{"id":30401,"date":"2013-10-21T17:33:43","date_gmt":"2013-10-21T17:33:43","guid":{"rendered":"http:\/\/www.labri.fr\/perso\/barla\/blog\/?p=30401"},"modified":"2013-10-21T17:44:48","modified_gmt":"2013-10-21T17:44:48","slug":"the-optics-of-life","status":"publish","type":"post","link":"https:\/\/www.labri.fr\/perso\/barla\/blog\/?p=30401","title":{"rendered":"The Optics of life"},"content":{"rendered":"<p id=\"top\" \/><em>by Sonke Jonsen<\/em><\/p>\n<h3>Units and geometry<\/h3>\n<ul>\n<li>A spectrum generated using bins with equal frequency intervals would look different from one with equal wavelength intervals. So which one is correct? Neither and both. The take-home message is that the shape of a spectrum depends on whether you have equal frequency bins or equal wavelength bins. p.14<\/li>\n<li>While it looks like we might have to make an arbitrary decision, there is actually a clear-cut division between extended objects, the radiance of which is conserved, and point objects, where the radiance drops with the inverse square of distance. It all depends on the detector. If the angular size of the object is below the resolution limit of the detector, then it is a point source; if it is above this limit, then it is an extended object. p.23<\/li>\n<li>Since area equals length squared, the solid angle of objects viewed underwater increases by a factor of m2, where m is the ratio of the indices of the two media (1.33^2 == 1.75 for water viewed from the air behind your mask). But the total amount of photons that reach your eye from the object has not changed (it can&#8217;t without violating fundamental conservation laws). Radiance is the total number of photons divided by solid angle, so the radiance of the object decreases by a factor of m2. In other words, objects viewed under water are larger and dimmer. p.24<\/li>\n<li>Even within the visible range, human eyes are much more sensitive to 555 nm light than to 410 nm light, which is why yellow looks brighter than violet to us. Therefore, while I would toss the units (unless you plan to work in human visual psychophysics), I would still keep this general principle. p.26<\/li>\n<\/ul>\n<h3>Emission<\/h3>\n<ul>\n<li>If an object absorbs more than it emits, it gets hotter. If it absorbs less than it emits, it gets colder. The equality of absorptivity and emissivity has some interesting consequences. For example, a perfectly reflective object absorbs no radiation-all of it just bounces away. So its absorptivity is zero, which means that its emissivity is also zero. This is why tea ketles and thermoses are often shiny. The high reflectivity means a low emissivity, which reduces radiative (but not conductive) heat loss, though I am sure style has something to do with it as well. It also explains those flimsy metallic space blankets that you find in camping stores. They are lousy insulators, but are good at reducing radiative heat loss to the sky. Since absorptivity equals emissivity, a 100% emissive object absorbs all radiation that hits it and, because of this, is called a &#8220;black body radiator.&#8221; A black body radiator in this formal sense emits the maximum amount of radiation possible for its temperature. p.32-33<\/li>\n<li>While fairly transparent and colorless in the visible range, at shorter and longer wavelengths it rapidly becomes opaque. Since most organisrVs live in water and all organisms are full of water, this narrow window of low absorptivity is the only thing between us and blindness. p.34<\/li>\n<li>The color temperature of a light is simply the temperature of a black body radiator at which the black body spectrum best matches the light&#8217;s spectrum. p.38<\/li>\n<li>Therefore, while the radiance of the sky is far less than that of the sun, its relatively enormous size means that the sky&#8217;s contribution to down-welling irradiance is substantial at certain wavelengths. As you might guess from the color of the sky, its contribution is greatest at the lower wavelengths. The high contribution of skylight to the irradiance in the ultraviolet pardally explains why people get so sunburned in open places like beaches, fields, and lakes. In all these habitats, there is substantially more visible sky than is found in typical urban and suburban landscapes. The importance of the skylight to total UV also suggests that sitting under a small umbrella at ; the beach may not save you from sunburn, even if you are out of the direct sun. You will feel cooler, because the skylight contributes almost nothing to the infrared irradiance, but your dose of UVB photons (those responsible for sunburn) may only be cut by half. p.40-41<\/li>\n<li>When the sun is below the horizon though, its light takes a long raking pass across the ozone layer (figure 3.13). By the time the light reaches the overhead sky and is scattered back down to us, at least half of the orange, yellow, and red light has been removed, leaving blue and violet. &#8216;\\nother way to distinguish the blue onate twilight from that of daytime is to look at an overcast twilight sky. Overcast daylight is gray, but overcast rwilight is still a deep blue. This is because the absorption of the longer wavelengths occurs at altitudes far above the clouds. p.48<\/li>\n<li>The moonlit sky is just a sunlit sky with an economy lightbulb. p.50<\/li>\n<li>The average radiance of the lit portion of the moon is not constant. In other words, when less of the moon is lit, the part that is lit is also dimmer. p.50<\/li>\n<li>The moon is like a retroreflector on a bicycle, sending a great deal oflight directly back to where it carne from. As the moon becomes less full, these craters and mountains get lit more and more from the side, which means that they cast longer shadows. p.51<\/li>\n<li>The less full a moon is, the closer it is to the sun. 3ecause it is so close to the sun, a crescent moon is highest around noon, when the sun is also highest. Half moons are 90\u00b0 from the sun and so are highest at dawn and dusk. During the night, both crescent and half moons are otten either below the horizon or so close to it that they are obscured by the landscape. Full moons rise when the sun sets, set when the sun rises, and are highest in the middle of the night. Because they are directly opposite the position of the sun, their peak height matches that of the sun in the opposite season. Therefore, the full moon is highest in the winter and lowest in the summer. If you ignore starlight, the total dose (irradiance X time) of moonlight over the course of a winter night is about 5000 times greater for a full moon than for a crescent moon. If you include the background starlight the difference is still in the range of 50- 200-fold.\u00a0 p.51<\/li>\n<li>So the irradiance on moonless nights is a sum of these three components-integrated starlight, zodiacal light, and airglow-whose relative contributions depend on the sunspot cycle and where the Milky Way is in the . tributions depend on the sunspot cycle and where the Milky Way is in the sky. On average though, it is significantly red-shifted compared\u00a0 to daylight or moonlit nights and massively red-shifted compared to deep twilight. Therefore, for those animals that can see it, the sky goes through some impressive color changes every time the moon or sun rises or sets. p.54<\/li>\n<li>This shift from white to red to white to blue to purple to white to red can happen over a period of an hour or two, depending on where you live. It is of i course influenced by clouds and other factors, but the bottom line is that, whenever the sun or moon rise or set, the brightness and color of the sky go through some serious gyrations. p.55<\/li>\n<li>The aurora can in some ways be thought of as the ultimate airglow. As with the latter, the aurora is due to light emission by excited molecules in the upper atmosphere. In fact, the same molecules are involved. Both the green and red auroras are due to emission by excited oxygen acorns. Where the two phenomena differ is in the source of tbe exciting energy. In the case of airglow, the atoms are excited by solar photons. In rhe case of the aurora, the atoms are excited by charged particles ejected from the sun. p.55<\/li>\n<li>The interaction between the solar wind and the earth&#8217;s magnetic field is such that these particles can usually only brc:lk through in regions that ar about 2000 miles from a magnetic pole. p.56<\/li>\n<li>The reason we don&#8217;t pay much attention to bioluminescence is because over 80% of the light-producing genera (and the vast majority of the lightproducing individuals) are marine. For unknown reasons, bioluminescence is rare on land, limited to fireflies, glow worms, and a grab bag of random arthropods, mushrooms, and worms. Bioluminescence is enzyme-mediated chemiluminescence. p.59<\/li>\n<li>The upshot of all this is that, even if we make great allowances for what is possible, thermal radiation simply cannot be used for visual signaling in animals. There are, of course, animals that can detect the thermal radiation of objects at physiologically reasonable temperatures, but only in the infrared. p.60<\/li>\n<li>The reaction itself occurs in one of three places. In many fish and a few cephalopods, the light is produced by symbiotic bacteria that live in a pouch. The bacteria produce light continually, so the animal uses a shutter to control whether the light actually exits the body. In many other cases, the light is produced by the animal itself in5:ide cells known as &#8220;photocytes.&#8221; These cells are otten part of beautifully complex organs that contain filters, mirrors, lenses, and other apparatus for controlling the color and direction of the light. Finally. some marine species perform the chemical reaction outside of their bodies, essentially vomiting the necessary reactants into water. p.62<\/li>\n<li>Thus, many animals cover their ventral surfaces with photophores and make sure that they match the intensity of the down-welling light, even keeping up with passing douds. Some species of squid even modulate the color of the light to better match the down-welling light. p.64While bioluminescence is a far more efficient way of producing light than thermal emission (at least for organisms), proclueing enough light to be easily seen under daylight conditions is expensive. The central problem is not energy expenditure, but the fact that most luciferins are derived from diet and can be used only once. p.64<\/li>\n<li>Certain materials emit light when you deform them. This is known as &#8220;piezoluminescence.&#8221; More common and potentially biologically relevant is triboluminescence, which is light emission caused by breaking something. p.69<\/li>\n<\/ul>\n<h3>Absorption<\/h3>\n<ul>\n<li>This is a shame, because oscillation and resonance show up everywhere in biology. Running, walking, flying, and swimming are all examples of harmonic oscillation, and many animals take advantage of resonance to improve the efficiency of their locomotion. Indeed, much of biomechanics is the study of harmonic oscillators. But it doesn\u2019t stop there. Nearly every aspect of organismal physiology is cyclical, from heartbeats co circadian rhythms to seasonal migrations to breathing to the cell cycle itself \u2013 p.78<\/li>\n<li>As the electric field from a light wave passes an atom, the negatively charged electron cloud gets pulled one way and the positively charged nucleus gets pulled the other way. Then, very quickly, both the cloud and the nucleus get pulled in the opposite direction. This repeats over and over, and the relative positions of the electron cloud: and nucleus (together called a \u201cdipole\u201d) oscillate at about a hundred trillion times a second. In all cases, the oscillating dipole emits an electromagnetic wave. We call this scattering. However, just like dock pendulums, water in bathtubs, and Dodge Darts, atoms and molecules have resonant frequencies. At these frequencies, the incident visible light wave does a lot of work on me dipole and thus loses energy. In other words, the light gets absorbed. \u00a0\u2013 p.81<\/li>\n<li>The energy levels for vibrations and rotations are closer to each other than those between different electron orbitals, so they absorb light with lower energies and longer wavelengths. In most organic molecules, the differences between vibrational and rotational energy levels correspond to infrared radiation. In these same molecules, the differences between the energy levels of different electron orbitals correspond ultraviolet radiation. There is a window of wavelengths of light that are too energetic to match vibrations and rotations and too weak to kick electrons into higher orbits. In this range (from about 350 nm to 750 fill), there is little absorption by water and most organic molecules. This relatively tiny window between these two highly absorbing regions is what we call visible light. It is a complete accident that we can see at all. \u2013 p.83-85<\/li>\n<li>A wonderful example of the effect of environment and assembly is the difference between diamond and coal. Both are made of nearly pure carbon, but the former is transparent while the latter is strongly absorbent. The difference lies in how the carbon I atoms are bonded together. In fact, both the classical and quantum explanations of absorption are a bit like the Monty Python skit where the entire lesson on how to play the flute is \u201cblow into the big hole and run your fingers up and down the other holes.\u201d \u2013 p.85<\/li>\n<li>In other words, the light gets absorbed exponentially. This is not a property that is unique to light, but is found in many situations where events happen randomly, for example the half-lives of radioactive materials. Just like a radioactive nucleus has a constant probability of decaying at any moment, a photon traveling through a homogeneous material has a constant probability of striking a molecule and being absorbed. \u2013 p.86<\/li>\n<li>Without scattering, all pigmented substances are black when viewed via the light they reflect. If you bought some paint that was pure pigment and tried to paint over a black wall, the wall would stil1look black. Of course, if you shine light through the pigment, you will see that it is colored, but the vast majority of biologically relevant objects are viewed via reflected light, not transmitted light. Even most transparent species are primarily detected by reflected light, not by light transmitted through their bodies. The most common counterexamples I know of are leaves viewed from below via transmitted sunlight. \u2013 p.91<\/li>\n<li>White paint scatters light so well that much of the incident light comes right back out of the painted wall, giving it a high reflectance. Add it to pure pigment and you will actually be able to see the color. Add too much, though, and the color looks washed out. As with much of life, there is a balance. You need enough white paint that a significant amount of light is scattered back to your eye (otherwise the wall will look black), but not so much that light is scattered back out of the wall before it has had time to be signifycantly modified by the absorbing pigment. With the exception of rare backlighting situations, color in nature requires both absorption and scattering in the right balance. Too much of one or the other and you only have black or white. \u00a0\u2013 p.92<\/li>\n<li>So what you see, as you add \u201cyellow\u201d dye to the bowl, is that it goes from dear to light, unsaturated yellow to more saturated, darker yellow to orange to red to dark saturated red. As path length increases, the reflectance at all wavelengths drops, the saturation of the color increases, and the hue shifts from yellow to red. The same thing also happens if you increase the concentration of the dye. What truly matters is not the path length, but the number of dye molecules that the light intercepts. \u00a0\u2013 p.96 <i>Does the same phenomenon occur with inter-reflections from a same matte surface?<\/i><\/li>\n<li>Just because something changes color or has a color with two reflectance peaks, does not mean that it has two pigments. It is possible that everything is due to changing concentrations of one pigment. \u2013 p.97<\/li>\n<li>As you may know, clear water absorbs bIue-green light the least. The difference in absorption is fairly small over short wavelengths, which is why a glass of water has no color, but, because light is attenuated exponentially, these differences become enormous after you\u2019ve gone through a few hundred meters of water. So, by the time you get to depths of 500 m or so, the remaining light is a blue-green with almost laserlike purity. \u2013 p.100<\/li>\n<li>While you might think that boosting the gain in the photoreceptor would help, this boosts the noise just as much as it boosts the signal. So, all the solutions come down to the same thing: absorb more photons in the photoreceptors. \u2013 p.109<\/li>\n<li>In animals, pupil size is limited by two things: (l) a big pupil requires a big eye and thus probably a big head, and (2) a big pupil requires a big lens, which is hard to make and more likely to have aberrations. So an animal of a given size can only vary A by so much, though this does explain why nocturnal and deep-sea animals usually have big eyes for their size. \u2013 p.109<\/li>\n<li>Photoreceptors are already crammed to the gills with visual pigments, so the easiest way to increase the amount of light absorbed is to increase the path length 1. This can be done by increasing the length of the photoreceptor or by sending light through the cell twice. This latter trick is accomplished by putting a mirror behind the photoreceptor, so that light has two chances to get absorbed-once on the way in and once on the way out. These mirrors, known as \u201ctapeta,\u201d are found in the eyes of many animals: shine a flashlight at your cat\u2019s face and you\u2019ll see two of them. \u2013 p.111<\/li>\n<\/ul>\n<h3>Scattering<\/h3>\n<ul>\n<li>Optics is primarily photons interacting with electrons, atoms, and molecules. If the energy of the photon matches the difference in energy between two levels of excitation, then the photon vanishes, its energy converted into other forms (heat, etc.). However, if the photon\u2019s energy doesn\u2019t find a match among the differences in possible energy states, a new photon will quickly be emitted. The new photon is usually the same energy (and thus the same wavelength) as the old one, making it appear as if the original photon bounced, which is why this nonabsorptive interaction is known as \u201cscattering.\u201d It is much as if you were at the bottom of a deep wishing well throwing a penny out every time someone threw one in. From the viewpoint of the people above, it would look like the penny they tossed in bounced off the bottom of the well. \u2013 p.116<\/li>\n<li>In wave language, one can model light scattering using a simple rule; divide the object into small regions, each of which emits a train of spherical waves. The height and phase of the waves depend on the intensity and phase of the incident light in that region. Every part of an illuminated object emits these waves, not just regions where the refractive index changes. My image of this is a pond in an advancing rainstorm. As the storm moves across the pond, each raindrop creates a circular wave that interferes with the thousands of other raindrop waves, Your job is to see what the final wave looks like at the other side of the pond. &#8211; p.117-118<\/li>\n<li>Light is like a man running through a maternity ward, bumping into beds as he goes. By the time the babies wake up and start crying, he has already left the room. Similarly, by the time the incident light wave has started accelerating charges in matter, its wave front has aIready moved on. This wave from is weak but detectable. \u2013 p.118<\/li>\n<li>There is some variation with wavelength (which is considerable once you go far into the ultraviolet or infrared) and with material, but to a good approximation the density map of a tissue correlates with its spatial variation in refractive index. This makes life simpler. \u2013 p.119<\/li>\n<li>Imagine a parallel beam of light aimed at a nonabsorbing particle. A fraction of this light will be scattered and thus removed from the beam. Now consider an opaque disk whose cross-sectional area is such that, if placed in the beam, it would block me same fraction of light as the particle scattered from me beam. The area of this disk is the scattering cross section of the particle (usually called \u201cCsca\u201d). Dividing the scattering cross section by the real cross-sectional area of the particle gives another useful term, the scattering efficiency (called \u201cQsca\u201d). This tells you how good the particle is at scattering light. \u2013 p.121<\/li>\n<li>Because <i>g<\/i> tells you roughly how much light is scattered forward and backward, it is used in all sorts of simple, but fairly accurate models of light propagation through dense messy substances like tissue. \u2013 p.122<\/li>\n<li>While the details can get complex, only three things affect how efficiently a single isolated particle scatters light: (I) its size relative to the wavelength of light hitting it, (2) its refractive index relative to the index of surrounding medium. And-to a much lesser extent-(3) its shape. The absolute size and index of the particle don\u2019t matter. \u2013 p.122<\/li>\n<li>So for small particles-proteins, ribosomes, etc.-scattering is proportional to the square of the volume. You have probably noticed that I have been acting as if the particle were a sphere. This is because, unless the relative refractive index is quite high (higher than what you\u2019ll find in biology), the shape of small particles has only a moderate effect on scattering. Small particles scatter blue light more than red light, which is at least partially responsible for our blue sky and red setting sun, which will be discussed later. Remember, it\u2019s not that blue light scatters more easily, but mat a given partied looks larger relative to blue wavelengths, and larger particles scatter more light. \u2013 p.123<\/li>\n<li>All large particles remove as much light from the beam as an opaque disk of twice the cross-sectional area. You might think that a scattering efficiency of one would make more sense, but it turns out that large partides affect the light that passes near them as well, just like waves are bent as they pass around an island. Therefore they actually remove more light from the beam than their shadow would suggest. Of course, the actual amount of light scattered depends on the orientation of the particle relative to the light beam. A rod oriented parallel to the beam scatters much less light than one perpendicular to it, because it has a smaller cross section. But what if as so often happens in biology, we don\u2019t know the orientation? Or have a collection of particles that are found in all orientations? In this case we would need to know the cross-sectional area of the particle averaged over all orientations Aavg (in other words, the average area of its shadow). Therefore, objects with a lot of surface area relative to their volume (e.g., flat disks, long rods) scatter a lot of light per volume. \u2013 p.125<\/li>\n<li>A lot of important cellular components are in this range-nuclei, vacuoles, mitochondria, lysosomes and so forth. The sad truth is that scattering is difficult to determine in this range, because different parts of the particle experience different phases of the incident light, and because the incident light is altered by its trip through the particle. The combination of these two effects means mat the different regions of the particle are emitting scattered light at different phases and amplitudes and so a lot of complicated interference occurs. \u2013 p. 126<\/li>\n<li>It turns out that, while refractive index doesn\u2019t affect how much light gets scattered in large particles, it does affect where it gets scattered. As particles get larger, they scatter more and more of their light in a forward direction. This is due to wave interference. \u2013 p. 128<\/li>\n<li>Large particles scatter a lot of light, but scatter most of it forward, which means that g is close to one, making log small. Also, r gets large and is in the denominator, so volume-specific hiding power is low. Small particles scatter light in every direction (g = 0), but are incredibly inefficient, so volume-specific hiding power is again low. So where is the peak? Nicely enough, it is for particles with diameters mat are between one half and equal to the wavelength of light, regardless of refractive index. Break up any nonabsorbing substance into particles of this size and it will make a good white substance. Suppose you add a certain volume of ice to a glass of water. If the ice is broken up into microscopic crystals, the total scattering will be small (you need to ignore that fact that small crystals would melt right away). If you just add two big ice cubes, you can still sec through the glass fairly well. Each cube scatters a fair bit of light. But there are only two of them, so they don\u2019t scatter much light in coral. However, if you crush the cubes, you can barely see through the glass at all. You have the right combination of high scattering per particle and number of particles. This same process explains why you can see through rain, but not through fog. The volume density of water in fog is actually quite low, but if you have ever driven in it, you know it scatters light well. However, you can easily drive in a moderately heavy rain, even though the volume density of water in air in this case is far higher. This is because the volume-specific hiding power of fog-sized water droplets is much higher than for rain-sized water droplets. \u00a0\u2013 p.130<\/li>\n<li>If they are more than about ten wavelengths of light apart, then life is easy. This is because, unless they are regularly arranged, light scattered from particles separated by at least this distance interferes randomly and you have what is called \u201cincoherent scattering.\u201d In these cases-which include rain, fog, clouds, pollution, and even dense blooms of phytoplankton-the total scattered light is just the sum of the light scattered by all the particles. \u2013 p.132-133<\/li>\n<li>A scattered photon is merely knocked out of the way, which means there is always the chance that it can be knocked back to your eye by a second scattering event, especially if absorption is low. This is what creates the halos you see around lights in fog or mist. A photon that was not originally on the way to your eye gets scattered toward it. Your eye thinks the photon originated from the second scattering position and so you see a fuzzy halo around the light. This is one reason that heavily absorbing waters, like tannin-laced swamps or iced tea, look so clear. Any multiply scattered light gets absorbed long before it gets to your eye. \u00a0&#8211; p.139<\/li>\n<li>For the ocean to look blue from above, we need both absorption and scattering. The blue color we see from a ship is due to photons that entered the ocean and then were scattered so many times that they turned 180 degrees and came back out. How many times? Even if every scattering event is in the same direction and plane, the photon has to be scattered seven times to completely turn around and exit the water. These seven scattering events will take it through an average 70m of water, a trip that only blue photons will survive. Nearly all the photons of other wavelengths will get absorbed before they make it back out. \u2013 p.141<\/li>\n<li>If the light gets scattered in a more forward direction, it takes even longer for it to get back out of the material, leading to many more chances for absorption. Craig Bohren invoked this principle to explain why many wet things are dark. The difference between the refractive indices of sand and air is fairly large, so light is scattered over large angles and exits the sand relatively quickly. This gives the light relatively little chance of being absorbed by the pigments in the sand grains and so the sand looks light. Now suppose a wave washes over the sand. The water sinks in, replacing the air. The water has a higher index than the air and so the difference between it and the index of the sand is smaller. Thus the light is scattered in a more forward direction, takes longer to get back out, and has a greater chance of being absorbed. So the wet sand is darker. \u2013 p.142<\/li>\n<\/ul>\n<h3>Scattering with interference<\/h3>\n<ul>\n<li>If two beams start out with the same phase (for example, one laser beam made into two via a beam splitter), then their relative phases when they meet again (assuming they do) depends on how far they each traveled. It also depends on what the two beams traveled through, since the wavelength of a beam of light is affected by the refractive index of the medium in which it travels. This is wave interference in a nutshell. &#8211; p.153<\/li>\n<li>The important thing is that, due to the summing of large numbers of paired waves of opposite random phase, a substance can appear to not interact with light at all, even though it scatters light at every location. Even electrons take some time to accelerate. Due to this delay in acceleration, the forward scattered wave has a definite phase relationship with the main wave like a kid walking exactly three steps behind its mother. Again, after a short delay, the main wave induces a scattered wave in the second molecule. Because both the first and second molecule experienced the same delay (the delay depends on the substance, which is all water) and because the scattered waves have this defined relationship with the main wave, the two scattered waves are in phase with each other. These two scattered waves and the main wave move on to the third molecule, and so on, until eventually the main wave is trailing a clutch of scattered waves like baby ducks, except that, instead of being in a line, they are all in phase. This means they interfere constructively, and you, as the diver viewing the flashlight, see a bright light. \u2013 p.155-156<\/li>\n<li>The bottom line though is that there is a difference between scattering and observable scattering. Transparent objects do scatter a lot of light, but only the forward scattered light is observable (or at least it\u2019s by far the most dominant). The interference between this forward-scattered light and the main beam is what changes the wavelength and phase velocity of the detected beam, leading to what we call refractive index. \u2013 p.156<\/li>\n<li>A good rule of thumb, which both lenses and corneas follow, is that, for a substance to be clear, it has to look uniform at all size scales larger than half a wavelength of the incident light. The white of the eye is made of the exact same fibers as those in the cornea, but those in the former are larger and randomly oriented, leading to observable scattering and thus opacity. &#8211; p.158<\/li>\n<li>However, when the angle of incidence equals the angle of reflection, something special happens. For this particular set of angles, the total distance from bulb to eye via the mirror is minimal. More importantly, the change in distance for paths right near this minimal configuration is also minimal. Just like a rounded hill is flattest near the top. And a rounded valley is flattest near the bottom, the phase changes the least when the angle of incidence is near what we call the angle of reflection. Because the phases are similar, the light waves add constructively and it appears that light is only reflected when the angles are equal. \u2013 p.160-161<\/li>\n<li>The ecological functions of biological mirrors are as varied as their structures and locations. Perhaps the most common function-proven or hypothesized-is signaling. [\u2026] Broad-band biological mirrors are seldom used for signaling, perhaps because the one great advantage of saturated structural colors is their invariance under different optical environments (a blue butterfly wing looks blue under almost any natural light). Instead, broad-band reflectors seem to be used for camouflage, especially in pelagic habitats. \u2013 p.166-167<\/li>\n<li>However, the paths that are close to the one given by Snel\u2019s Law have nearly the same travel time, and thus nearly the same phase. So they constructively interfere and it looks like the light bends at the surface of the water. The fact that it\u2019s the minimum time doesn\u2019t matter. Again, what matters is that the change in time for paths near this minimum is also minimal. \u2013 p.172<\/li>\n<li>In other words, a lens is not bending individual rays of light so much as it is making sure that all the in-phase light from one point makes it to another point still in phase. \u2013 p.174<\/li>\n<\/ul>\n<h3>Fluorescence<\/h3>\n<ul>\n<li>First and foremost, fluorophores are pigments-meaning mat they absorb light. However, unlike most pigments that convert all the absorbed energy into other forms, fluorophores only convert part of it. The remaining energy is emitted as a photon. Usually this photon has a lower energy and longer wavelength than the incident photon, the difference in energies being known as the \u201cStokes shift\u201d. \u2013 p.182 <i>This sounds like it\u2019s sort of in-between absorption and scattering\u2026<\/i><\/li>\n<li>Regardless of how much light it converts, though, any molecule\u2019s fluorescence is described by three properties. The first is its excitation spectrum (figure 7.2). Though it\u2019s given a different name, this is essentially the absorption spectrum of the molecule. This makes sense, since in order for a photon to excite fluorescence. It needs to be absorbed first. The second defining property of a fluorescent molecule is its emission spectrum. You will notice that the emitted photons do not all have the same wavelength, despite the explanation of it all being due to the last jump between the lowest excited state and the ground state. This is due to additions and subtractions of vibrational energies and other bells and whistles that give you a hint of the true underlying complexity. The emission spectrum essentially gives you the probability that the emitted photon will have a certain wavelength. The final property of a fluorescent molecule is the efficiency. This could be defined in two ways. Energy emitted divided by energy absorbed. Or photons emitted divided by photons absorbed. \u2013 p.184-185<\/li>\n<li>Perhaps more importantly, fluorescence can be used to create light of wavelengths that are rare in a given environment. Even clear ocean water absorbs long-wavelength light strongly; leaving little orange, yellow, and red light below even a few meters and little green light below a few tens of meters. The color of an object depends as much on the light illuminating it as on the actual reflectance, so red objects a few meters below the surface look black (lips and tongues at scuba depths look ghastly). However, a fluorescent substance can convert light of dominant wavelengths (for example, blue light in the ocean) to wavelengths that are rare, creating a highly conspicuous and specific signal Viewers can accentuate this effect by using ocular filters that screen out the dominant wavelengths and thus increase the contrast even further. \u2013 p.188<\/li>\n<li>While the cost ofbiolumi- j nescence is clearly not prohibitive, since it is so common in the ocean, it does lppear to limit the phenomenon to environments with low light levels. ;luo- j lppear to limit the phenomenon to environments with low light levels. Fluorescence, in contrast, can be used in the bright near-surface waters of the tropics, which have led some of us in the visual ecology community to refer to it as the poor man\u2019s bioluminescence. \u2013 p.189<\/li>\n<li>Second, fluorescence is most likely to have an ecological purpose when it occurs in an environment with bright, nearly monochromatic, short-wavelength light. [\u2026] Longer wavelength monochromatic illumination works as well, but because the fluorescence has an even longer wavelength , and should be fairly separated from the excitation to be of any real benefit, you run out of visible spectrum sooner. Third, for fluorescence to be seen with any real contrast, the viewer has to be less sensitive to [he shorter wavelengths of light in the environment. As we discussed ill chapter 4, visual pigments have broad absorption spectra, so the best way to limit short-wavelength sensitivity is to have a filter that blocks short wavelength light. Therefore, look for cases in which the viewer of the fluorescence has a filter, which will most likely look yellow. \u2013 p.192<\/li>\n<\/ul>\n<h3>Polarization<\/h3>\n<ul>\n<li>In many cases, direct sunlight being a good example. The e-vector averaged over any biologically relevant time period is random. Light of this type called \u201cunpolarized\u201d and is easy to deal with \u2013we just forget that polarization even exists. However, the e-vectors of many important Sources of Iight-skylight, reflected light, underwater light-are not entirely random. Light from these sources is said to be partially polarized. Because. While the e-vector isn\u2019t on one axis or cyclic, it is statistically biased toward certain angles and\/or handedness. \u2013 p.207<\/li>\n<li>This is where things get even easier, because it turns out that any beam of l&gt;artially polarized light can be considered the sum of two beams, one completely polarized and one completely unpolarized. The e-vectors in the unpolarized beam are random and those in the polarized beam trace out the simple ellipse we discussed above. \u2013 p.209 <i>Due to which property of a vector field?<\/i><\/li>\n<li>Direct sunlight, light in murky water, airglow, fire, lightning, fluorescence, triboluminescence, soniluminescence, and direct light from the full moon and the outer planets arc unpolarized. The solar corona; skylight; rainbows; halos; zodiacal light; comets; underwater light in the ocean; reflected light; the red aurora; emission from hot metals, lava and tourmaline; and the light from the non-full moon and the inner planets are sometimes polarized. With two exceptions (which we\u2019ll get to in a moment), what distinguishes the second group from the first is the presence of scattering. \u2013 p.212<\/li>\n<li>It is often stated that multiple scattering dillute the degree or polarization, the implication being that each scattering event is a corrupting process leading a once idealistic and highly polarized photon down a decadent path to unpolarized ruin. However, each scattering event with a small particle creates highly polarized light, but only at 90\u00b0 to the direction that the photon was first traveling. What we lose in a hazy atmosphere is directionality. When we look at a patch of clear blue sky that is 90 away from the sun, most of the light we see came directly from the sun and then was scattered 90\u00b0 toward our eye. However, if the sky is hazy, then the patch we look at includes light that came from other places, and thus did not scatter by 90\u00b0.lt is the addition of this light, along with the reduction of direct sunlight reaching our sky patch, that lowers the polarization. \u2013 p.214<\/li>\n<li>The only natural exceptions I know to the \u201cscattering is the cause of polarization\u201d rule are the red aurora (polarized by the earth\u2019s magnetic field) and light transmitted through dichroic minerals like tourmaline (one of the few dichroic substances of any real size in nature). \u2013 p.216<\/li>\n<li>All substances have different refractive indices for different wavelengths of Iight. [\u2026] Birefringent materials themselves do not polarize light, they only transmit the two polarization components at different phase velocities. This difference, though, changes the phase relationship between the two components and thus can convert one type of polarization into another. As we discussed earlier, in linearly polarized light, the two perpendicular polarization components have the same phase. If you send this light through a birefringent substance, however, one of the polarizations will get through faster than the other and the phase relationship will no longer be zero. In other words, you will have changed linear polarized light to elliptically polarized light. \u2013 p.218<\/li>\n<li>Form birefringence is due to the organization of the molecules, in particular long parallel bands of fibers or molecules. Substances like muscle and connective tissue therefore can have substantial form birefringence. Stretching certain plastics can also cause the molecules to line up. Creating a form birefringence that is called \u201cstress birefringence.\u201d \u2013 p.219<\/li>\n<li>The ability to sense the polarization of light has been divided into two types. One is known as \u201cpolarization sensitivity.\u201d Animals that have polarization sensitivity are not much different from humans wearing Polaroid sunglasses. Polarization affects the intensity of what they see-but without a lot of head-turning and thinking, they cannot reliably determine the angle or , degree of polarization or even separate polarization from brightness. The other type is known as \u201cpolarization vision.\u201d Animals with polarization vision perceive the angle and degree of polarization as something separate from simple brightness differences. Behaviorally, this means that they can distinguish two lights with different degrees and\/or angles of polarization regardless of their relative radiances and colors. This is much like the definition of color vision, which involves the ability to distinguish two lights of differing hue and\/or saturation regardless of their relative radiances. \u2013 p.220<\/li>\n<li>A simpler navigational use of polarized skylight is as a guide for traveling in a straight line. This sounds trivial until you actually try it. Swim the backstroke under an open sky or hike in a featureless terrain and you will soon find yourself going in circles. \u2013 p.230<\/li>\n<li>Because the sun often hits water at intermediate angles, the reflected glare is-usually polarized. This glare is annoying to humans and obstructs the view into the water, which is why fishermen\u2019s sunglasses are nearly always of the polarizing type. However, it can also be a way for an animal of little brain to identify water. \u2013 p.230<\/li>\n<li>Transparent tissues may depolarize the transmitted background light, or, if they are birefringent, alter the polarization from linear to elliptical. This latter possibility is especially problematic because, as we discussed, form birefringence is found in muscle, connective tissue, and other nutrient-rich substances. So your birefringent material, if visible, is more or less saying, \u201cI\u2019m good to eat.\u201d Mirrors have the related problem of potentially changing the polarization of the reflected light. \u2013 p.234<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>by Sonke Jonsen Units and geometry A spectrum generated using bins with equal frequency intervals would look different from one with equal wavelength intervals. So which one is correct? Neither and both. The take-home message is that the shape of a spectrum depends on whether you have equal frequency bins or equal wavelength bins. p.14 &#8230; <a title=\"The Optics of life\" class=\"read-more\" href=\"https:\/\/www.labri.fr\/perso\/barla\/blog\/?p=30401\" aria-label=\"Read more about The Optics of life\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[621],"tags":[],"class_list":["post-30401","post","type-post","status-publish","format-standard","hentry","category-books"],"_links":{"self":[{"href":"https:\/\/www.labri.fr\/perso\/barla\/blog\/index.php?rest_route=\/wp\/v2\/posts\/30401","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.labri.fr\/perso\/barla\/blog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.labri.fr\/perso\/barla\/blog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.labri.fr\/perso\/barla\/blog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.labri.fr\/perso\/barla\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=30401"}],"version-history":[{"count":3,"href":"https:\/\/www.labri.fr\/perso\/barla\/blog\/index.php?rest_route=\/wp\/v2\/posts\/30401\/revisions"}],"predecessor-version":[{"id":30421,"href":"https:\/\/www.labri.fr\/perso\/barla\/blog\/index.php?rest_route=\/wp\/v2\/posts\/30401\/revisions\/30421"}],"wp:attachment":[{"href":"https:\/\/www.labri.fr\/perso\/barla\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=30401"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.labri.fr\/perso\/barla\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=30401"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.labri.fr\/perso\/barla\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=30401"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}