Back to the on-screen lesson ·
Boundary conditions give Snell's law and the Fresnel equations: reflectance $((n_1 - n_2)/(n_1 + n_2))^2$, Brewster's angle, total internal reflection and anti-reflection coatings.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute reflectance and transmittance at boundaries, find critical and Brewster angles, and design quarter-wave anti-reflection coatings.
You know Snell's law and total internal reflection from Physics 2, plane waves in vacuum, and the boundary conditions on fields at interfaces from the lessons on dielectrics and magnetic materials. You know that in a medium waves travel at $c/n$. This lesson derives the rules of reflection and refraction from Maxwell's equations and finds how much light each surface reflects.
| Term | What it means |
|---|---|
| Refractive index | $n = c/v = \sqrt{\varepsilon_r\mu_r}$, about $\sqrt{\varepsilon_r}$ for nonmagnetic materials. |
| Plane of incidence | The plane containing the incident ray and the normal to the surface. |
| Reflectance | $R$, the fraction of incident intensity reflected. |
| Transmittance | $T$, the fraction transmitted; $R + T = 1$ without absorption. |
| Brewster's angle | $\tan\theta_B = n_2/n_1$, where light polarized in the plane of incidence is not reflected. |
| Critical angle | $\sin\theta_c = n_2/n_1$, beyond which light from the denser medium is totally reflected. |
| Anti-reflection coating | A thin film, a quarter wavelength thick, that cancels reflections by interference. |
At an interface between two media with no free charges or currents, Maxwell's equations require the tangential components of $\vec{E}$ and of $\vec{H} = \vec{B}/\mu$ to be continuous. For a wave hitting the boundary, the incident, reflected and transmitted waves together must satisfy those conditions at every point of the boundary and at every instant.
Matching at every instant forces all three waves to have the same frequency. Matching at every point along the boundary forces their phases to agree there, which gives the law of reflection, $\theta_R = \theta_I$, and Snell's law,
$$n_1\sin\theta_1 = n_2\sin\theta_2.$$
Matching the amplitudes gives the Fresnel equations. At normal incidence, from index $n_1$ to $n_2$,
$$\frac{E_R}{E_I} = \frac{n_1 - n_2}{n_1 + n_2}, \qquad R = \left(\frac{n_1 - n_2}{n_1 + n_2}\right)^2, \qquad T = \frac{4n_1n_2}{(n_1 + n_2)^2} = 1 - R.$$
For air and glass, $R = 4$ percent. At oblique incidence the answer depends on polarization: for light polarized in the plane of incidence, the reflection vanishes at Brewster's angle, $\tan\theta_B = n_2/n_1$. And going from a denser to a less dense medium, beyond the critical angle $\sin\theta_c = n_2/n_1$ no wave can be transmitted: total internal reflection.
Another way: picture
Picture a wave on a light rope tied to a heavy rope. At the knot, part of the wave continues into the heavy rope and part bounces back, inverted. The bigger the mismatch, the bigger the reflection. Light at a boundary does the same: the refractive index plays the role of the rope's heaviness, and the fraction reflected depends only on the mismatch, $(n_1 - n_2)/(n_1 + n_2)$.
Another way: steps
Checks. $R + T = 1$ for lossless materials. $R$ must be the same whichever way light crosses the boundary at normal incidence. At $n_1 = n_2$, $R = 0$: an invisible boundary. The reflected wave is inverted when light goes from low to high index, which matters for thin-film interference. And critical angles exist only going toward lower index.
The incident, reflected and transmitted waves each have a phase $\vec{k} \cdot \vec{r} - \omega t$. For the sum to satisfy a condition at every point of the boundary plane and every time, the phases must match there, so all three have the same $\omega$ and the same component of $\vec{k}$ along the boundary. With $|k| = n\omega/c$ in each medium, equal tangential components give $n_1\sin\theta_1 = n_2\sin\theta_2$.
Snell's law therefore holds for any wave obeying linear boundary conditions: sound, water waves, seismic waves in the Earth, and the matter waves of quantum mechanics. The amplitudes, by contrast, depend on the particular conditions, which for light come from continuity of tangential $\vec{E}$ and $\vec{H}$. Those give the Fresnel equations, derived by Augustin-Jean Fresnel before Maxwell from a mechanical theory of the ether, later confirmed by Maxwell's.
Four percent per surface sounds small. A camera lens with ten glass elements, twenty surfaces, would lose over half its light to reflections, and the reflected light would bounce around inside as haze and ghost images. Early multi-element lenses suffered exactly this, which limited designs to a few elements until coatings appeared in the 1930s and 1940s.
High-index materials reflect much more. Diamond, $n = 2.42$, reflects $17$ percent per surface, part of its sparkle. Silicon, $n \approx 3.5$ in the visible, reflects over $30$ percent, so bare solar cells would waste a third of the sunlight hitting them. Germanium lenses for infrared cameras, $n = 4$, reflect $36$ percent per surface and would be nearly useless without coatings.
A thin film on the surface creates two reflections, from its top and bottom. If the film is a quarter wavelength thick inside, $t = \lambda/4n$, the second reflection travels an extra half wavelength and cancels the first. The cancellation is complete if the film's index is the geometric mean of its neighbors, $\sqrt{n_1n_2}$: about $1.22$ for glass, where magnesium fluoride, $1.38$, is the practical choice.
A single layer cancels one wavelength well and others partly, which is why coated eyeglasses show a faint purple or green tint: the residual reflection at the edges of the spectrum. Multilayer coatings of alternating materials reduce reflection below half a percent across the whole visible range. Solar cells use silicon nitride, $n \approx 2$, which matches silicon to air well and gives cells their characteristic blue.
For light polarized with $\vec{E}$ in the plane of incidence, the reflected amplitude passes through zero at Brewster's angle, where the reflected and refracted rays would be perpendicular. For glass it is $56°$. Unpolarized light reflecting near this angle comes off mostly polarized perpendicular to the plane of incidence — horizontally, for reflections from a lake or road.
Polarized sunglasses exploit this. Laser designers use it too: windows on gas laser tubes are set at Brewster's angle so that one polarization passes with no reflection loss at all, making the laser's output polarized. Photographers use polarizing filters to remove reflections from water and glass, and to darken skies, whose scattered light is also polarized.
Going from glass into air, Snell's law has no solution when $\sin\theta_1 > 1/n$: the transmitted angle would need a sine greater than one. All the light reflects, with no loss at all — better than any metal mirror. The critical angle is $41.8°$ for glass, $48.8°$ for water, and only $24.4°$ for diamond, which traps light inside and sends it out through the top facets, the source of a well-cut diamond's brilliance.
The field does penetrate a short distance into the air, decaying exponentially over a fraction of a wavelength, the evanescent wave. Bring a second glass surface within that distance and light tunnels across, frustrated total internal reflection, the optical analog of quantum tunneling, used in fingerprint scanners and some beam splitters.
An optical fiber is a glass core surrounded by cladding of slightly lower index, typically $1.47$ and $1.46$. Light traveling nearly along the fiber meets the core–cladding boundary beyond the critical angle, about $83°$ from the normal, and is totally reflected, zigzagging along with almost no loss at the boundary. Losses come instead from the glass itself: about $0.2$ dB per kilometer at $1550$ nm, so a signal keeps half its power over $15$ km.
Charles Kao showed in 1966 that purified glass could make such fibers practical, and Corning Glass Works in New York produced the first low-loss fiber in 1970. Kao shared the 2009 Nobel Prize. Today undersea cables carry more than $95$ percent of intercontinental internet traffic through fibers that rely on the critical angle.
A metal's free electrons respond to an incoming wave and re-radiate it, giving an effectively huge, complex refractive index. The reflectance formula then gives $R$ close to one: silver and aluminum reflect over $90$ percent of visible light. Copper and gold reflect red and yellow strongly but absorb blue, because their electrons can absorb blue photons in interband transitions — which is why they look colored.
The next lesson treats waves inside conductors directly. Mirrors combine both ideas: an aluminum or silver film for high reflectance, protected by a thin dielectric layer whose thickness is chosen, like an anti-reflection coating in reverse, to boost reflection at the wavelengths that matter. The mirrors of the James Webb Space Telescope are coated with gold, which reflects infrared best.
Nearly all eyeglasses sold in the United States today can be ordered with anti-reflection coatings, and most camera lenses and phone camera modules carry them. A single quarter-wave layer of magnesium fluoride, about $100$ nm thick for green light, cuts each surface's reflection from about $4$ percent to about $1.3$ percent; multilayer coatings reach $0.5$ percent across the visible spectrum, eliminating the distracting reflections that others see in uncoated lenses.
Solar cells need coatings even more. Bare silicon reflects over $30$ percent of sunlight. A silicon nitride layer about $75$ nm thick, deposited on every cell, drops the reflection to a few percent at the wavelengths where sunlight is strongest, and texturing the surface into tiny pyramids gives reflected light a second chance to enter. The deep blue of solar panels is the color that the quarter-wave coating fails to cancel.
The internet runs on light guided by total internal reflection. A single-mode fiber's core, about $9$ μm across, has an index about $0.3$ percent higher than its cladding; light traveling nearly along the core strikes the boundary beyond the critical angle and stays trapped. Modern fibers lose only about $0.2$ dB per kilometer at the $1550$ nm wavelength, and optical amplifiers every $80$ km or so restore the signal across oceans.
Hundreds of undersea cables link the continents, and new cables funded by American technology companies each carry hundreds of terabits per second. The same principle brings fiber to homes through the broadband programs expanding across rural America, and guides light in medical endoscopes that let surgeons see inside the body through a bundle of fibers thinner than a pencil.
Because glass is transparent, it is easy to assume it passes all incident light. Every change in refractive index reflects some light, about four percent at each air–glass surface at normal incidence and more at steep angles. That is why you see your reflection in a window at night, when the outside is dark, and why the reflections of many surfaces in a lens must be suppressed with coatings.
A second misconception is that total internal reflection needs a mirror or coating. It needs only a boundary to a lower index and a steep enough angle, and it is more perfect than any metal mirror, reflecting essentially all the light.
Light strikes water ($n = 1.33$) at normal incidence. Write the amplitude ratio.
$\dfrac{E_R}{E_I} = \dfrac{1 - 1.33}{1 + 1.33} = -0.142$
Negative: inverted on reflection.
Find the reflectance.
$R = 0.142^2 = 0.020$
Two percent.
Find the transmittance.
$T = 1 - 0.020 = 0.980$
Almost all enters the water.
Check with the formula for $T$.
$T = \dfrac{4 \times 1 \times 1.33}{(2.33)^2} = 0.980$
Consistent.
Explain why a lake reflects the sky strongly at sunset.
$R \to 1 \text{ as } \theta \to 90°$
At grazing incidence the Fresnel reflectance rises toward one.
Write the condition.
$\tan\theta_B = \dfrac{n_2}{n_1} = \dfrac{1.5}{1.0}$
From air to glass.
Find the angle.
$\theta_B = \arctan 1.5 = 56.3°$
Measured from the normal.
Find the refracted angle.
$\sin\theta_2 = \dfrac{\sin 56.3°}{1.5} = 0.555 \quad\Rightarrow\quad \theta_2 = 33.7°$
Snell's law.
Check the geometry.
$\theta_B + \theta_2 = 90°$
Reflected and refracted rays are perpendicular.
Explain the vanishing reflection.
$\text{dipoles in the glass cannot radiate along their own axis}$
The would-be reflected ray lies along the induced dipoles' oscillation.
State the polarization of the reflected light.
$\vec{E}_R \perp \text{plane of incidence}$
Only that polarization reflects at Brewster's angle.
A glass lens ($n = 1.52$) gets a magnesium fluoride coating ($n = 1.38$) for $550$ nm. Find the ideal index.
$n_{\text{ideal}} = \sqrt{1.0 \times 1.52} = 1.23$
The geometric mean cancels the two reflections exactly.
Find the thickness.
$t = \dfrac{\lambda}{4n} = \dfrac{550}{4 \times 1.38} = 99.6\ \text{nm}$
A quarter wavelength inside the film.
Find the first reflection's amplitude.
$r_1 = \dfrac{1 - 1.38}{1 + 1.38} = -0.160$
Air to coating.
Find the second reflection's amplitude.
$r_2 = \dfrac{1.38 - 1.52}{1.38 + 1.52} = -0.048$
Coating to glass.
Account for the half-wave delay.
$r_{\text{total}} \approx r_1 - r_2 = -0.160 + 0.048 = -0.112$
The round trip flips the second reflection's sign.
Find the residual reflectance.
$R \approx 0.112^2 = 0.013$
About $1.3$ percent, down from $4.3$ percent uncoated.
Compare with the uncoated lens.
$R_0 = \left(\dfrac{0.52}{2.52}\right)^2 = 0.043$
A threefold improvement from one layer.
Note the color dependence.
$\lambda \ne 550\ \text{nm}: \ \text{partial cancellation}$
The faint colored sheen of coated glasses.
Write the condition.
$\sin\theta_c = \dfrac{n_2}{n_1}$
Refracted ray at $90°$.
Substitute the indices.
$\sin\theta_c = \dfrac{1}{1.5} = 0.667$
Glass to air.
Find the angle.
Light strikes a glass surface ($n = 1.5$) head-on from air. What fraction of it is reflected?
Complete the worked solution: light passes through a plate of index $3$ in air at normal incidence. Find the reflectance of one surface, the transmittance of one surface, and the fraction lost at both surfaces together.
Find the reflectance of one surface.
$R =$ r
Normal incidence.
Subtract it from one.
$T = 1 - R =$ t
No absorption.
Find the loss through both surfaces.
$1 - T^2 =$ l
Nearly twice one surface's reflectance when $R$ is small.
Match each law of reflection and refraction to its expression.
| $n_1\sin\theta_1 = n_2\sin\theta_2$ | $\tan\theta_B = n_2/n_1$ | $\sin\theta_c = n_2/n_1$ | $((n_1 - n_2)/(n_1 + n_2))^2$ | |
|---|---|---|---|---|
| Snell's law | ||||
| Brewster's angle | ||||
| the critical angle | ||||
| normal-incidence reflectance |
Light enters a material of index $n = 1.5$ from air at normal incidence. Fill in the amplitude ratio $|E_R/E_I|$, the reflectance and the transmittance.
| value | |
|---|---|
| amplitude ratio | |
| reflectance $R$ | |
| transmittance $T$ |
Light travels in glass ($n = 1.5$) toward a boundary with air ($n = 1$). Beyond what angle of incidence is it totally reflected, in degrees?
Answer: °
Light passes straight through window glass with refractive index $1.5$, surrounded by air. Ignoring absorption and multiple reflections, what fraction of the light gets through both surfaces?
Answer:
An anti-reflection coating of magnesium fluoride on a camera lens ($n = 1.38$) is designed to cancel reflections at $500$ nm. How thick should it be, in nm?
Answer: nm
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Light enters a material of index $n = 4$ from air at normal incidence. Fill in the amplitude ratio $|E_R/E_I|$, the reflectance and the transmittance.
| value | |
|---|---|
| amplitude ratio | |
| reflectance $R$ | |
| transmittance $T$ |
You can predict how light behaves at boundaries. Explain to someone why a window reflects your image at night.
19. Your turn: what is the critical angle for light leaving glass ($n = 1.5$) into air?, step 3
$\theta_c = 41.8°$
Why a $45°$ prism reflects totally.