How do 80 weak surfaces make a 74% mirror?
Module 2 lined up two reflections and got 0.61%. A real EUV mirror stacks 40 pairs of molybdenum and silicon layers: 80 surfaces, each as weak as before. Six ideas build the answer, five of them with an experiment. Open any of them, in any order.
Module 2 showed what two surfaces can do together. This module puts 80 of them together and gets the mirror the rest of the project is about: its period, its number of pairs, its 74.2%, and the method that computes them. Modules 4 and 5 use that method again and again.
Two quick counts show that the answer is not obvious. Adding intensities, forty times the 0.61% of one film would be about 25%: that count ignores interference and repeated reflections, so it predicts nothing. Adding the 80 arrows of 0.04 end to end would give a reflected amplitude of 3.2 against an incoming amplitude of 1, a reflectance of 3.2² = 10.24: more than ten times the light that arrives, impossible for a mirror that only returns what it receives.
Lining up all the arrows: the Bragg condition
Module 2 ended with one film reflecting 0.61% when its two reflections line up. A mirror needs many more reflections lined up at once. This step finds the condition that does it for every pair, and with it the first design number of the mirror: the period.
Call one molybdenum layer plus one silicon layer a pair, and its total thickness the period D. Each pair repeats the one above it, one period deeper. Light reflected one pair deeper travels an extra round trip through one period.
If that extra round trip is exactly one full turn, the reflection from every pair arrives with its arrow pointing the same way as that of the pair above, so the crests of the two waves arrive together and the contributions of the pairs add up. If it is not, each pair arrives a little out of step with the one above, and the sum is smaller.
In the formula, n̄ is the average index of the pair, θ the angle of the light from the perpendicular and λ its wavelength in vacuum. The left side is the length of the extra round trip as the light counts it. When it equals one wavelength, the arrow has made one full turn.
The period must be about half a wavelength. Then going one pair deeper and back adds one full turn, and all the pairs agree. This is the Bragg condition, the same rule that governs X-rays reflecting from the planes of a crystal.
The estimate gives 7.00 nm. The full calculation peaks at 6.95 nm, where the readout shows 0.99 turns instead of 1. The count of phase leaves out how absorption and the repeated reflections change the sizes and phases of the waves inside the stack. Absorption is the larger part: with every k set to zero the best period would be 7.01 nm. Each single surface is still weak; what the simple count misses is their combined effect. A design has to come from the full calculation.
Within a pair there are two surfaces that reflect with opposite signs, as in module 2. How the period is shared between molybdenum and silicon decides how well those two families of arrows agree with each other. Module 4 looks at that.
- Where it comes from
- From module 2, crossing a layer twice turns the arrow by 4π·n·d·cos θ/λ. One pair has two layers, so its round trip is 4π·(nMo·dMo + nSi·dSi)·cos θ/λ. Write the bracket as n̄·D, with n̄ the average index weighted by thickness, and ask for one full turn, 2π.
- Assumptions
- Identical pairs. Real indices and weak reflections, so that every arrow can be treated on its own. One average index and one angle for the whole pair: the angle θ is the one inside the stack, close to the angle in vacuum because n̄ is close to 1. The exact phase of a pair adds the two layers separately, each with its own index and angle.
- The terms
- D = dMo + dSi is the period. n̄ is the thickness-weighted average of the two real indices: 0.4 × 0.924 + 0.6 × 0.999 = 0.969. θ is the angle from the perpendicular. λ is the wavelength in vacuum.
- Check
- With 13.5 nm and 6° the formula gives 7.00 nm. In the experiment the reflectance peaks at 6.95 nm, 0.8% lower: the right neighbourhood, and a reminder that this is a first estimate.
Why the sum levels off
The period is set. The next design number is how many pairs to deposit. If every pair added as much as the first ones, more pairs would always be better. This step shows why they do not, and why 40 is a sensible place to stop.
With the arrows lined up, adding pairs should keep adding reflection. Add them and watch two things: the reflectance, and how much light gets down to each pair.
Two things starve the deep pairs of light. The first is reflection itself: light sent back by the upper pairs never reaches the lower ones. In a stack tuned to the Bragg condition, as this one is, that alone would level the curve off near 100%: with 40 pairs and no absorption the mirror would reflect 98.7%. A stack that is out of tune does not approach 100%, however many pairs it has.
The second is absorption. Light that gets deep has crossed a lot of molybdenum, and must cross it again to come out. The real curve stops near 75%: 28 pairs give 70.6%, 40 pairs give 74.2%, and 20 more add less than one point.
The bars show both. At 40 pairs, switching absorption off changes the light going down only a little: in this tuned stack, reflection alone already keeps most of the light out of the deep pairs. The light coming back up grows at every depth, and the upward bar of the first pair is the reflectance. Each bar is the power of one travelling wave. Down minus up is the net power going down, apart from a small interference term inside the absorbing layers, at most 0.2% of the incoming power here.
This answers the question module 1 ended on. Eighty arrows of 0.04 cannot add up to 3.2, because the lower arrows are shorter than the upper ones: the light that would make them has already been reflected or absorbed.
A tool you need now: matrices
The arrows of the last two steps treat each reflection on its own. The curves shown beside them came from a calculation that does not, and that calculation needs one mathematical tool. It is introduced here with plain numbers, before any physics is attached to it.
Following every bounce through 80 layers by hand is hopeless. The method that replaces it needs one piece of mathematics: a 2×2 matrix. It is a table of four numbers that turns one pair of numbers into another pair.
Each row of the table produces one number of the result: multiply across and add. For example, the matrix with rows (2, 1) and (0, 3), applied to (1, 2), gives (2·1 + 1·2, 0·1 + 3·2) = (4, 6).
Applying one matrix and then another is itself a matrix, called their product. Matrices let a long chain of steps be collapsed into a single step. That is exactly what a stack of layers needs.
With matrices, order matters. Swapping two layers of a stack changes the mirror, and the mathematics agrees: the product of two matrices depends on which comes first.
Take A, which turns (x, y) into (y, x), and B, which doubles the first number only. Starting from (1, 3), A then B gives (6, 1), while B then A gives (3, 2).
Everything else works as with ordinary numbers. The matrix that changes nothing has rows (1, 0) and (0, 1) and is called the identity. The entries can be complex numbers, as they will be here.
One matrix per layer
With matrices in hand, this step builds the one that describes a layer. It is the core of the transfer-matrix method: the calculation behind every reflectance curve in these lessons, and behind the absorption and the heat of module 5.
Describe the light at any depth by two numbers: the total electric field E and the total magnetic field H, both parallel to the surfaces. Module 1 showed these are exactly the two quantities that cannot jump at a boundary. So crossing a boundary leaves the pair (E, H) unchanged. Only crossing a layer changes it.
Inside a layer the light is one wave going down, of amplitude a, and one going up, of amplitude b. Then E = a + b and H = η·(a − b), where η = ñ·cos θ says how much magnetic field a wave carries per unit of electric field. Here H is measured on a scale where a wave travelling in vacuum, perpendicular to the layers, has H = E. Crossing the layer turns a by eiδ and b by e−iδ, with δ = 2π·ñ·d·cos θ/λ, the phase of one crossing from module 2.
Write the pair at the top of the layer in terms of the pair at the bottom. The result is a matrix:
A stack is the product of its layers' matrices, in order from the top. Under the last layer is the substrate, where light only goes down. Choose its electric amplitude as 1 for now, so that there (E, H) = (1, ηs); the common scale cancels in the end. Multiplying through gives the pair (B, C) at the top surface, and from it the reflection:
The matrix method contains everything from modules 1 and 2. With no layer it is the Fresnel formula. With one layer it is the Airy formula, all bounces included, and the two readouts agree to every digit shown. With 80 layers it is the same recipe repeated: this is the transfer-matrix method.
At an angle, each layer has its own cos θ, fixed by Snell's law starting from the angle in vacuum: ñ·cos θ = √(ñ² − sin² θ₀). The 6° quoted for a scanner is the angle in vacuum. Inside molybdenum the ray travels at 6.5°.
A warning that has cost many people an afternoon. The −i in the matrix goes with the convention ñ = n + i·k used in these lessons. Books that write ñ = n − i·k have +i instead. Mixing the two gives reflectances above 100%.
- Where it comes from
- At the top of the layer E = a + b and H = η·(a − b). At the bottom the two waves have turned: E = a·eiδ + b·e−iδ, H = η·(a·eiδ − b·e−iδ). Solve the bottom pair for a and b, put them into the top pair, and use eiδ = cos δ + i·sin δ.
- Assumptions
- A uniform layer with flat, parallel, sharp faces. One wavelength and one polarization. E and H are the parts of the fields parallel to the surfaces. As written, η = ñ·cos θ is for s polarization; for p polarization η = ñ / cos θ. For p this bookkeeping gives r with the opposite sign to the formula of module 1, and the same R.
- The terms
- δ is the complex phase of one crossing. η is the ratio of magnetic to electric field for a wave travelling one way in that material. η₀ and ηs are the same ratio in vacuum and in the substrate.
- Check
- d = 0 gives the identity matrix: a layer with no thickness changes nothing. With no layers at all, B = 1 and C = ηs, and r becomes (η₀ − ηs)/(η₀ + ηs): the Fresnel formula. With one layer the method reproduces the Airy formula: compare the two readouts in the experiment at any thickness.
Forty pairs: 74%
Everything is in place to compute the full mirror: the period from the Bragg step, the number of pairs from the saturation step, one matrix per layer. This step runs the calculation and reads its three outputs, the numbers the rest of the project starts from.
Now the full mirror. From vacuum down: a molybdenum layer 2.78 nm thick, a silicon layer 4.18 nm thick, repeated, on a silicon substrate. Light of 13.5 nm arrives at 6° with s polarization.
These two thicknesses are a choice, kept for the rest of the lessons. Their sum is a period of 6.96 nm, with molybdenum taking 40% of it. That is 0.015 nm more than the best period of the Bragg step: a rounded reference design, not the exact optimum. It costs a tenth of a point, 74.2% here against 74.3% there. Module 4 shows what that small difference in period does to the peak.
The stack reflects 74.2%; a single vacuum–molybdenum surface reflects 0.16%. Of every 100 photons, 74 come back, 25 are absorbed in the layers and 1 reaches the substrate.
The order matters, as the matrices said it would. With silicon on top the light crosses 4.18 nm of silicon twice before meeting the first strong surface, which costs 1.4% of it: 74.2% × 0.986 = 73.2%, close to the 73.0% of the full calculation. The factor alone is an estimate and not the whole story: the small reflection at the vacuum–silicon surface also adds, as a wave, to the reflection of the stack below.
The number also depends on the conditions. For p polarization at 6° the same stack gives 73.5%. At perpendicular incidence it gives 72.8%, because this period was chosen for 6°. Whenever a reflectance is quoted, ask for the stack order, the angle, the polarization and the wavelength.
- Where it comes from
- Above the stack, an incoming wave of amplitude 1 and a reflected wave r give E = 1 + r and H = η₀·(1 − r). The matrices say that this pair must be proportional to (B, C). Dividing the two conditions removes the unknown constant and leaves r. The constant itself is the amplitude t of the wave that enters the substrate: adding the two conditions gives t = 2·η₀ / (η₀·B + C).
- Assumptions
- Those of the layer matrix, for every layer. A substrate thick enough that nothing comes back from its far side.
- The terms
- R = |r|² is the reflected fraction of the power. T is the fraction that enters the substrate: T = Re(ηs) / η₀ · |t|². The ratio of η turns the squared amplitude into a ratio of power flows, because a field of the same strength carries a different power in silicon and in vacuum. A = 1 − R − T is what is left: the fraction absorbed inside the layers. The substrate eventually absorbs T as well.
- Check
- With one pair the stack is almost the single film of module 2 and R is 0.54%. At 40 pairs the three parts are 74.2%, 24.9% and 0.9%, which add to 100%. With absorption switched off in the previous step, A would be zero and R + T = 1.
What the method assumes, and when it applies
A number such as 74.2% is only as good as the model behind it. Before the method is used in modules 4 and 5, this step states what it assumes, what it leaves out, and what to keep from this module.
The transfer-matrix method is exact for the stack it is given. Whether that stack is the real mirror is a separate question.
An exact method for an ideal stack. Its strength is generality: any layers, any thicknesses, any angle, absorption included. Its limit is the picture of the stack you feed it. Module 4 uses it to see how the mirror responds to wavelength, period and the share of molybdenum.
The Bragg condition is not among its requirements: it only estimates where the contributions of successive pairs reinforce one another, and the exact optimum comes from the full calculation. Small absorption is not one either: absorption is carried by the complex index inside δ and η, and no approximation is made about it. If the mirror warms up and every layer grows thicker, change the thicknesses and multiply the matrices again.
What the ideal calculation leaves out is the shape of the boundaries. In a real mirror, molybdenum and silicon mix over a few tenths of a nanometre at each one. The complex index describes absorption inside a uniform layer, not that: 74.2% is the value for ideal boundaries. Mixing can be included, for example by slicing each boundary into very thin uniform layers.
What to carry forward
- 2·D·n̄·cos θ = λ. Use it for a first estimate of the period, or of the wavelength a given period reflects. It is not a design value: the full calculation moves it by about 1%.
- One matrix per layer, multiplied from the top down. Use it for any stack of uniform layers with sharp, flat boundaries. It is exact for that stack and says nothing about roughness or mixing.
- R + T + A = 1. Whatever is not reflected is absorbed in the layers or passed to the substrate. Use it to check any calculation.
To retrace the reasoning
Why must the period be about half a wavelength?
Going one pair deeper and back must add one full turn, so that every pair reflects in step with the one above.
Why can 80 arrows of 0.04 not add up to 3.2?
The deeper arrows are shorter: the light that would make them has already been reflected by the pairs above, or absorbed.
Why describe the light by E and H instead of following each bounce?
They are continuous across every boundary, so only the layers need a matrix, and all the bounces are included automatically.
Where does the layer matrix come from?
From one wave going down and one going up, each turned by the phase of one crossing, rewritten in terms of E and H at the two faces.
What has to be stated together with "74.2%"?
Forty pairs with molybdenum on top, 2.78 and 4.18 nm, silicon substrate, 13.5 nm, 6°, s polarization, ideal boundaries.
Sources
The indices of molybdenum and silicon used in every calculation of these lessons come from the tables of the Center for X-Ray Optics (CXRO): B. L. Henke, E. M. Gullikson and J. C. Davis, X-ray interactions: photoabsorption, scattering, transmission, and reflection at E = 50–30,000 eV, Z = 1–92, Atomic Data and Nuclear Data Tables 54, 181–342 (1993), with later updates at henke.lbl.gov/optical_constants. Module 1 shows how n and k are read from them.