Reflection, Refraction and the Fresnel Equations
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You should be comfortable with the electromagnetic boundary conditions at a dielectric interface, specifically the continuity of tangential electric and magnetic fields, and the definition of the Poynting vector for time-averaged power flow.
Why this matters
The Fresnel equations quantify exactly how much light is reflected or transmitted at any interface, forming the foundation for optical design and polarization control.
While geometric optics gives us the directions of reflected and refracted rays via Snell's law, it says nothing about the intensity of these rays. In reality, a portion of the incident wave energy is always reflected back into the first medium, while the rest is transmitted into the second. The relative amounts depend critically on the angle of incidence and the polarization state of the light. Understanding this dependence is essential for designing anti-reflection coatings, polarizing filters, and analyzing interference in thin films. This lecture derives the precise amplitude coefficients that govern these processes, revealing a deep symmetry between the two polarization states and identifying the special Brewster angle where reflection vanishes for one polarization.
The Physical Picture
Imagine a plane wave of light traveling through air and striking a glass surface. The electric field of the wave exerts forces on the electrons in the glass, causing them to oscillate. These accelerating charges act as secondary sources of radiation, emitting waves that propagate back into the air (the reflected wave) and forward into the glass (the transmitted wave). The total field in each region is the superposition of the incident wave and these secondary waves. The boundary conditions at the interface demand that the tangential components of the electric and magnetic fields match perfectly across the surface. This matching requirement forces the frequencies to be identical in all three waves and dictates the specific amplitudes of the reflected and transmitted fields relative to the incident field. The result is a set of coefficients, known as the Fresnel coefficients, which determine the fraction of energy reflected and transmitted.
Geometry and Polarization States
To analyse the boundary conditions, we must define a coordinate system. The plane of incidence is the plane containing the incident wave vector and the normal to the interface. We distinguish two independent polarization states. For s-polarization (from the German senkrecht, meaning perpendicular), the electric field vector is perpendicular to the plane of incidence. For p-polarization (parallel), the electric field vector lies within the plane of incidence. Because Maxwell's equations are linear, these two cases can be treated independently. The magnetic field direction is determined by the right-hand rule, ensuring that the Poynting vector points in the direction of propagation. Note that for p-polarization, the magnetic field is perpendicular to the plane of incidence, while for s-polarization, it lies within the plane. This distinction is crucial when applying the boundary conditions to the magnetic field components.
Deriving the Fresnel Coefficients
Applying the boundary conditions to the tangential electric field components yields the first set of equations. For s-polarization, the electric field is entirely tangential, so the condition is simply \( E_i + E_r = E_t \). For p-polarization, the tangential component is \( E \cos\theta \), leading to \( E_i \cos\theta_i - E_r \cos\theta_r = E_t \cos\theta_t \). We must also apply the boundary condition to the tangential magnetic field. Using the relation \( H = \frac{n}{\mu_0 c} E \) for a dielectric medium, we can express the magnetic field in terms of the electric field amplitude. For s-polarization, the magnetic field lies in the plane of incidence, and its component along the interface is \( H\cos\theta \). With the reflected wave travelling back, the condition is \( H_i \cos\theta_i - H_r \cos\theta_r = H_t \cos\theta_t \). Substituting \( H \propto nE \) and \( \theta_r = \theta_i \), we can solve the system of linear equations for the ratios \( r_s = E_r/E_i \) and \( t_s = E_t/E_i \). The derivation for p-polarization follows a similar logic but involves the cosine of the angles due to the projection of the electric field onto the interface. The final results are the standard Fresnel equations listed in the derivation card.
Brewster's Angle and Energy Conservation
A particularly interesting feature of the p-polarization reflection coefficient \( r_p \) is that it becomes zero when the numerator vanishes. This occurs when \( n_2\cos\theta_i = n_1\cos\theta_t \). Using Snell's law, this condition can be rewritten as \( \tan\theta_B = n_2/n_1 \), where \( \theta_B \) is the Brewster angle. At this specific angle of incidence, the reflected and transmitted rays are perpendicular to each other. Physically, the dipoles in the second medium oscillate in a direction such that they do not radiate along the direction of the reflected ray, resulting in zero reflection. This principle is used in polarizing filters to eliminate glare from horizontal surfaces. It is also important to verify that energy is conserved. The reflectance \( R \) and transmittance \( T \) are defined as the ratio of the normal component of the Poynting vector to the incident power. For non-absorbing media, one can show that \( R + T = 1 \). For s-polarization, \( R_s = |r_s|^2 \) and \( T_s = \frac{n_2 \cos\theta_t}{n_1 \cos\theta_i} |t_s|^2 \). The factor involving the refractive indices and cosines accounts for the change in the cross-sectional area of the beam and the change in the wave impedance.
The derivation
Reflection, Refraction and Fresnel Equations
For a monochromatic plane electromagnetic wave incident from a linear, isotropic, non-absorbing dielectric of refractive index \( n_1 \) onto a sharp planar interface with a second such dielectric of index \( n_2 \), the electromagnetic boundary conditions force (i) the reflected and transmitted waves to share the incident frequency and tangential wavevector, giving the law of reflection \( \theta_r = \theta_i \) and Snell's law \( n_1 \sin\theta_i = n_2 \sin\theta_t \); and (ii) the amplitude ratios to take the Fresnel forms, for s-polarization (electric field perpendicular to the plane of incidence) \( r_s = \dfrac{n_1\cos\theta_i - n_2\cos\theta_t}{n_1\cos\theta_i + n_2\cos\theta_t} \), \( t_s = \dfrac{2 n_1\cos\theta_i}{n_1\cos\theta_i + n_2\cos\theta_t} \), and for p-polarization (electric field in the plane of incidence) \( r_p = \dfrac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t} \), \( t_p = \dfrac{2 n_1\cos\theta_i}{n_2\cos\theta_i + n_1\cos\theta_t} \), for non-magnetic media (\( \mu_1 = \mu_2 = \mu_0 \)).
A common source of confusion is the sign convention for the p-polarization reflection coefficient. Depending on whether the magnetic field or the electric field is used to define the positive direction, the sign of \( r_p \) may appear flipped. However, the physical reflectance \( R_p = |r_p|^2 \) is invariant under this sign change. When comparing results from different textbooks, always check the definition of the positive direction for the p-polarized electric field to ensure consistency. C
Worked example
Problem. A linearly polarised monochromatic plane wave with wavelength \( 500\ \mathrm{nm} \) in vacuum is incident from air (\( n_1 = 1.00 \)) onto a glass slab (\( n_2 = 1.50 \)) at an angle of incidence \( \theta_i = 45^\circ \). Calculate the reflectance for s-polarization and determine the Brewster angle for p-polarization.
- Determine the angle of transmission \( \theta_t \) using Snell's law: \( \sin\theta_t = \frac{n_1}{n_2} \sin\theta_i = \frac{1.00}{1.50} \sin(45^\circ) = \frac{0.7071}{1.50} = 0.4714 \). Thus, \( \theta_t = \arcsin(0.4714) \approx 28.1^\circ \).
- Calculate the s-polarization reflection coefficient \( r_s \). First, find the cosines: \( \cos\theta_i = \cos(45^\circ) = 0.7071 \) and \( \cos\theta_t = \cos(28.1^\circ) = 0.8819 \). Then, \( r_s = \frac{1.00(0.7071) - 1.50(0.8819)}{1.00(0.7071) + 1.50(0.8819)} = \frac{0.7071 - 1.3229}{0.7071 + 1.3229} = \frac{-0.6158}{2.0300} = -0.3033 \).
- Calculate the reflectance \( R_s = |r_s|^2 = (-0.3033)^2 = 0.0920 \). This means 9.20% of the incident power is reflected.
- Determine the Brewster angle \( \theta_B \) for p-polarization using \( \tan\theta_B = \frac{n_2}{n_1} = \frac{1.50}{1.00} = 1.50 \). Thus, \( \theta_B = \arctan(1.50) \approx 56.3^\circ \).
Units check. The reflectance is a dimensionless ratio of powers, which is correct. The Brewster angle is an angle, measured in degrees, which is consistent with the input data.
Three quick tests
- What is the physical significance of the Brewster angle?
Answer
It is the angle of incidence at which p-polarized light is not reflected, resulting in completely polarised reflected light. - Why does the transmittance formula include a factor of \( \frac{n_2 \cos\theta_t}{n_1 \cos\theta_i} \)?
Answer
This factor accounts for the change in the effective cross-sectional area of the beam and the difference in wave impedance between the two media, ensuring energy conservation. - If \( n_1 = n_2 \), what happens to the reflection coefficients?
Answer
Both \( r_s \) and \( r_p \) become zero, meaning there is no reflection at the interface, as expected for a uniform medium.
Problem set for PU-204 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.