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core36 min studyLesson 16 in Physics

Interfaces and Fresnel coefficients

Derive reflected and transmitted amplitudes from boundary conditions and distinguish amplitude ratios from power fractions.

Before you begin
On this page

The question for today

What you’ll understand

  • Apply tangential field continuity at an ideal dielectric boundary.
  • Calculate normal-incidence amplitude and power coefficients.
  • Explain the assumptions behind Brewster incidence and total internal reflection.

A light wave meeting glass generally creates both reflected and transmitted waves. Their strengths are not arbitrary material labels: the fields on both sides must fit the same Maxwell boundary conditions. The calculation also exposes a common mistake. An electric-amplitude transmission coefficient is not a transmitted-power fraction, because the two media can carry different power for the same electric amplitude.

Match the total fields at the boundary

Consider a flat interface between homogeneous, isotropic, lossless, nonmagnetic dielectrics with no free sheet charge or current. Shrinking a Faraday loop across the interface enforces continuity of tangential electric field. The corresponding Maxwell–Ampère loop enforces continuity of tangential magnetic field. The conditions apply to the sum of incident and reflected fields on one side, rather than separately matching each traveling wave to the transmitted wave.

The reflected magnetic field changes its relation to the electric direction because its propagation direction reverses. At normal incidence this sign produces two linear equations for the two unknown amplitudes. An apparent coefficient error often comes from missing that reversal. At oblique incidence, first define the plane of incidence and separate s polarization, perpendicular to that plane, from p polarization within it.

1+r=t,n1(1−r)=n2t,r=n1−n2n1+n2,t=2n1n1+n21+r=t,\qquad n_1(1-r)=n_2t,\qquad r=\frac{n_1-n_2}{n_1+n_2},\quad t=\frac{2n_1}{n_1+n_2}
r and t are signed electric-field amplitude ratios at normal incidence, using the same tangential axis. These expressions assume equal vacuum permeability in both media and real positive refractive indices n₁ and n₂.↔ Scroll wide equations horizontally.

Energy conservation includes wave impedance

In these dielectrics, a traveling wave with peak electric amplitude E0E_0 carries average intensity nϵ0cE02/2n\epsilon_0cE_0^2/2. Consequently, transmitted power depends on both the squared amplitude ratio and the refractive-index ratio. A transmission amplitude smaller than one can still correspond to almost complete power transmission. At oblique incidence, normal power flux also includes a cosine of propagation angle on each side.

For a lossless single interface, reflected and transmitted normal power fractions sum to one. This provides a strong check on the algebra. Absorbing materials require complex optical constants and a careful definition of flux; a finite slab adds repeated internal reflections and thickness-dependent interference. The single-boundary equations are therefore a controlled starting point, not a universal formula for every glass window or coating.

R=∣r∣2,T=n2n1∣t∣2,R+T=1(normal incidence, lossless media)R=|r|^2,\qquad T=\frac{n_2}{n_1}|t|^2,\qquad R+T=1\quad\text{(normal incidence, lossless media)}
R and T are dimensionless power fractions, whereas r and t are electric-amplitude ratios. The index factor assumes the same nonmagnetic constitutive model used to derive the amplitude coefficients.↔ Scroll wide equations horizontally.

Angle introduces special cases, not exceptions to Maxwell

Matching phase along the interface gives Snell's law. Under the same simple dielectric assumptions, the p-polarized reflected amplitude vanishes at Brewster incidence. The s-polarized reflection generally remains nonzero, so unpolarized light is not completely transmitted by that condition. Real absorption, magnetic response, anisotropy, and multilayers can alter or remove this simple zero.

When a wave travels from higher to lower refractive index above the critical angle, no propagating transmitted plane wave has the required tangential wave number. An evanescent field nevertheless extends into the second medium. For an isolated lossless interface its time-averaged normal transmitted power is zero, but another nearby interface can couple to that field. Total internal reflection does not mean that the field is exactly zero beyond the boundary.

n1sin⁡θi=n2sin⁡θt,tan⁡θB=n2n1,sin⁡θc=n2n1(n1>n2)n_1\sin\theta_i=n_2\sin\theta_t,\qquad \tan\theta_B=\frac{n_2}{n_1},\qquad \sin\theta_c=\frac{n_2}{n_1}\quad(n_1>n_2)
Angles are measured from the interface normal. θB applies to p polarization for nonmagnetic, lossless isotropic dielectrics; θc exists only for incidence from the higher-index medium.↔ Scroll wide equations horizontally.

Work it through

Why an amplitude of 0.8 transmits 96% of power

A normally incident plane wave travels from air, approximated by n₁ = 1.00, into a lossless dielectric with n₂ = 1.50. Find the signed reflected amplitude and both power fractions.

  1. Solve the amplitude relation: r=(1.00−1.50)/(1.00+1.50)=−0.20r=(1.00-1.50)/(1.00+1.50)=-0.20. The minus sign means a π phase reversal of the reflected electric field in the chosen tangential coordinate.

  2. The transmitted amplitude ratio is t=2(1.00)/(1.00+1.50)=0.80t=2(1.00)/(1.00+1.50)=0.80. Squaring it alone gives 0.64, which is not yet a power fraction across media with different indices.

  3. Calculate R=0.202=0.040R=0.20^2=0.040 and T=(1.50/1.00)(0.80)2=0.960T=(1.50/1.00)(0.80)^2=0.960. Their sum is one, as required for this lossless boundary.

The result

The reflected field has amplitude −0.20 times the incident field, with 4% reflected power and 96% transmitted power. These values describe one interface and omit any back surface.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Amplitude sign is a phase relation

Change phase difference and amplitude ratio in two co-propagating scalar waves and inspect their sum. A 180° phase shift reverses one contribution's sign. This supports boundary-amplitude bookkeeping, but the lab contains neither a dielectric interface nor counter-propagating reflected waves and does not calculate Fresnel coefficients.

Amplitude sign is a phase relationTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forSigned field amplitudes must be combined before the resulting power is calculated.

Pause & reason

Check your understanding

At normal incidence from n₁ = 1 into n₂ = 1.5, why is transmitted power not simply |t|²?

Your turn

Try it yourself

Estimate the Brewster angle for air-to-glass incidence with indices 1 and 1.5, and the critical angle for glass-to-air incidence. Explain why the two phenomena are different.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use arctangent for Brewster incidence and arcsine for the critical angle.

Compare with the solution

Brewster incidence is θB=arctan⁡(1.5)=56.3∘\theta_B=\arctan(1.5)=56.3^\circ from air, where p-polarized reflection vanishes in the ideal model. The glass-to-air critical angle is θc=arcsin⁡(1/1.5)=41.8∘\theta_c=\arcsin(1/1.5)=41.8^\circ. Above it there is no propagating transmitted wave for either polarization, although an evanescent field remains. The different directions of incidence and different boundary conditions prevent identifying these as the same angle.

Keep the idea

Take it with you

  • Fresnel coefficients follow from continuity of total tangential fields.
  • Amplitude transmission requires impedance factors before it becomes a power fraction.
  • Brewster reflection and total internal reflection have distinct polarization and incidence conditions.

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