _{\text{t}}}}\right)^{2}{\frac {\,Y_{2}\,}{Y_{1}}}\,{\frac {\cos \theta _{\text{t}}}{\cos \theta _{\text{i}}}}={\frac {4Y_{1}Y_{2}\cos \theta _{\text{i}}\cos \theta _{\text{t}}}{\left(Y_{1}\cos \theta _{\text{i}}+Y_{2}\cos \theta _{\text{t}}\right)^{2}}}}$

(27)

for the s polarization, and

${\displaystyle T_{\text{p}}=\left({\frac {2Y_{1}\cos \theta _{\text{i}}}{Y_{2}\cos \theta _{\text{i}}+Y_{1}\cos \theta _{\text{t}}}}\right)^{2}{\frac {\,Y_{2}\,}{Y_{1}}}\,{\frac {\cos \theta _{\text{t}}}{\cos \theta _{\text{i}}}}={\frac {4Y_{1}Y_{2}\cos \theta _{\text{i}}\cos \theta _{\text{t}}}{\left(Y_{2}\cos \theta _{\text{i}}+Y_{1}\cos \theta _{\text{t}}\right)^{2}}}}$

(28)

for the p polarization. The last two equations apply only to lossless dielectrics, and only at incidence angles smaller than the critical angle (beyond which, of course, T = 0 ).

For unpolarized light:

${\displaystyle T={1 \over 2}(T_{s}+T_{p})}$

${\displaystyle R={1 \over 2}(R_{s}+R_{p})}$

Where ${\displaystyle R+T=1}$.

Equal refractive indices

From equations (4) and (5), we see that two dissimilar media will have the same refractive index, but different admittances, if the ratio of their permeabilities is the inverse of the ratio of their permittivities. In that unusual situation we have  θ_t = θ_i‍ (that is, the transmitted ray is undeviated), so that the cosines in equations (13), (14), (21), (22), and (25) to (28) cancel out, and all the reflection and transmission ratios become independent of the angle of incidence; in other words, the ratios for normal incidence become applicable to all angles of incidence.^[35] When extended to spherical reflection or scattering, this results in the Kerker effect for Mie scattering.

Non-magnetic media

Since the Fresnel equations were developed for optics, they are usually given for non-magnetic materials. Dividing (4) by (5)) yields

$${\displaystyle Y={\frac {n}{\,c\mu \,}}\,.}$$

For non-magnetic media we can substitute the vacuum permeability μ₀ for μ, so that

$${\displaystyle Y_{1}={\frac {n_{1}}{\,c\mu _{0}}}~~;~~~Y_{2}={\frac {n_{2}}{\,c\mu _{0}}}\,;}$$

that is, the admittances are simply proportional to the corresponding refractive indices. When we make these substitutions in equations (13) to (16) and equations (21) to (26), the factor cμ₀ cancels out. For the amplitude coefficients we obtain:^[5][6]

${\displaystyle r_{\text{s}}={\frac {n_{1}\cos \theta _{\text{i}}-n_{2}\cos \theta _{\text{t}}}{n_{1}\cos \theta _{\text{i}}+n_{2}\cos \theta _{\text{t}}}}}$

(29)

${\displaystyle t_{\text{s}}={\frac {2n_{1}\cos \theta _{\text{i}}}{n_{1}\cos \theta _{\text{i}}+n_{2}\cos \theta _{\text{t}}}}\,}$

(30)

${\displaystyle r_{\text{p}}={\frac {n_{2}\cos \theta _{\text{i}}-n_{1}\cos \theta _{\text{t}}}{n_{2}\cos \theta _{\text{i}}+n_{1}\cos \theta _{\text{t}}}}}$

(31)

${\displaystyle t_{\text{p}}={\frac {2n_{1}\cos \theta _{\text{i}}}{n_{2}\cos \theta _{\text{i}}+n_{1}\cos \theta _{\text{t}}}}\,.}$

(32)

For the case of normal incidence these reduce to:

${\displaystyle r_{\text{s0}}={\frac {n_{1}-n_{2}}{n_{1}+n_{2}}}}$

(33)

${\displaystyle t_{\text{s0}}={\frac {2n_{1}}{n_{1}+n_{2}}}}$

(34)

${\displaystyle r_{\text{p0}}={\frac {n_{2}-n_{1}}{n_{2}+n_{1}}}}$

(35)

${\displaystyle t_{\text{p0}}={\frac {2n_{1}}{n_{2}+n_{1}}}\,.}$

(36)

The power reflection coefficients become:

${\displaystyle R_{\text{s}}=\left|{\frac {n_{1}\cos \theta _{\text{i}}-n_{2}\cos \theta _{\text{t}}}{n_{1}\cos \theta _{\text{i}}+n_{2}\cos \theta _{\text{t}}}}\right|^{2}}$

(37)

${\displaystyle R_{\text{p}}=\left|{\frac {n_{2}\cos \theta _{\text{i}}-n_{1}\cos \theta _{\text{t}}}{n_{2}\cos \theta _{\text{i}}+n_{1}\cos \theta _{\text{t}}}}\right|^{2}\,.}$

(38)

The power transmissions can then be found from T = 1 − R.

Brewster's angle

For equal permeabilities (e.g., non-magnetic media), if θ_i and θ_t are complementary, we can substitute‍ sin θ_t‍ for‍ cos θ_i‍, and‍ sin θ_i‍ for‍ cos θ_t‍, so that the numerator in equation (31) becomes‍ n₂‍sin θ_t − n₁‍sin θ_i‍,  which is zero (by Snell's law). Hence  r_p = 0  and only the s-polarized component is reflected. This is what happens at the Brewster angle. Substituting‍ cos θ_i‍ for‍ sin θ_t‍ in Snell's law, we readily obtain

${\displaystyle \theta _{\text{i}}=\arctan(n_{2}/n_{1})}$

(39)

for Brewster's angle.

Equal permittivities

Although it is not encountered in practice, the equations can also apply to the case of two media with a common permittivity but different refractive indices due to different permeabilities. From equations (4) and (5), if ϵ is fixed instead of μ, then Y becomes inversely proportional to n, with the result that the subscripts 1 and 2 in equations (29) to (38) are interchanged (due to the additional step of multiplying the numerator and denominator by n₁n₂). Hence, in (29) and (31), the expressions for r_s and r_p in terms of refractive indices will be interchanged, so that Brewster's angle (39) will give  r_s = 0  instead of  r_p = 0‍,‍ and any beam reflected at that angle will be p-polarized instead of s-polarized.^[36] Similarly, Fresnel's sine law will apply to the p polarization instead of the s polarization, and his tangent law to the s polarization instead of the p polarization.

This switch of polarizations has an analog in the old mechanical theory of light waves (see § History, above). One could predict reflection coefficients that agreed with observation by supposing (like Fresnel) that different refractive indices were due to different densities and that the vibrations were normal to what was then called the plane of polarization, or by supposing (like MacCullagh and Neumann) that different refractive indices were due to different elasticities and that the vibrations were parallel to that plane.^[37] Thus the condition of equal permittivities and unequal permeabilities, although not realistic, is of some historical interest.