Where does the light that does not come back go?
The mirror of module 3 reflects 74.2% of the light at 13.5 nm. The other 25.8% stays in the mirror. Five ideas follow it, four of them with an experiment. Open any of them, in any order.
Up to here the mirror has been judged by what it sends back. From here on the project is about what it keeps: a quarter of the power, turned into heat a few hundred nanometres below the surface. This module finds how much is absorbed and where. Module 6 takes those two answers and turns them into a temperature.
In front of the mirror: a standing wave
Absorption happens where the field is strong, so finding where the light is absorbed starts with finding the field. The simplest place to look is just outside the mirror, where only two waves meet. The pattern found here continues inside.
Above the mirror two waves overlap: the incoming one, of amplitude 1, and the reflected one, of amplitude √R, travelling the opposite way. Module 2 gave the rule for adding two waves: the result depends on their phase difference.
Here that difference depends on where you look. Moving a height z away from the mirror shortens the path of one wave and lengthens the path of the other. So the two reinforce at some heights and oppose at others, in a fixed pattern that does not travel: a standing wave. The quantity plotted is |E|², the squared size of the total field, compared with the incoming wave alone.
A good mirror makes a deep pattern. At 74% the field swings between 3.5 times the incoming value and 0.02 times. The bright and dark planes repeat every λ/(2·cos θ) = 6.8 nm, almost exactly the period of the stack. That is the Bragg condition seen from the side of the field.
A value of |E|² above 1 does not mean that more power is there than came in. Power travels down with the incoming wave and back up with the reflected one. |E|² says how strong the field is at a point, and that is what decides how much a material placed there absorbs.
- Where it comes from
- It is the two-wave formula of module 2, A² = a₁² + a₂² + 2·a₁·a₂·cos(phase difference), with a₁ = 1 and a₂ = √R. Going up by z changes the phase of each wave by 2π·z·cos θ/λ, in opposite senses, so the difference changes by twice that.
- Assumptions
- One wavelength, a flat mirror, s polarization: the two electric fields point along the same line, so they add like numbers.
- The terms
- z is the height above the mirror. R is the reflectance. φ is the phase of the reflection, which fixes where the first bright plane sits. |E|² is in units of the incoming wave's own |E|².
- Check
- R = 0 gives 1 everywhere: no reflected wave, no pattern. R = 1 gives values between 0 and 4. Averaged over height the cosine vanishes and |E|² = 1 + R: the two intensities simply add, as they should on average.
Inside the mirror
The pattern in front of the mirror goes on below the surface, where the absorbing layers are. This step looks at where its bright and dark planes fall in the molybdenum and in the silicon, because that decides how much each of them absorbs.
The standing wave does not stop at the surface. The matrix method of module 3 gives the pair (E, H) at every boundary, and from it the field at any depth. Here is |E|² through the top ten pairs, with the molybdenum layers shaded.
Near the peak the standing wave locks onto the layers. Its period matches the period of the stack, so each pair sees nearly the same pattern, with a dark plane near the lower face of the molybdenum layer in the top pairs shown. Averaged over a layer, |E|² is lower in molybdenum than in silicon: by 20% at 13.5 nm, by almost 30% at the top of the peak. Molybdenum is the absorber, so this arrangement loses less light than the same material would under a uniform field.
The pattern also fades with depth. Its bright planes drop by a factor e about every seven pairs, 50 nm. Below pair 20 the field is less than a tenth of its value at the surface: the lower half of the stack works in near darkness, as module 3 found from the outside.
The two sides of the peak are not alike. Move to 13.0 nm: the reflected wave is weak, most of the light travels on into the stack, and the average |E|² in the molybdenum is twice that in the silicon. The thin layers then absorb about two thirds of the light, and another 22% enters the substrate and is absorbed there. On the long-wavelength side the field stays out of the molybdenum.
Where the energy is left
The field says where absorption is likely. This step counts it: how much power each layer keeps. The result, how much and at what depth, is what a thermal calculation needs as its heat source.
To find where the light is absorbed, follow the energy. The net power crossing a plane inside the mirror, what goes down minus what comes up, is S = Re(E·H*)/η₀, as a fraction of the incoming power. The star is the complex conjugate: the same number with the sign of its imaginary part reversed. Re means: keep the real part.
Whatever enters a layer through its upper face and does not leave through its lower face has been absorbed in that layer.
Half of the absorption happens in the top five pairs, 35 nm. By pair 20, 94% of the absorption in the stack is done. Of every 100 photons, 74.2 are reflected, 15.8 are absorbed in molybdenum, 9.1 in the silicon layers, and 0.9 reach the substrate.
Molybdenum takes 1.7 times more than silicon, although the silicon layers are thicker. A simple count of absorbing strength times thickness, n·k·d, would say 2.2 times. The difference is the standing wave, which keeps part of the field out of the molybdenum.
- Where it comes from
- Energy is conserved: under steady light, what flows into a layer and does not flow out is absorbed there. The power carried by an electromagnetic field is the product of its electric and magnetic parts. For two quantities that oscillate, the average of their product over time is one half of Re(E·H*). The same one half, and the constant that fixes the scale of H, appear for the incoming wave too, so they cancel when one is divided by the other. What is left is the fraction S.
- Assumptions
- Steady illumination. E and H are the parts of the fields parallel to the layers, with H on the scale defined in module 3. For s polarization their product gives the power flowing perpendicular to the layers. Nothing is scattered sideways.
- The terms
- S is the net downward power through a plane, as a fraction of the incoming power. η₀ = cos θ is the ratio H/E of the incoming wave in vacuum, for s polarization. Alayer is the fraction absorbed in one layer.
- Check
- At the top surface S = 1 − R = 25.8%. Entering the substrate S = T = 0.9%. The 80 differences in between add up to 24.9%, the A of module 3, and each of them is positive: no layer creates energy.
From absorbed light to heat
Absorbed light becomes heat. This step turns the fractions of the last one into watts, first for one mirror and then for a chain of them, and shows which mirror carries the largest load.
Everything that is not reflected ends as heat in the mirror: the 24.9% absorbed in the layers, and the 0.9% that reaches the substrate and is absorbed there. The heat is 1 − R of the incoming power, and about 95% of it is left within the first 210 nm below the surface.
Now consider a chain of such mirrors, as in a scanner. Each one passes on only what it reflects, so each receives less than the one before.
Take 250 W on the first mirror and a reflectance of 74.2%: that mirror keeps (1 − 0.742) × 250 = 64.5 W.
The first mirror takes a quarter of the power, the last one about a hundredth. After ten reflections only 5% of the light is left, so the eleventh mirror absorbs 1.3% of what the first one received.
Small gains in reflectance matter a great deal. Raise R from 74.2% to 75.2%. Each mirror then keeps a smaller fraction of what reaches it, the first mirror takes less heat, and the chain as a whole absorbs less. Mirrors further along can receive slightly more heat than before, because more light reaches them. The light that survives all eleven rises from 3.75% to 4.35% of the start: 16% more light at the end from one point of reflectance.
One caution. The value 74.2% is for light inside the mirror's band. A real source also emits other wavelengths, and each has its own R. At 14.2 nm this mirror reflects 6% and turns the rest into heat.
- Where it comes from
- The balance R + A + T = 1 of module 3, plus the fact that the substrate absorbs T. So A + T = 1 − R of the incoming power is turned into heat.
- Assumptions
- A substrate thick enough to absorb all the light that reaches it, and no light scattered out of the mirror. The equation holds at any wavelength, with R taken at that wavelength, angle and polarization. For a source with several wavelengths, add the heat from each one, each with its own R: 74.2% applies only inside the band.
- The terms
- Pin is the power arriving at a mirror and Pheat the part it keeps. R is its reflectance. m counts the mirrors along the chain, starting from 1.
- Check
- Add the heat of all eleven mirrors and the light that leaves the last one: the total is the starting power. With R = 74.2% the eleven mirrors keep 96.2% and 3.8% leaves. The three readouts of the experiment show it.
Reading these numbers correctly
Field strength, power flow and absorbed fraction are three different things with similar names. Before they are handed to the thermal modules, this step separates them and says what the optical model leaves out.
Three quantities appeared in this module that are easy to confuse: the field |E|², the power flow S, and the absorbed fractions. Each answers a different question.
Just above the mirror |E|² is 3.1. That is the strength of the field where two waves overlap. It does not mean that three times the incoming power is flowing there: the net power going down is 1 − R.
The heat load of the whole mirror, substrate included, is 1 − R = 25.8%. A = 24.9% counts only the 80 thin layers, and T = 0.9% is the small part that reaches the substrate and is absorbed there.
A real mirror has slightly rough boundaries that scatter some light into other directions, out of the mirror. The ideal calculation has no scattering, so that light appears nowhere in R + A + T = 1: a real balance needs a separate term for scattered light that escapes.
A mirror is also a heater. It returns 74% of the light and keeps 26%, about 95% of that in the top 210 nm. What that heat does to the mirror, how warm it gets and how its surface moves, is a question of heat flow and mechanics. The optics you have built in these five modules supply its starting point: how much power, and where. Module 6 takes it from there.
What to carry forward
- |E|² = 1 + R + 2·√R·cos(…). Use it for the field in front of a mirror. It measures field strength, which sets local absorption. It does not measure power flow.
- Alayer = Stop − Sbottom. Use it to find where energy is absorbed in any stack. It needs E and H at each boundary, which the matrix method provides.
- Pheat = (1 − R)·Pin. Use it for the heat load of a mirror on a thick substrate, with R taken at the wavelength of the light. Light scattered out of the mirror is not included.
To retrace the reasoning
Why is there a standing wave in front of a mirror?
The incoming and reflected waves overlap. Their phase difference changes with height, so they reinforce at some heights and oppose at others, in a pattern that stays put.
Why can |E|² be 3.5 when only one unit of power arrives?
Amplitudes add before squaring: 1 + 0.86 = 1.86, squared 3.5. It is field strength at a point, and it averages to 1 + R over height.
How do you find the light absorbed in one layer?
Compute the net power S crossing its upper and lower faces. The difference is what the layer absorbed.
Why does molybdenum absorb less than n·k·d suggests?
Near the peak the standing wave is locked to the layers, and at 13.5 nm |E|² averaged over molybdenum is 20% lower than over silicon.
Which mirror of a chain has the largest heat load, and why?
The first. Every mirror keeps the same fraction, 1 − R, but the first receives the most light. How warm it gets depends on how it is cooled, which is not computed here.