How hot does the mirror get?
Module 5 found that the mirror keeps a quarter of the light as heat, within a few hundred nanometres of its surface. This module follows that heat into the body of the mirror and turns it into a temperature. Five ideas, three of them with an experiment. Open any of them, in any order.
This is the bridge between the optics and the mechanics of the project. The optical modules end with one number, the fraction of light absorbed. The mechanical ones, deformation and wavefront error, start from a temperature at every point of the mirror. In between sits the conduction of heat, and one awkward fact: the glass chosen for mirrors because it barely expands is also a poor conductor of heat.
From a stack to a sheet of heat
A thermal calculation needs a heat source: how much power, and where it is released. Module 5 gave both answers in the language of optics. This step puts them in the form a heat calculation uses, and makes the simplification that everything after it relies on.
The multilayer is 40 pairs of 6.96 nm: 278 nm in all. The body of the mirror under it is some 50 mm of glass, about 180,000 times thicker. Module 5 showed that about 95% of the heat is released in the top 210 nm.
Seen from the glass, then, the stack is a sheet: all the heat enters through the top face. A calculation of the mirror's temperature does not need the layers at all. It needs one quantity at each point of the surface: the absorbed power per unit area, called the heat flux q, in watts per square metre.
Here I is the intensity of the light arriving at that point, also in watts per square metre, and 1 − R = 25.8% is the fraction that module 5 found to end as heat.
The optics hands over one map: watts per square metre on the surface. Everything the first five modules computed, indices, phases, matrices, fields, reaches the thermal calculation through the single factor 1 − R.
Is it safe to forget the thickness of the stack? Two checks say yes. First, the temperature difference across it. Molybdenum and silicon conduct heat well, and in the base case pushing the flux through 278 nm of them takes a few hundred-thousandths of a kelvin. Even if such thin layers conducted a hundred times worse than the bulk materials, the difference would be a few thousandths of a kelvin there, and about a tenth of a kelvin at the most intense setting of the experiment. This leaves out the thermal resistance of the interfaces between the layers. Second, the time. Heat crosses 278 nm in far less than a millionth of a second, so the stack keeps up with the light on the slow scales of this steady model. The individual pulses of the source are a different question, left out here.
- Where it comes from
- It is Pheat = (1 − R)·Pin of module 5, applied to each small patch of the surface instead of to the whole mirror.
- Assumptions
- The same reflectance everywhere on the mirror. Light of one wavelength inside the band of the mirror. A stack thin enough to count as a surface, as checked above. Nothing scattered out of the mirror.
- The terms
- q is the heat flux entering the glass, in W/m². I is the intensity of the light on the surface, in W/m². R is the reflectance.
- Check
- A perfect mirror, R = 1, gives q = 0: no heat. With R = 74.2%, a point that receives 1 W/cm² of light takes in 0.258 W/cm² of heat.
The light arrives as a spot
The flux is not the same everywhere: a beam is brighter at its centre. It is the peak of the flux, more than the total power, that decides how hot the hottest point gets. This step gives the beam a shape and finds that peak.
The beam on a mirror is modelled here as a Gaussian spot: brightest at the centre, fading smoothly outwards. Its size is the radius w, the distance from the centre at which the intensity has dropped to 1/e² of the peak, 13.5%.
The peak flux is the power divided by the area of the spot. Halve the radius and the same power lands on a quarter of the area: four times the flux. Double the power and the flux doubles everywhere.
To give the number a scale: 1.28 W/cm² is about thirteen times the midday sun on the ground, which delivers 0.1 W/cm². The 70 W are the example used by the project. For scale: in the chain of module 5, with 250 W on the first mirror, the fifth receives 76 W.
- Where it comes from
- Add up the intensity over the whole surface, ring by ring. The total of e−2r²/w² over the plane is π·w²/2, so for the total to be P the peak must be P divided by that. Multiplying by 1 − R turns light into heat.
- Assumptions
- A circular Gaussian spot under steady light. At 6° the spot on the mirror is stretched by 1/cos 6°, less than 1%, and that is ignored. The mirror is wide enough to hold the whole spot.
- The terms
- P is the power of the light arriving. w is the radius at which the intensity is 1/e² of its peak. r is the distance from the centre. q is the heat flux.
- Check
- With 70 W, w = 30 mm and 25.8%: I₀ = 2 × 70 / (π × 0.03²) = 49,500 W/m², and q at the centre is 0.258 times that, 12,800 W/m² or 1.28 W/cm². The width of the spot at half its peak is 1.18·w, and 99% of the power falls within 1.52·w.
Heat flows down a temperature slope
A heat flux is not yet a temperature. What links the two is how well the material conducts heat. This step introduces that property in the simplest geometry there is, and meets the awkward number of this project: the conductivity of the glass a mirror is made of.
Heat flows from hot to cold, and the flow is proportional to how steeply the temperature falls. The constant is the thermal conductivity k, in watts per metre per kelvin. This is Fourier's law.
Take a plate of thickness H, heated evenly on top with flux q and held at a fixed temperature at the bottom. Once things are steady the same flux crosses every depth, so the temperature falls in a straight line, and the top is hotter than the bottom by ΔT.
A good material for a mirror is a poor conductor of heat. ULE is a glass made to have almost no thermal expansion near room temperature: a mirror made of it keeps its shape when its temperature changes a little. Its conductivity, 1.31 W/(m·K), is about a hundred times lower than that of silicon or molybdenum.
The 489 K is not the answer for the mirror. It assumes the peak flux over the whole face, with the heat going straight down. The real spot is small, and around it there is cold glass into which the heat can also spread sideways. The next step lets it.
- Where it comes from
- Fourier's law is an observed fact about solids: the heat flux is proportional to the slope of the temperature. In a steady state no slice of the plate gains or loses energy, so the same flux crosses every depth. A constant flux means a constant slope, q/k, and over a thickness H that adds up to q·H/k.
- Assumptions
- A steady state. An even flux over a face much wider than the plate is thick, so that heat flows straight down. A conductivity that does not change with temperature.
- The terms
- q is the heat flux, in W/m². k is the thermal conductivity, in W/(m·K). H is the thickness. ΔT is the temperature of the heated face above the cooled one. The minus sign in Fourier's law says that heat flows towards lower temperature.
- Check
- Units: (W/m²) · m / (W/(m·K)) = K. Numbers: 12,800 W/m² × 0.05 m / 1.31 W/(m·K) = 489 K. Twice the thickness, or half the conductivity, doubles it.
The temperature of the real mirror
Now the three pieces meet: the Gaussian spot, the conductivity of the glass, and the real shape of the mirror, a disc cooled from behind. The result is the number the mechanical part of the project starts from.
The mirror of the project is a disc of ULE, 150 mm in radius and 50 mm thick. Its top face receives the Gaussian flux. Its back rests on a cooled plate, idealised as a face held at one fixed temperature. Its side is in vacuum and exchanges almost no heat, so it is treated as insulated. Inside, heat flows both down and sideways.
154 K above the cooled back, in a glass chosen for its very low expansion near room temperature. That is the result of this module, and the starting point of the mechanical part of the project.
Compare it with the last step. Heat going straight down gave 489 K. Letting it spread sideways brings that to 154 K. The surface is warm well outside the lit area: at the distance w, where the flux is down to 13.5% of its peak, the temperature is still 37% of its peak.
For a mirror much thicker and wider than the spot there is an exact formula, shown in the card below, which gives 183 K here. The cooled back, 50 mm down, takes about a sixth off that.
The formula also shows what counts. At a given absorbed power the temperature goes as 1/w and as 1/k. A spot twice as wide has a quarter of the peak flux, yet the formula only halves the temperature, because the heat of a wider spot has further to go before it can spread. In the disc of the experiment it falls a little more, to 64 K, because the cooled back helps a wide spot more.
With the 24.7% of the project's own calculation, which counts the layers alone, the same disc gives 147 K. That is the value the project finds with a finite-element model. The experiment here adds up the exact solution for a disc instead, and the two agree to about one part in a thousand.
Why it matters. The mirror sits at about 22 °C, and ULE is chosen because its expansion is very low near that temperature: its maker specifies a mean coefficient of 0 ± 30 parts per billion per kelvin between 5 and 35 °C. A rise of 154 K puts the centre of the spot near 176 °C, where that specification no longer applies, and this is with an ideal cooling plate. To estimate the deformation there, the next step needs how the expansion of the glass changes with temperature. A better conductor is not the easy answer: the same disc in silicon would warm by 1.3 K, but silicon expands with temperature far more than this glass does near room temperature.
- Where it comes from
- Inside the glass two facts hold at every point: Fourier's law, and conservation of energy, which in a steady state says that as much heat leaves any small volume as enters it. For a Gaussian flux on a body much thicker and wider than the spot the two can be solved exactly, a classic result of heat conduction (Carslaw and Jaeger): T = q₀·w·√(π/8)/k. Writing q₀ as 2·Pabs/(π·w²) gives the form above.
- Assumptions
- A steady state, a constant conductivity, heat entering only through the top face. The formula needs a body much larger than the spot. The experiment does not: it solves the disc of the stated radius and thickness, with insulated side and cooled back, by adding up 60 circular patterns of heat flow, each of which fades with depth in a known way.
- The terms
- Pabs = (1 − R)·P is the absorbed power. k is the conductivity. w is the spot radius. T is the temperature of the hottest point above the temperature far from the spot; in the experiment, above the cooled back.
- Check
- 18.06 W / (2.507 × 1.31 W/(m·K) × 0.03 m) = 183 K. In the experiment a thicker disc moves towards this value: 170 K at 100 mm. A thinner one moves away from it: 117 K at 20 mm.
What this number assumes, and what comes next
A temperature is the hand-over point of the project: optics before it, mechanics after. This step states what went into it, so that it can be trusted for what it is, and says where the chain goes next.
The temperature of the last step is as good as what went into it. Four things went in that a real mirror does not fully respect.
A steady state. These are the temperatures the mirror settles to under constant light. Glass this thick takes time to get there. Heat needs a time of order H²/α to cross a thickness H, where α is the thermal diffusivity, the conductivity divided by the heat a cubic metre stores per kelvin. For 50 mm of ULE that is close to an hour.
An ideal cooled back. A real cooling plate is not at one temperature, and its contact with the glass is not perfect. Both make the real mirror warmer than computed here.
Only the light inside the band. The 25.8% holds for light of 13.5 nm. Module 5 noted that a real source also emits other wavelengths, and the mirror turns most of those into heat.
A constant conductivity and an insulated side. The conductivity of the glass changes a little over 150 K. Vacuum removes convection, but not radiation, and the model leaves radiation out. An estimate on the temperatures computed for the base case, with surroundings at 22 °C, gives at most about 0.2 W lost from the side and at most about 5 W from the hot top face, against the 18 W that come in. Both figures are for a perfect emitter and scale down with the emissivity of the real surface. So the side matters little, while radiation from the top face is not negligible. These powers are estimated with the temperatures of the model without radiation. To know how much the temperature falls, conduction and radiation have to be solved together; here 154 K stays the reference, for conduction alone.
What comes next in the project follows from the result itself. The glass under the spot is hot and the rest is not, so it expands unevenly and the surface rises where the light falls. A mirror for lithography has to keep its shape to a fraction of a nanometre. How large the bump is, and what it does to the reflected wave, is the mechanical part of the project.
What to carry forward
- q₀ = 2·(1 − R)·P / (π·w²). Use it for the peak heat flux of a Gaussian spot. It needs the power on that mirror, the spot radius and the reflectance at the wavelength of the light.
- ΔT = q·H / k. Use it when heat flows straight through a plate. For a small spot on a wide mirror it overestimates, because it ignores the spreading sideways.
- T = Pabs / (√(2π)·k·w). Use it for the hottest point under a Gaussian spot on a body much thicker and wider than the spot. A cooled back close to the surface lowers the result.
To retrace the reasoning
Why can the stack be treated as a sheet of heat?
It is 278 nm thick on 50 mm of glass. In the base case the temperature difference across it is far below a hundredth of a kelvin, and heat crosses it in far less than a millionth of a second.
What does halving the spot radius do, at the same power?
The peak flux becomes four times larger, because the area is a quarter. The peak temperature of a thick mirror only doubles, because the heat of a smaller spot spreads out over a shorter distance.
Why is ULE used although it conducts heat poorly?
Because it barely expands near room temperature, which is what keeps the shape of the mirror. Its conductivity is the price paid for that.
Why is the real mirror cooler than the straight-down estimate?
The estimate puts the peak flux on the whole face. The real spot is small, and its heat also spreads sideways into cold glass.
What does a rise of about 150 K mean for the glass?
The centre of the mirror is far outside the narrow range around room temperature in which ULE barely expands, so the surface rises under the spot.