The Saga of Light, Part 3: Invisible Bones, Invisible Beads and Why Mirrors Aren’t White

Welcome back to our ongoing series of posts on light. Now, I’m going to be honest and say that this entry may, for some, be a little drier. We’re at a very natural point in our “Saga of Light” where we can start discussing what happens when light hits something like a pane of glass and, by doing so, moves from one medium to another. And the reality is that that is probably not the “sexiest” topic under the Sun. However, it’s also a subject that leads to a lot of misconceptions and confusion and that’s why I feel it needs to be talked about. If you’ve ever wondered about any of the following then this post is aimed at you: 1) why does the wavelength of light change in different media but its frequency does not?; 2) why does light reflect at all?; and 3) how is light different (or not different) from other types of waves?

Of course, I’ve managed to sneak in a few “cooler” topics that relate to what we’ll learn. For example, we’ll learn about how you can take something like a bone or mouse spleen (ewww) and make it transparent:

We’ll also talk about how those “invisible beads” work that you can buy at arts and craft stores and that disappear when submerged in water. Finally, we’ll correct a rather alarmingly common set of misconceptions (that are literally everywhere on the internet) over what is the difference between a mirror and something that is white. (Honestly, I don’t know if I found a single top hit on Google that gave a correct answer).

Light Ain’t So Special

I think it’s a very common thing when learning about the properties of light to get confused about what specifically is a property of light, an electromagnetic wave, and what is a property of waves in general. It’s something I see quite often where people get it into their heads that because this bizarre Frankensteiny thing (I know, I know, Frankenstein’s monstery thing) – with its oscillating electric field and perpendicular magnetic field – does a thing that that must be something very special about light. However, it’ll turn out that in reality what they’re talking about is something that, say, ocean waves do too, as do sound waves and seismic waves and so on. This point can be made pretty concretely by just taking a brief dive into the history of humankind’s understanding of light.

Starting in the mid 1650s, physicists started to become quite enamored with the idea that light was made of little billiard-ball-like particles, which they called “corpuscles”,which – for me – is right up there with the word “moist” in terms of words that just feel unpleasant to say. It was a corpuscular theory of light. CorPUScular. CorPUUUSSSScular… Nah, doesn’t work for me. Anyways, there were dissenters, like Dutch physicist Christiaan Huygens, but on the corpuscular (corPUSSScular) side, you had heavy-weights like Renee Descartes and Sir Isaac Newton.

Actually another big name against a corpuscular theory was the physicist Robert Hooke, who, when Newton suggested his ideas at the Royal Society in the UK, was so vehemently against them that Newton postponed publishing his work on light until after Hooke died… about 30 years later. So, ‘atta boy Newton, if you can’t refute the man you can at least out-live the jerk! I’m being glib of course, Robert Hooke’s contributions to physics are many and both he and Newton were deeply unlikable people whose pettiness and feuds certainly makes the history of 17th century physics an interesting read. My point being that I wouldn’t waste too much emotional energy feeling bad for Newton, he did plenty of similar things to other people.

Now, of course, Newton was no dummy – he is undoubtedly one of the greatest (and weirdest, like you seriously have no idea) minds of all time – and there were very good reasons why he came to this conclusion; that light was made of corpuscles. And it did improve upon the ideas that came before him. However, a century or so after his death, the tides started to shift away from a corpuscular theory.

The source of this change lay largely on the shoulders of the experimental work of British polymath Thomas Young followed by later work by French civil-engineer-turned-physicist Augustin-Jean Fresnel. Now, when I say Thomas Young was a polymath, I am not using that term lightly. He was a medical doctor. A linguist who spoke, at least, 7 languages. He was an accomplished Egyptologist who made contributions to the deciphering of the Rosetta stone. He produced work on the topic of musical theory, on solid mechanics (Young’s modulus), and of course on light. There is a biography about him that is literally called “The Last Man Who Knew Everything”. So, you know, you’re welcome for your new feelings of shame and inadequacy.

Now, Young’s work was first but I’ll leave that for your own curiosity to explore, but what I really want to talk about is Fresnel’s. He did quite a few things regarding light, but the specific thing I want to talk about is how he was able to derive precise mathematical statements dictating the fraction of light that is reflected off a boundary between two mediums.

He did this by assuming light was a wave. And he did it in 1821. It wasn’t until 1865 that James Clerk Maxwell discovered that light was an *electromagnetic* wave. In other words, Fresnel was able to work out how light reflects based solely on the assumption that it was “some” kind of wave, before anyone had any idea what actually was “waving”. My point in this historical aside is that Fresnel didn’t need to know anything about the physics of electromagnetic fields to understand the reflection of light, as it is an example of a property of waves in general. Put another way, the reason that you can see your reflection in a pane of glass isn’t because light has special properties but rather that is the natural result of it being a wave at all.

Let’s take a look at what he did. As I said, reflection of light may not seem like a super interesting topic but it will help us a lot in understanding how light communicates to us the nature of materials and objects around us. This understanding will ultimately lead us to some very interesting places moving forward.

Would A Wave By Any Other Name Not Do The Same Things?

To get the most out of these concepts going forward, I think I’m just going to lay out what I want to address and motivate in this post upfront before we move on in our discussion of light. There are going to be a few detailed discussion coming up so I think it’s helpful to lay down “the forest” before we go staring at some trees like some sort of dendrophiliac (why did I say that???)

So let’s see the forest; I specifically want to motivate the ideas that: 1) all waves carry energy, light included, 2) the speed of waves change when moving from one material (or medium) to another, 3) because of this, one cannot avoid the fact that some amount of the total intensity of the wave must always be reflected at the boundary and the remaining intensity will transmit through into the second medium and 4) as a result of the change of speed, at an interface the wavelength of the wave changes in different media but the frequency remains the same.

To start out, let’s ignore light and instead talk about a different type of wave: a skipping rope held taut. If you pluck such a rope or flick one end of it, you can send a single “wave pulse” down the length of it. One could also imagine producing a continuous wave, travelling down the length of the rope by flicking one of the ends up and down in a continual, oscillatory and periodic manner:

If I have someone else holding the other end of the rope stationary, as the wave pulse or sinusoidal travelling wave reaches them, it’s going to apply a force on their hand and try and move it up and down accordingly. In other words. the arrival of the oscillation will try to set your hand into motion. What this is telling us is that waves carry energy. This is probably not news to you, and is pretty clear for anyone who’s ever been knocked over by an ocean wave or felt intense pain in their ears from sound waves (like when you go to a *insert band you don’t like* concert!)

Things are, of course, no different for light. This is pretty obvious in everyday life, so I’m not planning to spend much time motivating this. Sunlight causes your skin to burn, but it can also power your home with solar panels and provide plants with the energy they need to grow. Lasers can cut metal and seriously hurt your eye when Martha in Mrs. Steeles’ 3rd grade class brought in a laser pointer for show-and-tell and hit ya right in the face with it (I mean, hypothetically speaking).

For our skipping rope at least, I probably don’t have to motivate the idea that the energy delivered every second (i.e. the power delivered) by a wave is proportional to its amplitude. In other words, how large your flicks are:

This too is true of light, its energy depends on its amplitude, which is the maximum value the electric field takes.

This is a note for those with a stronger physics background. This discussion is subtler than it may initially seem. The energy of a skipping rope actually depends on the amplitude and the frequency. However, the power of an EM wave (dictated by Poynting’s theorem) depends only on the maximum amplitude of the electric field, and has no frequency dependence. So this would suggest that maybe a skipping rope is very different than an EM wave. However, one can also go one step deeper and can actually re-derive Poynting’s theorem in terms of a vector potential A, rather than the fields E and B. Poynting’s theorem expressed in terms of a vector potential is actually dependent on $\omega^2 |A_0|^2 c$, where $\omega$ is the frequency and $c$ is the speed of the wave, just like our wave on a rope, and $A_0$ is the amplitude of the vector potential. From this perspective things look much more like our wave on a string whose energy is proportional to $v \omega^2 |y_{max}|^2$. So it’s interesting that thinking of EM waves in terms of an oscillation of the vector potential rather than the E and B fields leads to a more natural analogy with waves on a string.

Now let’s compare what happens if I have two ropes of different thicknesses and repeat our little game of “flick the skipping rope” (that could use a better name…). By having a thicker rope, effectively the rope is more resistant to being set into motion. This is exactly like trying to push a heavy object, because using a thicker rope means that the mass/weight of the rope per unit length is increased. When you push something, the heavier it is the less motion you get out of it for the same strength of push. So if you do this with a thicker rope, but keep the rope equally as taut (i.e. same tension in the rope) you will find that the wave moves slower.

The same is true for light. Light too can travel in a medium, like glass. Air, actually, is also a medium, though it’s so omnipresent in everyday life that we tend forget it’s there – that it’s an actual gas of stuff – but it is. And thus light can also move slower in an optically “heavier” (whatever that means) medium. The heaviness of a medium, in terms of light waves, is defined by something called the “index of refraction” but I think I’ll often just keep calling materials with high indices of refraction: “heavy-for-light” materials. I think it draws the mental picture back to this string example in a useful way.

Now, caveat time. (Which, I know, is no one’s favourite time) It’s actually, simultaneously, both very illuminating (puns!) and very treacherous to compare light to a, so-called, “elastic” wave like our skipping rope. Illuminating because, as I said, much of how light behaves is exactly the same, and we’re just seeing general wave properties. However, it’s treacherous because an elastic wave exists in a medium, like our skipping rope, but although light can exist in a medium as well, like glass or air, it can also exist in a vacuum. In fact, one of the most profound realizations of early 20th century physics was that light requires no medium. People thought it did. They thought the vacuum itself was secretly just another special type of medium which they called the luminferous aether, but it isn’t and light has no medium and that has profound consequences (time dilation, disagreements on simultaneity of events and even ordering of events, and other things). However, that would be a story for another time. For us, it’s okay to think of light in terms of our elastic wave.

Now, let’s look at one final phenomenon and consider a very funny type of rope: one where half of it is thin (i.e. light weight) and half of its is thicker and there’s a boundary or interface between the two “regions”:

and let’s say I take the end of the lighter side and start flicking the rope back and forth to make a sine wave. This, by the way, was pretty much exactly how Fresnel thought about light impinging on a boundary between materials (like light passing from air to the glass of a prism). He didn’t know about the electric or magnetic fields, all he assumed was that light was some kind of “transversely-polarized” wave. A transverse wave, like light or our string wave, is one where the “wiggling” is happening in a plane perpendicular to the way the light is moving. The opposite of a transverse wave is a longitudinal wave, like sound in air, where there isn’t “wiggling” up and down (or for an EM wave, an oscillation of the direction of the electric and magnetic fields in the perpendicular direction, remember, an EM wave isn’t physically moving up and down) but rather areas that are compressed and rarefacted (the fancy schmancy word for “less compressed than average”):

Fresnel’s treatment was a bit more complicated than ours, of course. He considered such a wave coming at a material at any angle, and also worked out how the angle of polarization and the angle of incidence were related. Like in our skipping rope, if the rope is laid in the x-direction we’ve assumed that the up-and-down-ness is only in, say, the y-direction. But in 3D it can really be any kind of ripple in the y-z plane. That wouldn’t matter when we only think about light hitting straight-on, but if it comes in at an angle it matters for what happens. But we’ll ignore all of that and just think of light coming at the interface straight-on. So we have light, which is a transverse wave, impinging on a boundary between two media/materials and we know that the wave has a different speed in the two materials. The question is: What happens at the interface?

Well, what happen at the boundary between the thin and thick rope in our “elastic” wave? The important thing, if you look into the physics, is that the rope can’t “kink”. Both its height and its slope have to be the same at the boundary. That’s basically the constraints that “the laws of physics” puts on the situation. If it “kinked” then energy and momentum would not be conserved, which is a big no-no in physics. Look at this diagram closely if you don’t understand what I’m saying:

If you’re suspicious of where these requirements come from, it’s maybe best to just take them by fiat. When you actually go into the math and the real physics (i.e. the equations of force and movement and disturbing a media and such) you’ll find that when a physical system can allow for “waves” (i.e. the surface of water, the pressure of air, the electric fields in space, etc.) that they can’t “kink” anywhere. That is, unless there’s something else present in the system that could steal energy and momentum, but that isn’t the case here.

So neither the height nor the slope can change at the boundary. Let’s consider both of these requirements separately, starting with the “height” having to be the same. For light this means that the electric and magnetic fields can’t suddenly change at the boundary. It’s not obvious, so don’t kick yourself if this doesn’t seem clear, but this actually enforces the fact that that when a wave moves from one medium to the next, where the speed of the wave is different in the two mediums, its wavelength can and will change but its frequency will not.

Take a look at this diagram:

It imagines a wave working its way from one medium (the white) to another (orange) under the assumption that I’m wrong about this, that the wavelength can remain the same and everything will be fine. And our guiding reality is the fact that the speed of the waves is different in the two media. Looking at a specific moment on the left, it might look like things are fine: no kinks. But if that’s the situation at one moment take a look what must happen a moment later, because the waves have different speeds. The light dotted blue shows where the wave was (i.e. the wave on the left, for reference) and then we see what the wave must look like now if the wave in the orange medium is moving slower than the one in the white. You get a big ‘ole kink.

I’ll draw for you what it really does look like in a moment after we consider our second condition. But let’s just re-iterate this crucial result: if the speed of a wave is different in different media, the wavelength of the wave will change as it moves from one to the other, but the frequency will not.

This actually creates a problem in terms of how we talked about waves in previous parts of this blog where I claimed that one can talk about frequency and wavelength in an interchangeable way. In reality, because the speed of waves is different in different media, the wavelength is also different. So previously when I said thing like “red light has a wavelength of 700 nm”, that’s being almost dangerously lazy (hey, I like being lazy!) as the wavelength of red light is different in different mediums. However, whenever I say “wavelength of blah nanometers” I always implicitly mean “in a vacuum”, in which case we know the medium and there is no ambiguity.

So let’s look at the second condition, that the slope of the wave can’t change at a boundary. Well, they say that a picture is worth a thousand words. Well, I made 3 gifs for you, so that’s gotta be worth, like… a million words? Is it a thousand words per frame? Or just some general multiplicative factor based on pedagogical value? Or…??? Annnyyywaaayyys. Hopefully looking at these gifs will make things much clearer than any verbal description. The first gif:

shows what happens if you do the math right, so that the wavelength of a wave changes at a boundary but the frequency does not. We see that it is possible to make it so that there is no kink in the height by doing this. Of course Fresnel figured out the actual formula for how the wavelength needs to change and that’s the formula I plugged in to make that gif, but we’ll stick with pictures.

So by having the wavelength change we can avoid a discontinuous kink in the height but clearly there is still a kink in the slope. Well, it turns out there is no way to fix this… unless you allow the amplitude of the wave to change at the boundary. As a first step toward fixing things you can demand that the wave in the denser material is of less intensity relative to the incoming wave.

However, the whole reason we need to worry about kinks is because they imply that energy or momentum is not conserved, so if we allow the amplitude of one of the waves to change then clearly we’re breaking this rule, intensity “went” somewhere. But there is another save. You can add a second wave reflecting back off the boundary. Now let me throw down this gif but it’s going to initially look like madness:

So now I have three waves of different intensity/amplitude: the income wave (blue) heading to the right; the transmitted wave in the medium (purple) that continues to the right but with a reduced intensity; and a reflected wave (red) heading to the left which also has an intensity different from both the incident and transmitted waves. Looking at this you probably think I’ve gone crazy. None of the waves line up any more and there’s all kinds of kinks. However, what Fresnel showed is that if they all have the right intensities, it will all work out. Let me simply add the incident and reflected waves and show you this last gif:

No kinks! Neither in height nor slope. If you have an incident wave of intensity I, and then a reflected and transmitted wave, whose total intensities add up to I ( i.e. R+T= I, where R and T are the intensities of the reflected and transmitted waves respectively), you can make it all work. And Fresnel showed this was the only way it could work and gave the exact equations for how R and T were related for a given medium.

Of course, looking at that gif things look a little funny. I have a travelling wave moving to the right but the final wave kind of look like it’s not moving at all, it’s more like a “standing-wave” held in place. This is because we’ve only discussed the situation of light impinging head-on, rather than at the angle and when you add up two sine-waves moving in opposite directions, but with the same wavelength, you really do get standing-waves:

and that’s what we’re seeing. The right moving incident wave plus the left moving reflected wave, because they’re on the same line of travel, add up to make an *almost* standing-wave. I say “almost” because for it to be a perfect standing-wave the intensities of the incident and transmitted waves would have to be the same but here the reflected wave is of lower intensity and thus they add up to make an “almost” standing-wave.

However, if we had light coming in at an angle then the reflected wave would not overlap with the incident wave and you’d have two separate travelling waves.

A reader with some familiarity with physics might find my claim suspicious that all you need to know to determine Fresnel’s equations is that the speed of an EM wave changes (according to $v = c/n$) in different media and that there are no discontinuities in the intensities/amplitudes and their first derivative at the interface. It’s likely if you’ve ever seen a derivation of Fresnel’s equations it was through the full machinery of Maxwell’s equations with $E$s and $B$s and maybe even polarization $P$s and displacement fields $D$. A very complicated calculation!

So let me prove it to you here for the simple case of p-polarization and also derive Snell’s (or Snellius’) law at the same time for fun. By p-polarization we mean that, if the interface is in the y-direction at x=0 and the direction of motion of the waves is in the x-y plane then the polarization is in the z-direction. Let’s say I have an incident wave of amplitude $E_i$ (which, remember, is entirely in the z-direction) moving in a direction given by the $k$-vector $k = (k_x,k_y)=(|k|\cos\theta,|k|\sin\theta)$. I also have a reflecting wave of amplitude $E_r$ with a $k$-vector that is the same angle as the incident wave from the x and y axes (angle of incidence equals angle of reflection, though we could prove that too here, but I’m trying to be brief). Thus its $k$-vector is the same as the incident except mirrored over the y-axis: $k' = (-k_x,k_y)=(-|k|\cos\theta,|k|\sin\theta)$. Finally we have a transmitted wave of amplitude $E_t$ who has an unknown $k$-vector, $k'' = (|k''|\cos\alpha,|k''|\sin\alpha)$ which is moving with an unknown angle (given by $\alpha$).

So our general wave equation is:

$E_i \cos(k_x x + k_y y - \omega_i t) + E_r \cos(-k_x x + k_y y - \omega_r t) = E_t \cos(k_x'' x + k_y'' y - \omega_t t)$.

In order to meet out first continuity condition we must have the left side equal the right side at x=y=0 and it must be equal at any and all times. Thus we have:

$E_i \cos(- \omega_i t) + E_r \cos(- \omega_r t) = E_t \cos(- \omega_t t)$

Since this equation must be true always it immediately follows that $\omega_i = \omega_r = \omega_t = \omega$ and we recover that frequency can’t change across the interface. Furthermore, since the frequencies must be equal, we also immediately recover $E_i + E_r = E_t$.

If we now take a derivative with respect to x and take the case of x=y=0 we get the following set of equations:

$k_x E_i \cos( - \omega t) - k_x E_r \cos( - \omega t) = k_x'' E_t \cos(- \omega t)$

substituting in $-E_r = E_i - E_t$ we get an expression for the transmission amplitude, “t”:

$t = (E_t/E_i) = 2k_x/(k_x + k_x'')$

So that’s one equation, based on demanding continuity of the x-derivative. If we do the same for continuity of the y-derivative we get (taking the derivative of our wave equation with respect to y and then taking the case x=y=0):

$k_y E_i \cos( - \omega t) + k_y E_r \cos( - \omega t) = k_y'' E_t \cos(- \omega t)$

which, subbing in $E_i + E_r = E_t$ as before, we get very simply:

$k_y = k_y''$

Now for the last bit. We know that the wave changes speed and the speed of a wave is given by $v = \omega/|k| = c/n$. Therefore, equating the two sides by using the fact that $\omega/c$ is a constant we get:

$\frac{|k|}{n_1} = \frac{\omega}{c} = \frac{|k''|}{n_2}$

With this we can rewrite our $k''$-vectors as

$(k_x'',k_y'') = (|k''|\cos\alpha,|k''|\sin\alpha) = (\frac{n_2}{n_1}|k|\cos\alpha,\frac{n_2}{n_1}|k|\sin\alpha)$

Now it’s just a matter of plugging in. Based on our derivative equation for the y-derivative we immediately recover Snell’s (Snellius’) law:

$k_y = k_y'' = |k|\sin\theta = |k''|\sin\alpha = \frac{n_2}{n_1} |k| \sin \alpha$

$\rightarrow n_1 \sin \theta = n_2 \sin \alpha$

And plugging into our x-derivative equation we get:

$t = \frac{E_t}{E_i} = \frac{2k_x}{k_x + k_x''} = \frac{2|k|\cos\theta}{|k|\cos\theta + \frac{n_2}{n_1}|k|\cos\alpha}$

or

$t = \frac{2n_1\cos\theta}{n_1\cos\theta + n_2\cos\alpha}$

Which is Fresnel’s equation for the transmission. The reflection term is just one minus this one.

So you see, you really don’t need to know anything about electromagnetic fields!

Shiny

What we’ve been talking about so far, in terms of reflection at a boundary, allows us to understand what happens at the surface of a pane of glass. We shine light at the glass, as light hits the air-to-glass boundary, because the “index of refraction” or the “heaviness to light” of glass is higher than air (it’s about 50% higher), there is some reflection and some transmission. In fact, you can take the actual value of the “index of refraction” of glass and air and plug them into Fresnel’s actual equations and you’ll find that you expect about 4% to be reflected and 94% to be transmitted. This is actually wrong by about half, it should be about 8% and we’ll see why in a minute (can you guess yourself why we’re off by about a factor of 2?), but for now just trust me, it’s a bit less than 8%.

So we understand why glass is both shiny and transparent and we really didn’t have to say diddly-squat about the complicated physics of electromagnetism. Once we said: “well, it’s clearly some sort of transverse wave and it goes different speeds in different media” we were set and could then understand not just the simple fact that is should partially reflect as it hits an interface but Fresnel was able to provide a whole set of equations based on those simple assumptions.

It’s actually non-obvious but this is actually also all we need to understand why metal is completely shiny/reflective as well. They end up being a very special case of Fresnel’s equations that’s too far afield to go into but for the very special type of index of refraction that metals have Fresnel’s equations predict almost perfect reflection. So Fresnel’s insights also lets us quantitatively understand mirrors. You shine light on them and it’s all reflected. (Though no real metal is a perfect reflector, something we’ll learn a bit about in a later post.)

A metal has an index of refraction that is a complex number with an imaginary component, which if it’s perfectly imaginary, which no metal is, yields a 100% reflection. So the complete phenomenology of reflective metals is hidden within Fresnel’s equations provided we allow for the index of refraction to be a complex number. In general usage the index of refraction is actually considered to be a complex number with the real component capturing the physics we’ve been talking about and the imaginary component capturing the absorptivity of the material (i.e. dissipation that light suffers within the material). This will actually come in to our next post where we discuss the paradoxical result that materials which absorb a given wavelength of light very strongly will actually also reflect it very strongly. Here we see why, because they have a large imaginary component to their refractive index which both means they absorb strongly but also Fresnel’s equations predicts strong reflection.

Index-Matching

Of course, another thing we’ve now learned is that the ultimate driver of reflection is the mismatch of the “index of refraction” or “heaviness-to-light” between two mediums. The flip-side to this is that if I have two mediums and even if they’re very, very different in nature, if they happen to have the same “index of refraction”/”optical heaviness” then light will not reflect as it moves from the first medium to the second. This is how things like these invisible beads work:

It’s called “index-matching” and basically – in this case – you have beads made of a polymer that, because of their high absorption of water, effectively end up with a refractive index that is very close to water (since they mostly *are* water). Of course this might seem like cheating. I mean you’re “index-matching” by just submerging something that contains a lot of water, in water. That’s a cop-out!

Well, you can actually do the same thing with glass by submerging it in certain engineered types of oil (engineered to have an index of refraction that matches glass). However, in this case, this isn’t just a party trick but actually of technological importance. In Part 2 we talked about how light is fairly insensitive to features that are smaller than its wavelength. Like if you have EM waves with a wavelength of 12 cm, like in your microwave oven, and you put holes in the metal door of your oven that are ~0.1 cm then the microwaves will treat that door as a solid hunk of metal and not “see” the small-detail holes.

For reasons that are somewhat similar to this, optical devices like microscopes cannot discern details of objects that are smaller than the wavelength of light you’re using. So if you’re looking into an optical microscope with your eyes, which can only see light of wavelengths between 400 nanometers and 700 nanometers, the smallest objects you can see are those that are a few hundred nanometers. If it’s smaller than that you’re out of luck and it can’t be resolved. Or can it…

There is a common technique used in microscopy called oil or water immersion which is… pretty much exactly what it sounds like. Light moves slower in oil and water than in air and, as we discussed, since its frequency does not change in moving from medium to medium, and its speed is equal to is frequency times its wavelength, its wavelength must decrease in a material with a higher index of refraction. So as, say, blue light moves from air into the oil or water its wavelength will be *less* than 400 nm and thus it can resolve smaller objects. Then it reflects and comes back out of the oil/water and it’s stretched back to a wavelength our eyes can see, but it still makes a “zoomed” image where no amount of real “zooming” would ever have been able to make an image.

Another place this pops up is in, so-called, “optical clearing” techniques used in medicine. When diagnosing disease or performing research it’s often necessary to visually inspect things like tissue samples in order to look for specific features. However, tissue is generally made of quasi-solid objects (i.e. cells, collagen, elastic fibers, etc.) suspended in a surrounding medium like cytoplasm or interstitial fluid which, relative to the solid objects, has a noticeably lower index of refraction. Because of this mismatch of refraction index, in tissue, between the solid-structures like cells and the solution they reside in, there is a substantial amount of reflection and scattering of light at the interfaces between them. As a result, tissues are quite opaque (we’ll talk about this more in the next post) and one cannot see their interior structure very well with a microscope.

This leads to the natural idea of “Tissue Optical Clearing” (TOC) which basically is exactly what we just talked about: refractive-index matching. The idea is to replace the liquid medium, which is not really what you’re interested in (you care about the cells) with a different liquid, one with a higher refractive index that more closely matches that of the cells. As a result of this, one can essentially make tissue that was fairly opaque mostly transparent or at least “more” transparent so that some of the interior structure can be discerned.

This leads to some pretty wild (and gross) images where things that aught-not to be transparent can be made transparent. I won’t walk you through a carnival show (you ever wanted to see a transparent mouse brain? What, no?) and maybe just leave that to you and Google (just put “Tissue Optical Clearing” into Google Image Search) but I’ll just show this one (sadly low resolution) image:

That’s a mouse bone (and spleen) that have been made transparent by replacing the liquid medium of the cell with one which is more refractive index matched to the cells themselves. That’s pretty wild. And seriously, if you’re going to Google for more I was serious about that NSFL (Not Safe for Life) warning.

I’d also add just a quick final point that this is actually the same reason why things like paper become somewhat transparent when they’re wet. In paper you have a bunch of solid white fibers and then plenty of air between them and thus the index difference between the paper fibers and the air is quite large. However, when water is wet the water will take the place of the air and fill in all the nooks and crannies with a liquid whose index of refraction is closer to that of the paper fibers. Thus, the paper becomes less opaque.

Specular… SPECular

Let’s now change gears and look at one final topic before we close out this post: what’s the difference between a mirror and something that is white, like white cotton or chalk? In order for cotton to be white, it must reflect (or mostly reflect) all the wavelengths of the visible spectrum, so that when it is illuminated with white light, the light my eyes see when I look at it is also white. (If you don’t fully understand this, don’t worry, we’ll come to understand what makes “color” much better in the next post.)

But, then… isn’t that what a mirror does? Reflects all visible light?

Yes!

So what’s the difference?

The direction it reflects light. Or put another way, how faithfully it preserves the angles of incidence and reflection. The difference between a mirror and white cotton is whether it reflects ”specularly” or ”diffusely”. These two words are statements about how smooth the surface of a material is. In a nutshell, a diffusely reflective material has a very rough surface and thus if I have a bunch of parallel rays of light coming in, those rays are reflected at random angles:

Conversely, a mirror is a very smooth surface and thus if I have a bunch of parallel rays coming in, they maintain their “shape” after reflection. In other words, if I say I see an object, really that object has been illuminated by an external – usually white – light source and those rays have reflected off the object and the rays have hit my eyes. However, if the reflection randomizes the direction of the rays, the “image” is lost.

So white cotton is a mirror, but with a really rough surface such that it randomizes any spatial “image” information that the incident light was carrying before.

Don’t believe me? Do an experiment. Take a black T-shirt, a white T-shirt, a small mirror and a flashlight and go into a dark place like a closet. Ignore the looks people give you. It’s not weird. They’re the ones who are weird. You’re doing SCIENCE!

Okay, now put the flashlight very close to the black T-shirt, and ball the shirt up so no light shines through it. Not too close, of course, we want to make sure that all light hits the black T-shirt, but if the light reflects it can escape elsewhere. Now turn on the flashlight. The room isn’t illuminated at all. You’re shining all the light but it’s all going directly to the black which is absorbing it. (We haven’t talked about absorption yet – we will in the next post – but hopefully this still makes sense to you without all the fine details.) Now do the same to the mirror. You will see the that the room is lit up, but also that there’s a fairly bright central spot which is along the line of direct reflection from how you’re holding the flashlight. Finally, do it with the white shirt. You’ll see that the room again lights up, but the light is more evenly spread in all directions.

I Just Lied To You! The Scattering of Subs (Yum!)

Now that maybe I’ve provided a bit of illumination (puns!) on this issue, let me pull the rug out a bit and say that, despite lots and lots of online “sources” I found saying that this is the case, what I just said about the difference between white cotton and a mirror is wrong. Well, okay, it’s not wrong. But it is dangerously incomplete. Let me pose to you this: if this is the whole story, how is it possible to have white gloss paint? Shouldn’t that just be a mirror? How can you have metal with a rough surface, like brushed aluminum or the back side of aluminum foil? Shouldn’t that be like our white t-shirt?

Well, to see what we’re missing let’s start by going back to our pane of glass. Fresnel’s equations says that I should see 4% reflection off an air-glass interface. But I claimed that if you check, you’ll be off by a bit less than a factor of two, and will get a bit less than 8%. What gives? Was Fresnel wrong?

No. He wasn’t. But in our pane of glass, there are *two* interfaces, the front-face which is an air-glass and a back-face which is a glass-air. Fresnel’s equations in the simple case we considered are actually symmetric so that a media1-media2 and media2-media1 interface see the same reflection. (Note! This is true in our simple case of straight-on reflection, it is not generally true for reflection at an angle.) So, if you shine light on a pane of glass – and I’d strongly recommend looking at the diagram below first, I’m about to get dense but what I’m saying really isn’t that complicated and hopefully the diagram makes it clear – you’re getting a reflection of 4% from the front surface and 4% of the 96% that transmitted past the front interface. (4% of 96% is 3.85%). So it’s 4% + 3.85% = 7.85% reflected? Well, that’s very nearly it. Buuuutttttt… if you diiiiiddddddd want to be a stickler and a smart-ass you’d point out: “but wait, can’t that 3.85% that makes it back for a second trip then be sent back again at the front-face?”. Yes. Yes it can. So we would have 96% of that 3.85% transmitted and 4% of the 3.85% sent back yet again. So 96% of 3.85% is 3.7%, so really you’d have 4% (the original front-face reflection) and only an additional 3.7% (for a total of 4% + 3.7%) from the back reflection, with 0.15% having been back reflected by the front-face on the second go and got turned back away, back towards the back-face (so many uses of the word front/back…). And of course a smart-ass would then say, “but wait, 4% of that 0.15% that was sent back to the back-face will be re-re-reflected to the front… and then when it hits re-re-re-hits the front-face it…”. Ya. We get the point. You’re kinda a jerk buddy!

So this is why I said “a bit less than 8%” because of this possibility of multiple bounces. Now figuring out the total final reflection assuming an infinite number of possible re-re-re-re…reflections to the front-face is actually pretty easy to do mathematically, but it actually ends up being just ~7.7% because the other bounces add so little that they only affect lower decimal places.

I apologize for walking through the tedium of that but there was a greater point to it. Most materials in life aren’t mirrors or glass and even if they are, no mirror or glass material is free of imperfection and defects. In reality, a material may be made of small grains, or have impurities or defects lodged in them:

Thus, at every interface, be it a grain boundary, a defect boundary or a boundary with air/vacuum (the case for a hole), there is a chance for the “reflection” of light. I put reflection in quotes because, since the distribution of these objects is going to be fairly random, a given ray of light may very well be sent in some totally bonkers direction other than the front-face.

There is one additional aspect of this, which is that transparent materials separate light (like prisms) due to what is called refraction (a very important aspect of light, but not something we’re going to explicitly talk about here).

[Greatly Exaggerated!]

We’ll talk more about how the light is changed as it travels in the material or moves from grain to grain in the next post, but here what is important is that this acts even more to scramble the light. These two effects – scattering and reflection from internal surfaces and refraction and (later we’ll talk about absorption) which occur as the light spends time in the material produce what is called sub-surface scattering.

Thus, we have a front-face which may reflect either diffusively or specularly but once light makes it into the material there are many internal surfaces that will also reflect and scatter it. These surfaces enhance the amount of diffusive reflection but also change the light based on the material (through absorption and refraction), something that doesn’t happen so much for just the surface reflection. It is this that makes white cotton different from matte or brushed aluminum. Brushed aluminum has a rough diffusive-reflecting surface but all reflection does happen at the surface and there is no sub-surface scattering. Conversely, in white you have light that is even more diffusively reflected but also “scrambled” by its time spent in the interior of the material. If “scrambled” sounds pretty vague, don’t worry, you’ll understand much better what happens to light as it travels through a material in the next post.

However, for now let’s wrap it up. We learned in this post that many properties of light are really just general properties of waves and can be understood as such. In other words, many properties of light aren’t really as mysterious as people might initially think, but rather people just simply might not be that familiar with how waves – in general – behave. We also found out about how surface and sub-surface reflection can make the difference between “shiny metal” and “cotton white”. We also learned how to make mouse brains transparent, which was… I’m going to say… “neat”?

In the next post we’re going to talk all about colour and explore this in much greater detail; understand what makes a leaf green, but also how a piano can be shiny and black (which, we will find, initially seems like an impossibility). We’ll also talk about how you can see THROUGH “opaque” objects without any of this business of index-matching. We’ll also talk about how NASA is tricking you… (Ever-so-slightly. Click-baits!! I have them!)

Moving forward we’ll then talk about photons and quantum mechanics and neon signs and predator-vision and Star Trek-esque medical tricorders and much, much more.

See ya then.

4 thoughts on “The Saga of Light, Part 3: Invisible Bones, Invisible Beads and Why Mirrors Aren’t White”

1. “Like if you have EM waves with a wavelength of 12 cm, like in your microwave oven, and you put holes in the metal door of your oven that are ~0.1 cm then the microwaves will treat that door as a solid hunk of metal and not “see” the small-detail holes.”

But the bars of metal are also less than 12 cm. What determines whether it passes through or reflects if both the solid metal and the holes are smaller than the wavelength? Does it just take the average?

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1. It’s not going to be anywhere as simple as simply averaging, unfortunately. There’s some fairly complicated analysis of the notion of holes in Faraday cages in advanced E&M textbooks, but it’s not terribly illuminating. But, the key point is that it’s not really a case of “no transmission” and “complete transmission”. A complete analysis would depend on many material parameters of the metal such as conductivity and dissipation parameters and in the end it would output some transmission and attenuation factors which you could then tune until you got the “blocking performance” you want. The “if the holes are much smaller than the wavelength” is a general heuristic rule of thumb that comes from these more complex considerations. People can and do do such calculations, but pragmatically, you’d probably just try a bunch of designs and just measure the transmission through the mesh in the microwave range until you found a design that meets your engineering specifications.

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2. Assuming an up-down transverse wave, is the amplitude proportional to the amount of charge or the distance the charge moves up and down?

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