How much EUV light does a single surface reflect?
Lithography scanners use light with a wavelength of 13.5 nm. Steering it takes mirrors, and the material that does the reflecting is molybdenum. A window reflects 4% of visible light. A smooth molybdenum surface reflects 0.16% of EUV light. Eleven ideas explain that number, nine of them with an experiment. Open any of them, in any order.
This is the starting point of the whole project. Everything that follows, the film of module 2, the 40-pair mirror of module 3, its heat in modules 5 and 6, is built from two numbers per material, n and k, and from one formula for one surface. This module introduces them from the ground up, together with the mathematics they need.
First, what a light wave is
Every result in these lessons is a statement about waves that add or cancel. Before any material appears, this step fixes the two ideas they all rest on: intensity goes as the square of amplitude, and one wave can be flipped with respect to another.
Light is an electromagnetic wave: an electric field and a magnetic field that swing together as the wave travels, at right angles to each other and to the direction of travel. The two are tied: in vacuum the magnetic field is the electric field divided by the speed of light, so one of them is enough to describe the wave. These lessons follow the electric field, because in the conditions considered here the electrons of a material respond almost only to it.
Two numbers describe the shape of the wave. The amplitude A is the height of the swing, measured from the centre line, where the field is zero, up to a crest: the field rises to +A at every crest and falls to −A at every trough. The wavelength λ is the distance from one crest to the next: 13.5 nm for EUV, about 500 nm for green light. The figure of the experiment below marks both.
A detector does not follow the field, which swings far too fast. It responds to the energy the wave brings. Three quantities need to be kept apart here. Energy is measured in joules. Power is the energy that arrives each second, in watts. Intensity, I, is the power that arrives on each square metre of the detector, in watts per square metre. These lessons use intensity, because it does not depend on how large the detector is.
The intensity of a wave is proportional to the square of its amplitude. For a wave in vacuum:
The two constants are properties of empty space. ε₀, the electric constant, says how much energy an electric field stores in each cubic metre: a field E stores ½ ε₀ E². c is the speed of light, the speed at which that energy travels. Their values, and the steps from these two ideas to the formula, are in the card below the experiment.
What matters in this step is the ratio between two intensities: for example, how much of the light that arrives at a surface is sent back. Take two waves, one of amplitude A₀ and intensity I₀, the other of amplitude A and intensity I. Divide the formula for the second by the formula for the first, and the constants cancel:
The experiment below does exactly this. The dashed curve is a wave that stays fixed, the reference wave, with amplitude A₀ = 1. The solid curve is the wave you change with the slider. The slider gives its amplitude as a fraction of the reference amplitude, and the large number is its intensity as a percentage of the reference intensity.
Half the amplitude, a quarter of the intensity. The intensity follows the amplitude squared: with the slider at 0.5 the intensity is 0.5 × 0.5 = 0.25, that is 25%. On small amplitudes the effect is stronger. A wave with 4% of the reference amplitude has 0.04 × 0.04 = 0.0016 of the reference intensity, that is 0.16%. This is the case of a single surface of the mirror, which comes later in this module: it sends back 4% of the amplitude of the light that arrives, and therefore only 0.16% of its intensity.
Now press Flipped. The solid wave turns upside down: where the reference wave has a crest, it now has a trough. Its intensity does not change, because a flipped wave swings just as high as before.
Flipping matters when two waves arrive at the same place, because their fields add. Two waves with their crests together make a taller wave: 1 + 1 = 2. A wave and its flipped copy cancel: 1 − 1 = 0. The line «sum with the reference wave» gives this number for the two waves in the figure.
To keep track of this, a flipped wave is given a negative amplitude: −A instead of A. The square is the same, (−A)² = A², so the intensity does not change, while the sum does. You will need this when two reflections meet.
- Where it comes from
- The formula I = ½ ε₀ c A² comes from three facts. The energy stored in each cubic metre of an electric field E is ½ ε₀ E²: proportional to the field squared, just as the energy of a stretched spring is proportional to the stretch squared. A light wave also carries a magnetic field, and that field stores as much energy as the electric one. So the energy of the light in each cubic metre, electric plus magnetic, is twice ½ ε₀ E², that is ε₀ E². At each point the field swings as a sine wave around zero, up to +A at a crest and down to −A at a trough, A being measured from the centre line of the figure, where the field is zero. So E² swings between 0 and A². It spends as much time above A²/2 as below it, and its average over one oscillation is A²/2. This energy travels at the speed c, so each second a square metre receives c · ε₀ · A²/2. The formula I / I₀ = (A / A₀)² follows from it: write the first formula for two waves and divide one by the other.
- Assumptions
- The formula I = ½ ε₀ c A² holds for a steady beam of a single wavelength that travels in vacuum. Inside a material the constant in front of A² is different; a later step of this module, on the reflection at the boundary between two materials, says how. The formula I / I₀ = (A / A₀)² holds when the two waves have the same wavelength and travel in the same medium, as in this experiment: the constant, whatever its value, is then the same for both and cancels.
- The terms
- A is the amplitude of the electric field, the height of its crests, in volts per metre. I is the intensity averaged over one oscillation: power per unit area, in watts per square metre, what a detector reads. ε₀ = 8.854 × 10⁻¹² F/m is the electric constant of vacuum. c = 2.998 × 10⁸ m/s is the speed of light. A₀ and I₀ are the amplitude and intensity of the reference wave.
- Check
- For I = ½ ε₀ c A², a worked case: a field of amplitude A = 1 V/m gives I = ½ × 8.854 × 10⁻¹² × 2.998 × 10⁸ × 1² = 0.00133 W/m², about one thousandth of a watt on each square metre. The units agree too: F/m × m/s × V²/m² is F·V² per second per square metre, and one farad times one volt squared is one joule, so the result is in joules per second per square metre, which is W/m². For I / I₀ = (A / A₀)², use the experiment: a wave with a fifth of the reference amplitude, A / A₀ = 0.2, should have 0.2 × 0.2 = 0.04 of the reference intensity. Set the slider to 0.20 and read 4%.
The real part: how different the material is from vacuum
A mirror reflects at boundaries between materials. To say how much, a material must first be turned into a number. This step introduces the first half of that number, the half that, as long as absorption is left out, decides whether a boundary reflects at all.
To light, every material is a number: the refractive index n. Vacuum has n = 1. Glass, in visible light, has n = 1.5.
What the number means: inside the material the wave keeps its rhythm, but its crests travel at speed c/n instead of c, so they sit closer together or further apart. The wavelength inside is λ/n.
The figure below shows it in two ways. «Wave» draws a single wave as a sine curve along its path, as in the first step. «Crests» draws a whole beam seen from the side, with its crests as lines across it. In both, some crests are marked, so you can follow them. A third view, «Triangle», stops the crests and marks the lengths used by the card of Snell's law, further down in this step. Start with «Head-on»: the crests reach the surface one after the other, at the same rhythm above and below it, but below it they move more slowly, so they bunch up: λ/n apart instead of λ.
Then switch to «At an angle»: the wave turns at the surface and goes on, shorter, in a new direction. «Crests» shows why. One end of each crest reaches the material before the other. That end slows down first, while the other end is still moving fast above the surface: the crest bends at the surface, and the whole beam turns. It is what happens to a row of people walking arm in arm from a pavement onto sand, at an angle: whoever steps onto the sand first slows down first, and the row turns. The ray is the direction the crests move in, always perpendicular to them, so it turns with them.
Move n and watch the reflected ray and the ray that enters. The angles are measured from the dashed vertical, the perpendicular to the surface.
Light reflects only where the two indices differ. With n = 1 the light does not notice the boundary: it goes straight through and nothing comes back.
Notice which way the ray bends too: with n above 1 it moves toward the vertical, with n below 1 it moves away. At 13.5 nm molybdenum and silicon are both below 1.
An index below 1 means the crests move faster than c. The speed c/n is the speed of the crests, called the phase velocity. This does not allow information to travel faster than c.
- Where it comes from
- Each crest is one unbroken line that bends at the surface, as the thick crest of the figure in this step shows. So the point where a crest arriving on one side meets the surface is also the point where the same crest leaves on the other side. Call that point the mark of the crest on the surface. All of this is for a beam that arrives at an angle, as with «At an angle» in the figure: head-on, the crests are parallel to the surface, they do not cross it at separate points, and the ray simply goes straight on. The distance d between two neighbouring marks, measured along the surface, can be worked out from either side, and the two results must agree. Take the side with index n₁. Its crests are parallel lines λ/n₁ apart, a distance measured at right angles to them, that is along the ray. A crest is perpendicular to its ray, and the surface is perpendicular to the dashed vertical, so the crest meets the surface at the same angle θ₁ that the ray makes with the vertical. Now press «Triangle» in the figure of this step. It draws two right-angled triangles on the same stretch of surface, one above it and one below. In the figure the side above is vacuum, n₁ = 1, and the side below has n₂ = n, so the two distances between crests are marked λ and λ/n. Look at the triangle above the surface. It has these three sides: the stretch of surface between two neighbouring marks, of length d; a piece of the first crest; and the shortest path from the second mark to the first crest, of length λ/n₁. The stretch of surface is the longest side, and the side of length λ/n₁ lies opposite the angle θ₁. The sine of an angle is the opposite side divided by the longest side, so sin θ₁ = (λ/n₁) / d, that is d = (λ/n₁) / sin θ₁. The triangle below the surface gives, in the same way, d = (λ/n₂) / sin θ₂: its angle is θ₂ and the side opposite it is λ/n₂. The marks are the same on both sides, so (λ/n₁) / sin θ₁ = (λ/n₂) / sin θ₂. Turn both sides upside down and multiply by λ: n₁ · sin θ₁ = n₂ · sin θ₂. With a different wavelength on each side, the only way to keep the marks equal is a change of direction. This is Snell's law.
- Assumptions
- A flat boundary, much wider than the wavelength, between two uniform materials.
- The terms
- θ₁ and θ₂ are the angles of the two rays, measured from the perpendicular to the surface. n₁ and n₂ are the indices on the two sides.
- Check
- With n₂ = n₁ the two angles are equal: the ray goes straight, as you found at n = 1. The experiment draws the incoming ray at 30°. With n = 1.5 the law gives sin θ₂ = 0.5 / 1.5, so θ₂ = 19.5°: closer to the vertical, as drawn.
A tool you need now: complex numbers
The index of a real material has a second half, and a reflection has a size and a shift. Both need a number made of two parts. This is the only new mathematics of the module, and it returns in every module after it.
The next step needs one piece of mathematics. A complex number is a pair of ordinary numbers written a + i·b, where i is a symbol with a single rule: i² = −1. Picture it as a point on a plane: a steps to the right, b steps up.
The same point can be given as an arrow from the origin: its length, called the modulus |z|, and its angle, called the phase φ. Physics uses complex numbers because one of them carries the two things a wave has: a size and a shift.
One piece of shorthand: eiφ stands for the arrow of length 1 at angle φ, that is cos φ + i·sin φ. Multiplying a number by it rotates the arrow by φ without changing its length. In formulas, angles are counted in radians: a full turn is 360° = 2π, a half turn is 180° = π.
Multiplying by −1 is a half turn. In the shorthand above, eiπ = −1. A wave whose amplitude is multiplied by −1 has its phase shifted by 180°: it is the flipped wave of two steps ago. A complex amplitude generalizes this. Its modulus scales the wave, its phase shifts it by any amount.
- Where it comes from
- Pythagoras. The steps a and b are the two legs of a right triangle and the arrow is its hypotenuse.
- Assumptions
- None. It holds for every complex number.
- The terms
- a is the real part, b the imaginary part, |z| the modulus.
- Check
- 3 + 4i has modulus 5, because 9 + 16 = 25. In the experiment, modulus 1 at phase 90° reads 0 + 1i: one step up, none to the right.
The imaginary part: how much the material absorbs
Molybdenum and silicon both absorb EUV light. That absorption decides how thin the layers must be, how many pairs are useful and, in module 5, how much heat the mirror keeps. This step introduces the number that measures it.
The full index is a complex number, ñ = n + i·k. Here is what the second piece does. A wave that travels a distance z through a material gets multiplied by ei·2π·ñ·z/λ. Read the exponent piece by piece: z/λ counts how many wavelengths fit in the distance, each one is a full turn of 2π, and the index stretches or shrinks the count. Now split the index into its two parts:
The first factor is an arrow of length 1 that only rotates: the wave oscillating, with wavelength λ/n. In the second factor the i is gone, because i · i = −1. What is left is an ordinary number that shrinks as z grows. The imaginary part of the index turns oscillation into decay.
The wave is an arrow too, as in the previous step: its length is the amplitude, its angle the phase. Each of the two factors multiplies that arrow. The first rotates it without changing its length, so it leaves the amplitude as it was. The second is an ordinary positive number, and multiplying an arrow by an ordinary positive number makes it longer or shorter without turning it: the arrow keeps its angle, and its length, the amplitude, is multiplied by that number. So after a distance z the amplitude is A · e−2π·k·z/λ, where A is the amplitude at the start. The crests are still λ/n apart, but they are lower.
Intensity goes as the amplitude squared. The square of ex is ex · ex = e2x, so the exponent doubles: the intensity is multiplied by e−4π·k·z/λ. The distance after which 37% of the intensity is left is called the attenuation length L.
The larger k is, the shorter the distance light travels. It is an inverse proportion: twice the k, half the L.
For molybdenum at 13.5 nm, k = 0.00644, so L = 13.5 / (4π × 0.00644) = 167 nm.
What an exponential means in practice: every stretch of length L multiplies the light by the same factor, 0.37. After 2L there is 0.37 × 0.37 = 14% left, after 3L 5%. Equal steps remove equal fractions. That is the signature of absorption.
- Where it comes from
- The amplitude decays as e−2πkz/λ. Squaring it for the intensity gives e−4πkz/λ. Giving the distance λ/(4πk) the name L turns that into e−z/L. So the 4π is 2π from the wave's phase, times 2 from the square.
- Assumptions
- A uniform material, light of a single wavelength, and a beam weak enough that the material treats every photon the same way.
- The terms
- λ is the wavelength in vacuum. k is the imaginary part of the index. z is the depth inside the material. I₀ is the intensity just inside the surface.
- Check
- Units: λ is a length and k is a pure number, so L is a length. Limit: k = 0 gives an infinite L, no absorption, as in vacuum. Numbers: silicon has k = 0.001826, so L = 13.5 / (4π × 0.001826) = 588 nm, the value the next steps display.
The two real materials: molybdenum and silicon
With n and k defined, this step reads their real values for the two materials of the mirror. These tabulated numbers are the only measured input of the optical modules: every later reflectance is computed from them.
These are tabulated optical constants from the CXRO database, interpolated here at the wavelength you choose. Move the wavelength and read n and k for Mo and Si.
Each one has a job. Molybdenum differs from vacuum, so it is the layer that reflects. Silicon has n ≈ 0.999 and absorbs little: it acts as a nearly transparent spacer.
At 13.5 nm both materials have n just below 1, while glass in visible light has 1.5. Here is a picture of how an index can end up on either side of 1. The electric field of the light pushes the electrons of the material back and forth. Electrons that move back and forth send out a wave of their own, and it adds to the wave that arrives. The index describes the result of that sum. An electron is held by its atom somewhat as a weight is held by a spring: left alone, it would swing at a rhythm of its own. When the light swings more slowly than that rhythm, the electron follows the push, and the wave it sends out delays the crests: n is above 1. This is glass in visible light.
Now let the light swing much faster than the rhythm of the electron. Shake quickly the hand that holds a spring with a weight: the weight cannot keep up, and it moves down while the hand moves up. In the same way the electron moves opposite to the push of the field. The wave it sends out now moves the crests ahead instead of delaying them, and that pulls n below 1. Far above its own rhythm the electron barely moves, so the pull is small. A real material has many groups of electrons, each with its own rhythm. At a given wavelength some of them pull n down and others push it up. The picture gives the direction of each contribution, not their total. The total is what the tables record: for molybdenum and silicon at 13.5 nm it comes out just below 1.
Drag the slider all the way left. Below 12.45 nm the k of silicon quadruples: there the photons have enough energy to pull out a deeper group of silicon electrons, and absorption jumps. EUV works at 13.5 nm to stay on the good side of that edge.
How much material can EUV light cross?
The values of k just read have a direct consequence for the design: they say how much material the light can cross. That rules out lenses, and it sets the thickness below which a layer costs almost nothing.
Now use the real k values at 13.5 nm. Pick the material, change the thickness and see how much light comes out the other side.
Only attenuation inside the material counts here. Reflections at the two surfaces are neglected.
It takes about 400 nm of silicon to halve the light, and only 116 nm of molybdenum. A lens is millimetres thick, thousands of times more: nothing would come out. That is why there are no EUV lenses and the whole beam path is in vacuum. Only mirrors are left.
There is good news too. Switch back to silicon and set the thickness to 4 nm: more than 99% gets through. Molybdenum at 4 nm still passes 97.6%. A thin layer costs little, so many of them can be stacked.
- Where it comes from
- The formula T = e−d/L is the result of the step on absorption, read at the far face of the slab. That step found the intensity at a depth z inside a material: I(z) = I₀ · e−z/L, where I₀ is the intensity that enters. A slab of thickness d ends at z = d, so the intensity that comes out is I₀ · e−d/L. T is the fraction of the light that comes out: the intensity that comes out divided by the intensity that enters. Dividing by I₀ leaves T = e−d/L. Now a question about this formula: which thickness lets out exactly half of the light? The answer is the second formula, d½ = L · ln 2, and it takes a few steps to find. Half of the light means T = 1/2, so e−d/L = 1/2. A minus sign in the exponent means one divided by: e−d/L is 1 / ed/L. So 1 / ed/L = 1/2, that is ed/L = 2. In words: d/L is the power to which e must be raised to give 2. That power has a name, the natural logarithm of 2, written ln 2, and its value is 0.693, because e0.693 = 2. So d/L = 0.693, that is d = 0.693 · L. This particular thickness is given the name d½.
- Assumptions
- The formula T = e−d/L holds under the assumptions of the step on absorption: a uniform material, light of a single wavelength, and a beam weak enough that the material treats every photon the same way. It adds two. First, the light crosses the slab at right angles to its faces: only then is its path inside the material as long as the thickness d. Second, the light sent back by the two faces of the slab is not counted, as if all the light that arrives entered and all the light that reaches the far face came out. A later step of this module shows that at 13.5 nm, for light that arrives at right angles, one face sends back 0.16% of the light or less, so the error is small. The formula d½ = L · ln 2 adds no assumption: it is the first formula solved for d.
- The terms
- d is the thickness of the slab. L is the attenuation length of the material, from the step on absorption: at 13.5 nm, 588 nm for silicon and 167 nm for molybdenum. The experiment shows it as «L of the material». T is the fraction of the light that comes out, a number between 0 and 1; the experiment shows it as a percentage. e is a fixed number, 2.718…, the base of the exponentials of this module. ln 2 = 0.693 is the power to which e must be raised to give 2. d½ is the thickness that halves the light.
- Check
- For T = e−d/L, two limits and a worked case. With d = 0 the formula gives e0 = 1: no material, all the light comes out. With d = L it gives e−1 = 0.37: the 37% that defines L. For 4 nm of molybdenum, d/L = 4 / 167 = 0.024 and T = e−0.024 = 0.976: the 97.6% of the experiment. For d½ = L · ln 2: silicon gives 0.693 × 588 nm = 408 nm, the «about 400 nm» of the text above, and molybdenum gives 0.693 × 167 nm = 116 nm. Press «Mo, 116 nm» in the experiment and read about 50%. Twice that thickness, 232 nm, leaves half of a half, that is 25%.
Where the Fresnel formula comes from
So far reflection has been a fact: it happens where the indices differ. A design needs the amount. This step derives it, and the two conditions it uses come back in module 3 as the basis of the matrix method.
You have seen the fact: light reflects only where the two indices differ. Now the reason, and the amount. Send a wave of amplitude 1 perpendicular onto a boundary. A reflected wave of amplitude r goes back, a transmitted wave of amplitude t goes on. Two things must match at the boundary.
The electric field. The part of the electric field parallel to the surface cannot jump across it. On the incoming side the field is the sum of the incoming and reflected waves, on the far side it is the transmitted wave.
The magnetic field. Every light wave carries a magnetic field alongside the electric one. Its part parallel to the surface cannot jump either, provided the two materials are non-magnetic and no current flows along the surface. In a plane wave the magnetic field is proportional to n times the electric field, and it changes sign when the wave travels backwards. So the reflected wave enters with a minus sign.
The same condition can be read as a statement about slope: measured along the perpendicular to the surface, the electric field must be equally steep on both sides. It is one condition seen in two ways, not an extra law.
Two equations, two unknowns. Put t = 1 + r from the first into the second: n₁ − n₁·r = n₂ + n₂·r. Move the terms with r to one side: n₁ − n₂ = r · (n₁ + n₂). Divide.
r is the reflected fraction of the amplitude, and it can be negative. R is the reflected fraction of the intensity, the one a detector measures.
Try it on a window. Air onto glass has n₁ = 1 and n₂ = 1.5, so r = (1 − 1.5)/(1 + 1.5) = −0.2 and R = r² = 0.04.
4%: the reflection you see on a window. The minus sign of r says the reflected wave is flipped. It disappears in the square, but two steps from now it takes the lead.
One more check, on energy. The first equation gives t = 1 + r = 0.8. The wave inside the glass carries n₂/n₁ times its amplitude squared, so the transmitted fraction is 1.5 × 0.8² = 0.96. Reflected plus transmitted: 0.04 + 0.96 = 1. Nothing is lost at a boundary between two transparent materials.
- Where it comes from
- Two matching conditions at the boundary, one for the electric field and one for the magnetic field, solved for r.
- Assumptions
- Light arriving perpendicular to the surface. A flat, perfectly sharp boundary. Two uniform, non-magnetic materials. A single boundary: nothing behind it sends light back.
- The terms
- n₁ is the index on the side the light comes from, n₂ the index on the far side. r is the reflected amplitude relative to the incoming one; its sign, or its phase when the indices are complex, says how the reflected wave is shifted. R is the reflected fraction of intensity.
- Check
- Equal indices give r = 0: nothing reflects, as in the experiment with n = 1. A very large n₂ gives r close to −1: a perfect mirror that flips the wave. Swapping the two materials changes the sign of r and leaves R the same.
Molybdenum at 13.5 nm reflects 0.16%
Formula and data are both in hand. This step puts the index of molybdenum into the Fresnel formula and gets the number that sets the problem for all the modules that follow.
With complex indices the formula stays the same, and the arithmetic is done with complex numbers. For vacuum onto molybdenum, ñ = 0.9238 + 0.0064i:
Then the modulus squared, as in the complex-number step: R = 0.0396² + 0.0035² = 0.0016, that is 0.16%. Twenty-five times less than a window.
Each dot is one photon out of 10,000 arriving. The lit ones bounce back.
At 13.5 nm the index of molybdenum is close to 1, and that of silicon is closer still. The Fresnel numerator, n₁ − n₂, is small, and squaring makes it tiny.
Your estimate without k gave 0.156% against 0.158% for the full calculation. At this wavelength the reflection of molybdenum comes from its n; absorption adds almost nothing.
A note on the data, at 6° and s polarization. This lesson interpolates the CXRO tables at exactly 13.500 nm and gives 0.1618%. Taking the nearest tabulated point instead, at 13.4948 nm, gives 0.1615%. Both are legitimate readings of the same tables; the choice moves the result in the fourth digit.
Direction matters: Si to Mo, or Mo to Si
Inside a mirror, boundaries come in two kinds: silicon onto molybdenum, and molybdenum onto silicon. Whether their reflections help or fight each other depends on a sign. Module 2 is built on it.
Inside the mirror, light does not meet vacuum: it goes from silicon to molybdenum and back, dozens of times. Swap the order of the two layers and watch r.
Same intensity, opposite sign. It follows from the formula in one line: swapping the materials swaps n₁ and n₂, which reverses the numerator and leaves the denominator alone.
An opposite sign means the reflected wave starts out flipped: a phase shift of 180°, the half turn of the complex-number step. In a layered mirror half the surfaces reflect upright and half flipped. Added as they are, they would cancel. The layer thickness is what brings them back into agreement. That is the subject of module 2.
What if the light arrives at an angle?
In a scanner the light does not arrive perpendicular to the mirrors, and the mirror of module 3 is computed at 6°. This step shows what an angle changes, and why tilting the mirror is not the way out of the 0.16%.
In scanners the light hits the mirrors at about 6° from perpendicular. At an angle the same derivation goes through with one change: only the motion of the wave perpendicular to the surface matters, so each index gets multiplied by the cosine of the angle in its own material.
The result depends on the polarization, the direction in which the electric field swings. Take the plane that contains the incoming ray and the perpendicular to the surface: the plane of incidence. In s polarization the field swings perpendicular to that plane, so it stays parallel to the surface. In p polarization it swings within that plane.
Move the angle on molybdenum at 13.5 nm. The two curves are the two polarizations: s solid, p dashed.
Near grazing, even EUV reflects well. This is the principle of grazing-incidence mirrors, used in collectors and X-ray telescopes.
Near 43° the reflection of p-polarized light reaches a very small minimum; absorption prevents it from vanishing completely.
Forming a sharp image over a wide field, however, takes mirrors hit almost head-on. There, between 0° and 6°, R stays pinned around 0.16%.
- Where it comes from
- The same two matching conditions as before, written for a wave that crosses the boundary obliquely. Snell's law supplies the angle θ₂ inside the material.
- Assumptions
- A flat, sharp boundary and one polarization at a time. The p formula is the s formula with the two indices exchanged in front of the cosines.
- The terms
- θ₁ is the angle of the incoming ray from the perpendicular. θ₂ is the angle of the ray inside, which becomes a complex number when the material absorbs.
- Check
- At θ₁ = 0 both cosines are 1 and the s formula becomes the perpendicular one. At normal incidence both polarizations give the same reflectance; with the p formula written above, the amplitude has the opposite sign. Two landmarks of the chart can be predicted by hand from the simpler case of a real index, with no absorption. The bend: with n = 0.924, Snell's law has no real solution once sin θ₁ exceeds n, that is beyond 67.5°, and without absorption all the light would be reflected from there on. Absorption smooths that transition: at 67.5° molybdenum reflects about 46% for s and 40% for p, rising toward 100% nearer grazing incidence. The dip of p: without absorption it would fall where tan θ₁ = n, at 42.7°.
The problem to solve, and when these formulas apply
One surface gives 0.16%, and a mirror must give tens of percent. This step states that gap as the problem the next two modules solve, and marks the limits of what has been built so far.
One surface reflects 0.16% in intensity, that is |r| ≈ 0.04 in amplitude. A real mirror has 40 Mo/Si pairs: 80 surfaces.
If all 80 reflections added up in agreement, the amplitude would be 80 × 0.04 = 3.2 and the reflectance would exceed 1000%.
That is impossible, and what prevents it is light itself running out: deep layers receive less, because part has already been reflected and part absorbed. The Fresnel formula keeps holding at every surface, buried ones included.
The sum saturates. The nominal model, an ideal stack of 40 pairs with perfectly sharp and smooth interfaces, reaches 74.2%. That number comes from a transfer-matrix calculation, the method the next modules build up to. For the same stack design and illumination conditions, interface imperfections can reduce the specular reflectance, the light in the main reflected beam, relative to this ideal model. Roughness scatters some light away from that beam, while mixing between Mo and Si makes the change in refractive index less abrupt. Absorption is already included in the nominal model.
The single-surface formula is the building block. What it cannot do alone is add reflections from several surfaces: for that you need the phase each wave picks up while crossing a layer.
Take a molybdenum film 3 nm thick on a silicon wafer. The film absorbs under 2% of the light, so its lower surface is fully lit and reflects too. The formula holds at each surface, and what is missing is how to add the two reflections.
The formula also assumes a flat, sharp boundary. Bumps 2 nm high are not small against a wavelength of 13.5 nm, and a surface that has them does not meet that assumption.
What to carry forward
- I(z) = I₀ · e−z/L, with L = λ/(4πk). Use it for light crossing a uniform material. It is not enough for a layer so thin that reflections between its two faces matter.
- r = (n₁ − n₂)/(n₁ + n₂), R = |r|². Use it for one flat, sharp boundary hit perpendicularly; at an angle, add the cosines. It is not enough for a rough surface or for several boundaries at once.
- Swapping the materials flips the sign of r. True at every single boundary, and the key to stacking layers.
To retrace the reasoning
What do n and k represent?
n sets the speed of the crests in the material, and so how much the material bends and reflects light. k sets how fast the wave decays: how much the material absorbs.
Where does the 4π in L = λ/(4πk) come from?
2π from the phase of the wave, and a factor 2 because intensity goes as the amplitude squared.
Which two conditions give the Fresnel formula?
At the boundary the electric field must be continuous, 1 + r = t, and so must the magnetic field, n₁(1 − r) = n₂t. Solving the pair for r gives the formula.
Why does EUV work at 13.5 nm and not at 12 nm?
Below 12.4 nm silicon starts absorbing much more. At 13.5 nm it is nearly transparent and acts as a spacer.
Why are there no EUV lenses?
Light is halved in about 400 nm of silicon and in 116 nm of molybdenum. A lens is thousands of times thicker.
How much does a molybdenum surface reflect, and why so little?
0.16%. Its n is only 0.076 away from 1, and Fresnel squares that difference.
What is the difference between Si→Mo and Mo→Si?
Same R, but r has the opposite sign: the reflected wave starts out flipped.