BrainFlux

Interfaces and thin films

Colour in white light

Drag the marker to read the colour at a thickness

The stack, and the light in it

Drag the layer's right edge (or the thickness slider) to change its thickness; click a layer to select it

Reflected fraction R against wavelength

Drag on the plot to change the wavelength

R against angle

Drag on the plot to change the angle of incidence

Light

s: E along the surface (solid lines); p: E in the plane of incidence (dashed). Unpolarised light is half of each.

Stacks to try

¼ wave: a quarter of the wavelength inside the layer, d = λ/4n (its two reflections then come back half a wave apart); ½ wave: twice that.

Design a coating

Write the stack as letters: H and L are quarter-wave layers of two materials (thickness λ₀/4n), 2L a half wave, (…)^N repeats. The light arrives from the left.

Layers

H L M

Show in the plots

White-light picture

Wavelength range

Experiments

Start here

Light comes in from the left and meets a stack of flat layers; the page shows how much of it is reflected and transmitted, and the colour that makes. It opens on a soap film in air, draining and thinning as a real one does.

Drag the layer's right edge in the stack picture (or the thickness slider) to make the film thicker or thinner, and watch the reflected fraction R and the colour swatch change together. Drag on a plot to change the wavelength or the angle; drag the marker on the colour column to read the colour at any thickness.

Look for this: each thickness picks out its own colours, and a film thinner than about 25 nm reflects almost nothing. The buttons under Stacks to try load the other experiments below.

1. A soap film: colours that move as it drains

Set it up: a film of water in air, draining: thinnest at the top, thicker downwards, and thinning as time passes. Watch the colour column: the bands move down as the film thins, and the top turns black. Drag the marker to any height and the stack picture, the plots and the swatches show that thickness.

Why: light reflected from the front and back surfaces interferes. The front reflection (air to water) flips the sign, the back one (water to air) does not, so a wavelength is reflected strongly when the round trip $2nd$ is a whole number of wavelengths plus a half, and each thickness picks out its own colours. Thinner than about 25 nm, the two reflections cancel for every colour and the film is black: real films do this just before they burst.

2. Reflection from glass: 4% head-on, more at a glancing angle

Set it up: bare BK7 glass in air, no layers. Read the readout: at normal incidence it reflects 4.2% of the light, for either polarisation. Now drag the angle up on the right-hand plot: $R_s$ (solid) rises steadily to 1 at grazing incidence, while $R_p$ (dashed) first falls to zero and then rises.

Why: at normal incidence $R=\left(\frac{n_1-n_2}{n_1+n_2}\right)^2$, which the glass's $n=1.52$ makes 4.2%. The rise at glancing angles is why a lake reflects the sky so well when you look across it. In front of the glass the incident and reflected waves make a weak standing wave: see the ripple in the stack picture.

3. Brewster's angle: no reflection of p light at 56.6°

Set it up: the same glass, with the angle set to 56.6°. The readout gives $R_p=0$: p-polarised light is not reflected at all, and the reflected light is purely s-polarised. Drag the angle a little either way and $R_p$ comes back.

Why: $\tan\theta_B=n_2/n_1$. At this angle the refracted and reflected rays would be at right angles, and the dipoles in the glass, driven along the p-field, do not radiate along their own axis. Polarising sunglasses, whose axis is vertical, cut the glare from water and roads because it is s-polarised.

4. Total internal reflection, and the wave that leaks out anyway

Set it up: the light is now inside the glass, going out into air at 50°. The readout says $R=1$ for both polarisations, and the stack picture shows the field reaching into the air and dying away. Drag the angle towards the critical angle, 41.2°: the field reaches farther (the readout gives its decay length, 147 nm at 50°, 224 nm at 45°). Below the critical angle, towards 0°, light gets out again.

Why: beyond $\sin\theta_c=n_2/n_1$ nothing can propagate in the air, but the field does not stop at the surface: it goes on as an evanescent wave, decaying as $e^{-\kappa z}$ with $\kappa=\frac{2\pi}{\lambda}\sqrt{n_1^2\sin^2\theta-n_2^2}$ (the plot shows $|\vec E|^2$, which falls twice as fast, by $1/e$ in 74 nm at 50°). It carries no energy away. The readout also gives the phase difference between the reflected s and p waves, 45° at 47.7° and at 55.3°: Fresnel cut a glass rhomb so that light is reflected twice inside it at such an angle, adding up to a quarter wave, which turns linear polarisation into circular.

5. Light tunnelling across an air gap

Set it up: two glass blocks with a 200 nm gap of air between them, at 45°, beyond the critical angle. The right-hand plot is now $T$ against the gap, on a log scale: a straight line. Drag the gap's right edge in the stack picture to widen it and watch $T$ fall; narrow it to nothing and everything gets through.

Why: when the gap is thinner than the evanescent wave's reach, light tunnels across it, $T\propto e^{-2\kappa d}$ (frustrated total internal reflection). This is exactly the mathematics of a quantum particle tunnelling through a barrier: for s polarisation $E$ obeys $E''+k_0^2\left(n^2-n_1^2\sin^2\theta\right)E=0$, the Schrödinger equation with the air gap as a region where the "energy" is below the "potential", and $T$ is the barrier formula $T=\left[1+\frac{(k^2+\kappa^2)^2}{4k^2\kappa^2}\sinh^2\kappa d\right]^{-1}$. Some beam splitters and fingerprint readers work this way.

6. Oil on water: paler colours than a soap film's

Set it up: a film of oil ($n=1.47$) on water ($n=1.33$), as a wedge thickening downwards from nothing to 1.2 µm. Compare the colour column with the soap film's: the colours are paler, and the top is not black. Drag the thickness down to nothing: the readout gives 2.1%, just what bare water reflects.

Why: the index rises at the top surface and falls at the bottom one, as for the soap film, so the two reflections have opposite signs; but the reflection from the oil–water surface is weak (the two indices are close), so the cancellation is only partial, and the colours, which repeat each time the thickness grows by half a wavelength in the oil, are washed out. On a wet road the dark asphalt underneath makes them stand out.

7. An antireflection coating: a quarter wave of MgF₂ on glass

Set it up: glass with a 100 nm layer of magnesium fluoride on it, the plots on a log scale. The readout gives $R=1.25\%$ at 550 nm instead of the bare glass's 4.2%, and the reflected swatch is purple. Drag the layer's edge: the dip in $R$ against wavelength moves with the thickness, and at 0 nm you have bare glass again. The colour column shows the coating thickening from nothing to 400 nm; 100 nm is the quarter wave.

Why: the waves reflected from the coating's two surfaces come back half a wavelength apart and nearly cancel. They would cancel exactly if $n_1=\sqrt{n_0n_2}$; with MgF₂ ($n=1.38$) on BK7, $R=\left(\frac{n_0n_2-n_1^2}{n_0n_2+n_1^2}\right)^2$ at the design wavelength. It works only near one wavelength, so blue and red are reflected more than green: the purple reflections of coated camera lenses and spectacles.

8. A Bragg mirror: many weak reflections that add up

Set it up: eleven alternate quarter-wave layers of TiO₂ and SiO₂ on glass, built by the Design a coating panel as (HL)^N H with $N=5$. The readout gives $R=99.8\%$ at 550 nm, and the plot against wavelength shows a band of high reflection, the stop band, from about 460 to 680 nm. Slide $N$ in the design panel: 1, 2, 3 and 5 pairs give 76.8%, 92.3%, 97.6% and 99.8%. Then drag the angle up: the band moves to shorter wavelengths.

Why: every reflection comes back in step, and with $N$ pairs the peak reflectance is $$R=\left(\frac{n_0n_L^{2N}-n_sn_H^{2N}}{n_0n_L^{2N}+n_sn_H^{2N}}\right)^2$$ for $(HL)^N$ on a substrate of index $n_s$, which approaches 1 quickly. The width of the band is set by the index contrast, $\frac{\Delta\lambda}{\lambda_0}\approx\frac{4}{\pi}\arcsin\frac{n_H-n_L}{n_H+n_L}$: choose BK7 glass ($n=1.52$) for H, close to SiO₂, and it shrinks to a sliver whose peak reflects only 15%. The shift with angle is why a beetle's shell or a peacock feather changes colour as you turn it. Inside the mirror the field dies away layer by layer: the band gap of a one-dimensional crystal.

9. A narrow-band filter: one sharp line through two mirrors

Set it up: two Bragg mirrors with a half-wave spacer between them, (HL)^N 2H (LH)^N. Inside the stop band, where each mirror alone reflects almost everything, a single sharp line gets through at 550 nm, 1.12 nm wide. Look at the stack picture: at that wavelength the field between the mirrors is 74 times the incoming $|E|^2$. Drag the wavelength a few nanometres off the line and it collapses. Slide $N$: each extra pair makes the line narrower, 3.75, 1.12 and 0.34 nm for $N=$ 3, 4 and 5.

Why: the two mirrors make a Fabry–Pérot cavity, a space in which light bounces back and forth. At the wavelength that fits it, the light builds up between the mirrors until what leaks out of the far side equals what comes in. Each extra pair raises the mirrors' reflectance and narrows the line by about $(n_H/n_L)^2$, 3.3 times.

10. A Fabry–Pérot etalon: a comb of transmission peaks

Set it up: two thin silver films with 2 µm of air between them. The plot against wavelength is a comb of peaks, and the right-hand plot, $T$ against the gap, shows the same comb as the gap grows: a peak every half wavelength, 275 nm. The readout measures the peaks' spacing, 63.8 nm at 514 nm, and the finesse, their spacing over their width: 16.4. Select each silver film in turn and thicken it to 55 nm: the finesse rises to 30.4, but the peaks drop from 41% to 16%.

Why: transmission peaks wherever the round trip is a whole number of wavelengths, $2d\cos\theta=m\lambda$ (nearly: the reflection from silver adds a phase of its own). So the peaks are evenly spaced in $1/\lambda$, by $1/2d$, which near a wavelength $\lambda$ is $\Delta\lambda\approx\lambda^2/2d$: 66 nm at 514 nm. The measured 63.8 nm is as if the gap were 2.07 µm: the light reaches a little way into the silver before it turns back. The peaks are sharp because the light bounces many times: $\mathcal F\approx\pi\sqrt{R}/(1-R)$ for mirrors of reflectance $R$; seen from inside the gap, each film here reflects $R=82.6\%$ at 514 nm, so $\mathcal F\approx16.4$. Thicker silver reflects 90.2% and sharpens the peaks, but absorbs a little at every bounce, so they are lower.

11. A silver mirror, and where it stops working

Set it up: bulk silver, with the plot from 280 to 900 nm. It reflects over 90% of visible light, nearly evenly (92% at 450 nm, 96% at 600 nm), so the reflected swatch is white. Drag the wavelength into the ultraviolet: the reflectance collapses, to 7% at 309 nm. Then drag the angle: $R_p$ dips to 91% at 71° but never reaches zero.

Why: silver's free electrons follow the field and re-radiate it. In the ultraviolet they can no longer keep up: at the plasma frequency the dielectric constant $\varepsilon=N^2$ passes through zero, here at 324 nm (3.8 eV; the free electrons alone would put it at 9 eV, but the d electrons beneath them pull it down). Thin silver lets that ultraviolet through. A metal has no true Brewster angle, only this "pseudo-Brewster" angle of least $R_p$. (Real silver reflects a few per cent more than this fit; see How it works.)

12. The colour of gold, in reflection and through

Set it up: a 40 nm gold film on glass. The reflected swatch is gold; switch the colour column to transmitted: the light that gets through is green-blue, the colour of gold leaf held up to a lamp. Drag the thickness down to 10 nm and up to 120 nm and watch both swatches.

Why: gold's electrons absorb blue and green light (an interband transition of its d electrons), so it reflects mostly yellow and red, and what it does not absorb or reflect, the green-blue, is what gets through a film thin enough to leak.

13. A surface plasmon: a gold film that swallows light at one angle

Set it up: a glass prism, a 50 nm gold film, water behind it, red light at 633 nm, arriving at 74.8°. The plot against angle shows $R_p$ dipping sharply, and the stack picture shows the field piled up at the gold's far side. The water is selected: move its index $n$ from 1.333 to 1.343 (as when molecules bind to the gold) and the dip moves from 74.8° to 76.5°. In the angle plot $R_s$ (solid) shows no dip: s light cannot do this.

Why: this is the Kretschmann geometry. The evanescent wave from the prism matches the wavevector of a surface plasmon, a charge-density wave running along the gold–water interface, and the light's energy goes into it and is absorbed. The match is $n_\text{glass}\sin\theta=\mathrm{Re}\sqrt{\varepsilon_m\varepsilon_d/(\varepsilon_m+\varepsilon_d)}$, with $\varepsilon_d=n^2$ the water's dielectric constant and $\varepsilon_m$ the gold's ($-9.8+1.96i$ at 633 nm in this fit): 75.4°; the gold film's coupling to the prism pulls the dip a little lower. The plasmon is a p wave, its $\vec E$ in the plane of incidence, so s light cannot excite it. At a fixed angle, the reflected light changes a lot for a tiny change of index: this is how SPR biosensors weigh molecules.

How it works

Light arrives from the left on a stack of flat layers. At every interface part of it is reflected and part goes on; inside each layer the waves bouncing back and forth add up with phases set by the layer's thickness, its index and the wavelength. The plots show the fractions reflected ($R$, blue), transmitted ($T$, orange) and absorbed ($A$, green), for s polarisation ($\vec E$ parallel to the surface: solid lines) and p ($\vec E$ in the plane of incidence: dashed). The stack picture draws the layers and the light in them: $|\vec E|^2$ in units of the incoming wave's, so the dashed line at 1 is the incoming wave alone. The colour column is the colour of the stack in white light, as the eye sees it, while the selected layer's thickness grows from top to bottom; the swatches under it are the stack as it is, brightened to show its colour, with the fraction of the light really reflected and transmitted.

Click a layer to select it and drag its right-hand edge to change its thickness, or use the slider. Drag across a plot to change the wavelength, the angle or the thickness. Any material can be typed as a formula $n(\lambda)$, with $\lambda$ in micrometres and $i$ for the imaginary part; every new symbol gets a slider.

How it is computed

The characteristic-matrix (transfer-matrix) method. In each layer the field is a wave going forward and one going back, $E=a\,e^{ik_zz}+b\,e^{-ik_zz}$ with $k_z=\frac{2\pi}{\lambda}\sqrt{N^2-n_0^2\sin^2\theta_0}$, and the tangential $\vec E$ and $\vec H$ are continuous at every interface; a layer turns them into ${(\begin{matrix} E \\ H \end{matrix})}_{front} = (\begin{matrix} \cos δ & - \frac{i}{η} \sin δ \\ - i η \sin δ & \cos δ \end{matrix}) {(\begin{matrix} E \\ H \end{matrix})}_{back}$ with $\delta=k_zd$, the phase a wave picks up crossing the layer, and $\eta=N\cos\theta$ (s) or $N/\cos\theta$ (p), the layer's ratio $H/E$ of tangential fields for a wave going forward (in units of vacuum's, $1/377\ \Omega$). Then $r=(\eta_0-Y)/(\eta_0+Y)$, where $Y=H/E$ at the front of the stack. The signs of $i$ are those of waves $e^{i(k_zz-\omega t)}$; books that write $e^{i(\omega t-k_zz)}$ (Macleod's, for one) have $+i$ and $N=n-i\kappa$. Indices are complex, $N=n+i\kappa$, so metals and evanescent waves need nothing special. Every number is exact for ideal flat layers and plane waves; the materials are the approximations.

Materials: water (Daimon and Masumura 2007), fused silica (Malitson 1965), BK7 (Schott), MgF₂ (Dodge 1984) and rutile TiO₂ (DeVore 1951) from their Sellmeier fits; silver, gold and aluminium from the Lorentz–Drude fits of Rakić et al. (1998), which are good to several per cent in the visible but not everywhere (their gold absorbs a little more than real gold, so its plasmon dip is shallower). Evaporated films differ from bulk: real TiO₂ coatings have $n\approx2.3$, not 2.6.

Colours: the CIE 1931 colour-matching functions (the analytic fit of Wyman, Sloan and Shirley, 2013) weight the reflected spectrum under daylight (D65) or equal-energy light, and the result is converted to your screen's sRGB, adapted so that a perfect mirror is white. The picture is brightened (the Brightness slider) so that faint reflections show their colour; the swatches say how much light is really reflected. The light is unpolarised and coherent across each layer: very thick layers would wash out to white in reality, as sunlight's coherence length is only a micrometre or so.

New in 2026 for BrainFlux. Source: github.com/gallicchio/brainflux, MIT licence.