How narrow is the peak, and what moves it?
Module 3 tuned 40 pairs so that their reflections line up at 13.5 nm. Change the wavelength a little and the arrows no longer point the same way. Five ideas show how selective the mirror is and what shifts its peak, four of them with an experiment. Open any of them, in any order.
A mirror that works at one wavelength, one period and one angle is a design on paper. A real one is lit by a source with a spread of wavelengths, coated with small errors in thickness, and hit at more than one angle. This module measures how much each of these matters. Its answers are the tolerances of the design. One of them, the shift of the peak when the period changes, is the first link between the optics and the heating studied in modules 5 and 6.
A mirror only for a band of wavelengths
Module 3 computed the mirror at one wavelength. A source emits a spread of them, so the first question about a real mirror is how much of that spread it returns. This step measures the width of the peak and explains what sets it.
Here is the reflectance of the mirror of module 3 against wavelength, computed with the matrix method. At each wavelength the indices of molybdenum and silicon are read again from the tables, because they change with wavelength too.
The width of the peak is measured at half its height: find the highest point, take half of that value, and measure the distance between the two wavelengths, one on each side, where the curve crosses it. This is the full width at half maximum, FWHM.
With 40 pairs the peak is 0.64 nm wide and sits at 13.53 nm. The reflectance stays above half its peak value between 13.14 and 13.77 nm. The peak is not symmetric: those two points lie 0.39 nm below it and 0.25 nm above.
The reason is in the arrows. Away from the best wavelength, each pair's arrow is turned a little against the previous one. Over a few pairs that hardly matters. Over many it adds up, the chain curls, and the sum collapses. The more pairs take part, the smaller the detuning that is tolerated.
With 5 pairs the simple estimate and the full calculation agree: 2.39 nm. With 40 pairs the estimate says 0.30 nm and the calculation 0.64 nm. The estimate counts all 40 pairs as equal partners, and module 3 showed they are not: the deep ones receive little light. Read through this estimate alone, 0.64 nm is the width that about 19 equally weighted pairs would give. The other pairs still contribute, only less and less, which is why adding more hardly narrows the peak.
- Where it comes from
- Add N arrows of length 1, each turned by the same small angle ε against the one before. This is a finite geometric sum, 1 + x + … + xN−1 = (1 − xN)/(1 − x), with x = eiε. Unlike the infinite series of module 2 it needs no condition on the size of x, only x ≠ 1. Its length is the ratio shown. Detuning the wavelength by Δλ turns each pair's arrow by ε = 2π·Δλ/λ. For small ε, sin(ε/2) ≈ ε/2, so the intensity goes as (sin u / u)² with u = N·ε/2, and falls to half at u = 1.39. That gives the half-width Δλ = 1.39·λ/(π·N). The full width is twice that: 0.886·λ/N.
- Assumptions
- N identical pairs that all contribute equally: weak reflections and no absorption. Indices that do not change over the width of the peak.
- The terms
- N is the number of pairs. ε is the extra turn from one pair to the next. λ is the wavelength of the peak. FWHM is the full width of the peak at half its height.
- Check
- With ε = 0 the ratio becomes N: all arrows in line. It first reaches zero when N·ε = 2π: the chain has closed into a full circle. For 5 pairs the estimate gives 2.39 nm, and so does the experiment. For 40 pairs it fails, as it should, because its first assumption fails.
The period sets the wavelength
The peak is narrow, so it has to sit in the right place. What a coating process controls is the period. This step finds how the peak follows the period: that is the tolerance on layer thickness and, later, the effect of thermal expansion.
The Bragg condition ties the best wavelength to the period: λ = 2·D·n̄·cos θ. A thicker period should move the whole peak to longer wavelengths, in proportion.
A period of 6.948 nm puts the peak at 13.50 nm. That is 2.78 nm of molybdenum and 4.17 nm of silicon. This is a nominal value. How closely a real coating has to match it depends on how much loss of reflectance, and how much optical error, can be accepted. The lessons keep the mirror of module 3, with 4.18 nm of silicon and a period of 6.96 nm, because the difference is small: its peak sits at 13.53 nm, and at 13.5 nm it reflects 74.2%, against 74.3% for the retuned one.
The proportion holds, roughly. Going from 6.96 to 7.03 nm is a 1% change, and the peak moves from 13.53 to 13.65 nm: 0.12 nm, or 0.9%, instead of the 0.135 nm of the simple rule. The difference comes from the materials. At longer wavelengths the index of molybdenum is lower, so n̄ falls as the peak moves, and the peak moves a little less.
There is also a constant offset. For D = 6.96 nm the Bragg formula predicts 13.41 nm, and the full calculation gives 13.53 nm. The formula is a guide to how the peak moves, and a first estimate of where it is. The design value has to come from the full calculation.
- Where it comes from
- In λ = 2·D·n̄·cos θ the wavelength is proportional to D. Multiply D by 1.01 and λ is multiplied by 1.01.
- Assumptions
- The share of molybdenum and the angle stay the same. The average index n̄ does not change between the old and the new wavelength.
- The terms
- ΔD is the change of the period and Δλ the resulting change of the peak wavelength. Both are taken relative to their starting values.
- Check
- From 6.96 to 7.03 nm the rule predicts +0.135 nm. The experiment gives +0.12 nm. The second assumption is the one that is not quite met.
The same mirror at another angle
The period is not the only quantity in the Bragg condition. A scanner has mirrors at different angles, and one mirror can see a range of angles. This step shows how far the peak moves when the angle does.
The Bragg condition contains the angle too. Light that crosses the stack obliquely has a shorter wave motion perpendicular to the layers, by the factor cos θ, so the same period fits a shorter wavelength.
A mirror is made for one wavelength at one angle. This stack reflects 13.5 nm best at about 7°. At perpendicular incidence its peak is at 13.60 nm, and 13.5 nm is reflected at 72.8%. At 15° the peak is at 13.16 nm, and 13.5 nm is almost not reflected at all: 9%.
In a scanner different mirrors are hit at different angles, and each needs its own period. Where the angle of incidence varies across a curved mirror, the period of the coating can be graded across the surface to keep it tuned.
- Where it comes from
- From λ = 2·D·n̄·cos θ with D and n̄ fixed. Module 2 gave the reason for the cosine: at an angle the phase gained crossing a layer is 2π·n·d·cos θ/λ.
- Assumptions
- θ is the angle inside the stack. Because n̄ is close to 1 it is nearly the angle in vacuum, and the formula uses that one. The average index does not change over the shift.
- The terms
- λpeak(0) is the peak wavelength at perpendicular incidence, 13.60 nm for this stack. θ is the angle from the perpendicular.
- Check
- At 10° the formula gives 13.60 × cos 10° = 13.39 nm, and the experiment 13.40 nm. At 20° it gives 12.78 nm against 12.83 nm: the approximations start to show.
How much molybdenum in each pair?
Period and angle place the peak. One design number is still free: how the period is split between the two materials. Module 3 used 40% molybdenum and left the reason for later. This step gives it.
The period fixes the wavelength. It does not say how to share the period between the two materials. Call Γ the fraction taken by molybdenum: Γ = dMo / D.
A pair has two surfaces that reflect with opposite signs. Module 2 found when two such reflections add best: when the round trip between them is half a turn. With one full turn per pair, half a turn means the molybdenum layer takes about half of the period.
A compromise between lining up the arrows and losing light. Shares near one half make each pair's arrow as long as it can be. But molybdenum is the absorber, and every nanometre of it is crossed twice by all the light that goes deeper. In the simple model, going from 0.5 to 0.4 gives up 5% of the arrow, since sin(0.4·π) = 0.951. With real materials it gains more than it costs: the peak rises from 73.2% to 74.3%.
The same curve shows how forgiving the choice is. Between 0.35 and 0.45 the peak stays within half a point of its best value. The share is far less critical than the period.
- Where it comes from
- Take the two surfaces of the molybdenum layer. Their reflections are +r and −r. The second has crossed the layer twice, a round trip of 2π·Γ when the whole pair makes one turn. The sum r·(1 − ei·2πΓ) has length 2·r·sin(π·Γ).
- Assumptions
- Weak reflections and no absorption, so that one pair can be looked at alone. The two indices are close enough that the share of thickness is also the share of phase.
- The terms
- Γ = dMo/D is the fraction of the period taken by molybdenum. r is the size of the reflection of one surface.
- Check
- In this simple model Γ = 0.5 gives 2: two quarter-wave layers, the best case of module 2. Γ = 0 or 1 gives 0: with a single material the repeated boundaries between molybdenum and silicon are gone, and with them the Bragg contribution. Only the outer boundaries of the coating remain. In the experiment without absorption the curve peaks near 0.5; not exactly, because with two different indices equal thicknesses are not quite equal phases.
What this means for a real machine
Width, position and height of the peak are now known. This step reads them as engineering facts: a chain of mirrors, a tempting idea that does not work, and a warm mirror. Then it says what to keep from the module.
The mirror reflects a band 0.64 nm wide, centred where the period and the angle put it. Three consequences, each a use of what this module built.
Consider eleven such mirrors in a row, each reflecting 74.2%. The light left after all eleven is 0.742 multiplied by itself eleven times: 3.75%.
Eleven mirrors pass less than 4% of the light. This is why every point of reflectance matters, and why the sources of EUV scanners have to be so powerful. It is also why the 26% that each mirror does not reflect matters: it ends up as heat in the mirrors. Module 5 follows that light.
More pairs do not buy a narrower band. With 200 pairs the simple estimate 0.886 · λ / N would promise 0.06 nm, but absorption and reflection keep the light out of the deep pairs, and the width stops near 0.6 nm.
Now suppose the mirror warms up and its period grows by 0.1%. The peak moves by about 0.013 nm, some 2% of its width.
For this small, uniform change the reflectance hardly moves. Heating can also distort the surface of a mirror and introduce wavefront errors, which is a separate optical problem.
What to carry forward
- FWHM ≈ 0.886·λ/N. Use it when all N pairs contribute equally: few pairs, weak reflections. For a full EUV mirror it underestimates the width by a factor of two.
- Δλ/λ ≈ ΔD/D and λpeak ∝ cos θ. Use them to estimate how the peak moves with period and angle. They are not accurate enough to place the peak: that takes the full calculation.
- Arrow of one pair ∝ 2·sin(π·Γ). Use it to see why equal shares line the arrows up best. It ignores absorption, which is what moves the real optimum to 0.4.
To retrace the reasoning
Why does the mirror reflect only a band of wavelengths?
Off the best wavelength each pair's arrow is turned a little against the previous one. Over many pairs the chain curls and the sum collapses.
How is the FWHM defined?
Take half the height of the peak and measure the distance between the two wavelengths, one on each side, where the curve crosses that value.
Why is the real band twice as wide as 0.886·λ/N says for 40 pairs?
The estimate assumes 40 equal pairs. Reflection and absorption give the deep ones less light, so the peak is as wide as about 19 equal pairs would make it.
What moves the peak, and which way?
A thicker period moves it to longer wavelengths, in proportion. A larger angle from perpendicular moves it to shorter wavelengths, as cos θ.
Why is the share of molybdenum 0.4 and not 0.5?
Equal shares give the longest arrow per pair, but molybdenum absorbs three and a half times more than silicon. Slightly less of it gains more than it costs.