What do two surfaces close together do?
Module 1 ended with a single molybdenum surface sending back 0.16% of the light. Now put a thin molybdenum film on silicon. The light meets two surfaces: the top of the film, and the boundary between film and silicon a few nanometres below. Six ideas show what the two do together, five of them with an experiment. Open any of them, in any order.
A mirror is made of many surfaces, and the step from one to many is taken here with the smallest case: two. The rule for adding two reflections, and the phase a layer adds between them, are all that module 3 needs to line up eighty.
A tool you need now: adding two waves
A film returns two reflections, and a detector sees one. This step gives the rule that combines them. It is the working principle of the whole mirror, used again for 80 reflections in module 3 and for the field inside the mirror in module 5.
When two waves of the same wavelength overlap, their fields add point by point. The result depends on how the two are shifted against each other: their phase difference φ. Crest on crest, they reinforce. Crest on trough, they cancel.
Module 1 gave the bookkeeping. Each wave is an arrow, with its amplitude as length and its phase as angle. Adding waves is adding arrows, tip to tail. The length of the resulting arrow is the amplitude A of the combined wave.
Amplitudes add; intensities do not. Two equal waves in step give four times the intensity of one, not twice. Two equal waves half a turn apart give nothing at all. No energy is destroyed: at a mirror, the light that does not come back goes forward instead.
This is interference, and it is the whole working principle of a multilayer mirror: arrange many weak reflections so that their arrows line up.
- Where it comes from
- Draw the two waves as arrows, a₁ along the horizontal axis and a₂ at angle φ. Their sum has a horizontal part a₁ + a₂·cos φ and a vertical part a₂·sin φ. Pythagoras on those two parts, together with cos² φ + sin² φ = 1, gives the formula.
- Assumptions
- Two waves of the same wavelength, travelling in the same direction, with a phase difference that stays fixed in time.
- The terms
- a₁ and a₂ are the two amplitudes. φ is the phase difference between them. A is the amplitude of the combined wave, and A² its intensity.
- Check
- φ = 0 gives (a₁ + a₂)²: the amplitudes add. φ = 180° gives (a₁ − a₂)²: they subtract. φ = 90° gives a₁² + a₂²: only in that case do the intensities simply add. Try all three in the experiment.
The phase a layer adds
The rule for adding needs the angle between the two arrows. Part of that angle comes from the thickness of the film, which is the thing a designer can choose. This step computes it.
A wave that crosses a layer comes out with its arrow turned. Module 1 gave the rule: crossing a thickness d in a material of index n turns the arrow by one full turn, 2π, for every wavelength-inside λ/n that fits in the layer. Call that angle β.
Light reflected at the bottom of a film crosses it twice, down and back up. Compared with light reflected at the top, it comes out turned by 2β.
A film a quarter of a wavelength thick turns the returning wave by half a turn. The wavelength meant here is the one inside the material, λ/n = 14.6 nm for molybdenum, so the quarter-wave thickness is 3.65 nm.
This step counts only the real part of the index. The imaginary part k does its own job on the same trip: it shortens the arrow. Over 3.65 nm down and back the amplitude drops by about 2%.
- Where it comes from
- One turn is 2π. The number of wavelengths that fit in one crossing is d divided by the wavelength inside the material, λ/n. Multiply the two, and double for the return trip.
- Assumptions
- A film with flat, parallel faces, and light travelling perpendicular to it. At an angle the phase is smaller: d is replaced by d·cos θ, with θ the angle of the ray inside the film.
- The terms
- n is the real part of the film's index. d is its thickness. λ is the wavelength in vacuum. β is the phase of one crossing.
- Check
- d = 0 gives no turn. A film one wavelength-inside thick, 14.6 nm, gives two full turns on the round trip: set 14.6 nm and read about 720°.
Two reflections from one film
Both pieces are ready: how arrows add, and how a layer turns them. This step puts them on the real film, with the reflections of module 1, and finds the best thickness. It is the first design choice of the project.
Now the two reflections of a molybdenum film on silicon. From module 1: the top surface, vacuum to molybdenum, reflects with r₁ = +0.040. The bottom surface, molybdenum to silicon, reflects with r₂ = −0.039. Almost equal in size, opposite in sign.
The second wave also makes the round trip through the film, so its arrow arrives turned by 2β and slightly shortened by absorption. The reflection of the film is the sum of the two arrows.
Two flips make the arrows line up. The minus sign of r₂ is one half turn, the round trip is the other. Together they bring the second arrow almost parallel to the first, and the amplitudes add: approximately 0.040 + 0.038 = 0.078. Squared, that is about 0.61%, almost four times a single surface.
The two-arrow sum of this step gives 3.86 times a single surface, and the exact formula of the next step 3.85 times. It is not 4 for small reasons you can see in the arrows. The bottom reflection is a little weaker than the top one. Absorption shortens it by another 2%. And at the best thickness the two arrows are almost, not exactly, parallel: they differ by less than one degree.
The peak sits at 3.61 nm, a little below the quarter-wave value of 3.65 nm. The quarter-wave rule uses only the real part of the index, so for an absorbing film it is a first estimate. The complex indices move the thickness at which the two arrows are exactly parallel to 3.63 nm, and absorption favours a slightly thinner film still. With all bounces included the peak is 0.608%. At the 6° used in scanners, for s polarization, it moves to 3.63 nm and 0.623%.
At d = 0 the two arrows nearly cancel, as they should. With no film the light goes from vacuum straight to silicon, which reflects almost nothing.
- Where it comes from
- Add the arrows of the two reflected waves. The first is r₁. The second has travelled down and back, so it carries the factor ei·2β, with the complex index inside β: the real part turns the arrow, the imaginary part shortens it.
- Assumptions
- Both surfaces are weak. Almost all the light reaches the bottom surface and comes back out, so the transmission through the top surface is taken as 1, and light that bounces more than once inside the film is ignored. Perpendicular incidence, flat and sharp surfaces.
- The terms
- r₁ is the reflection of the top surface and r₂ that of the bottom surface, each from the Fresnel formula. β = 2π·ñ·d/λ is the complex phase of one crossing.
- Check
- At d = 0 the formula gives r₁ + r₂ ≈ 0.0005 − 0.0009i, so R = 0.0001%: the tiny reflectance of bare silicon. Set the thickness to 0 and read it.
All the bounces: the Airy formula
The last step kept two reflections and ignored the light that keeps bouncing inside the film. Before building on that shortcut, this step checks what it leaves out. The answer explains why EUV mirrors can be reasoned about with simple arrows.
The two-arrow sum left something out. The wave coming back up meets the top surface from below. A small part of it is reflected down again, bounces off the bottom once more, and so on. Each extra round trip multiplies the wave by the same factor.
Adding infinitely many terms, each a fixed factor x times the one before, is called a geometric series. When x is smaller than 1 in size, the sum is finite:
Summing all the bounces this way gives the exact reflection of a single film, the Airy formula:
A quick check of the series: 1 + 0.5 + 0.25 + 0.125 + … has x = 0.5 and adds up to 1/(1 − 0.5) = 2.
In EUV, two arrows are almost the whole story. Each surface reflects about 4% of the amplitude, so a wave that bounces twice more is weaker by a factor 0.0015. The soap film shows the other case: each surface reflects 14% of the amplitude, and the exact curve peaks at 7.7% against 8.0% for two arrows only.
This is why EUV mirrors are simple to reason about, and also why they need so many layers: each surface does very little.
- Where it comes from
- Follow the wave, writing P for ei·2β. The first reflection is r₁. The part that enters, reflects at the bottom and leaves contributes t·t′·r₂·P. Each further round trip multiplies by −r₁·r₂·P, because seen from inside the film the top surface reflects with −r₁. Summing the geometric series, with t·t′ = 1 − r₁², gives r = r₁ + (1 − r₁²)·r₂·P / (1 + r₁·r₂·P). Put both terms over the same denominator: the numerator becomes r₁ + r₁²·r₂·P + r₂·P − r₁²·r₂·P, and the two middle terms with r₁² cancel.
- Assumptions
- One film with flat, parallel, sharp faces. Light of a single wavelength, perpendicular to the film. Nothing else: for this ideal film the formula is exact.
- The terms
- r₁ and r₂ are the Fresnel reflections of the top and bottom surfaces. t and t′ are the transmissions through the top surface, going down and coming up. β = 2π·ñ·d/λ.
- Check
- With d = 0 the formula becomes (r₁ + r₂)/(1 + r₁·r₂). Written out with the three indices, that reduces exactly to (n₁ − n₃)/(n₁ + n₃): the film has vanished and the Fresnel formula for the two outer materials is back. When r₁·r₂ is tiny, the denominator is 1 and the two-arrow sum is recovered.
A film too thick stops working
If one film gives 0.61%, an obvious idea is to use more molybdenum. This step shows why that fails. The failure is the reason the mirror is made of thin layers with silicon in between.
Keep adding molybdenum. Every extra 7.3 nm the round trip gains a full turn and the pattern repeats: peaks where the arrows line up, dips where they oppose. But the bottom reflection has to cross more and more absorbing material.
Past a few hundred nanometres, only the top surface is left. The returning wave fades as e−d/L, with the same L = 167 nm as in module 1.
The lesson for a mirror: more molybdenum does not mean more reflection. To use many surfaces the light must reach them and come back, so the absorbing layers have to stay thin, separated by something nearly transparent. That something is silicon.
- Where it comes from
- In module 1 the intensity fell as e−z/L over a path z, so the amplitude falls as e−z/(2L). The round trip has path z = 2d. The two factors of 2 cancel and leave e−d/L.
- Assumptions
- A uniform film and perpendicular light, as for L itself.
- The terms
- d is the film thickness. L = λ/(4πk) is the attenuation length of the film material. The left-hand side is the length of the second arrow.
- Check
- At d = 3.65 nm the factor is e−3.65/167 = 0.978: the 2% loss quoted two steps ago. At d = 167 nm it is 0.37. Set 167 nm and read 37%.
What one film cannot do, and when these formulas apply
A film reaches 0.61% and stops there. This step says what is still missing, which is what module 3 supplies, and marks the limits of the formulas of this module.
The best single film reflects 0.61%. That is almost four times one surface, and still a hundred times short of a useful mirror.
The way forward is in the arrows. One film lines up two reflections. A stack of films could line up dozens.
One film, two surfaces, a factor of almost four. The same idea repeated is the multilayer mirror: 40 pairs give 80 surfaces whose arrows can be lined up. Module 3 extends the bookkeeping to any number of layers.
Making the film thicker does not help: it has only two surfaces however thick it is, and a thick one hides the bottom surface behind absorption. Stacking does. Two films on a base already have three surfaces, and one application of the Airy formula is no longer enough: the bookkeeping has to be extended.
The Airy formula takes one value of thickness. A film whose thickness varies between 3 nm and 4 nm from place to place can still be described patch by patch, as an approximation, if the thickness changes slowly along the surface.
What to carry forward
- A² = a₁² + a₂² + 2·a₁·a₂·cos φ. Use it whenever two waves of the same wavelength overlap with a fixed phase difference. Amplitudes add, intensities do not.
- Round trip 2β = 4π·n·d/λ. Use it for the phase gained by crossing a flat film twice. At an angle, multiply by the cosine of the angle inside the film.
- r = (r₁ + r₂·ei·2β)/(1 + r₁·r₂·ei·2β). Exact for one ideal film. In EUV the denominator differs from 1 by about 0.2%. It is not enough for two or more films.
To retrace the reasoning
Why can two reflections give four times the intensity of one?
Amplitudes add before squaring: 1 + 1 = 2, and 2² = 4.
Where does 2β = 4π·n·d/λ come from?
One crossing turns the arrow by 2π for every wavelength-inside, λ/n, that fits in d. The reflected wave crosses twice.
Why is the best thickness a quarter of a wavelength?
The bottom reflection starts out flipped. A quarter wave down and a quarter back make half a turn, which flips it again and lines it up with the top reflection.
What does the Airy formula add to the two-arrow sum?
All the further bounces inside the film, summed as a geometric series. In EUV they change the result by about 0.3%.
Why does a thick film reflect like a single surface?
The wave from the bottom surface is absorbed on its way down and back, by the factor e−d/L.