Light wave: a perpendicular sinusoidally oscillating Electric Field (E) and Magnetic Field (B) that propagate (move) in yet a third perpendicular direction. The animation at right shows this pretty well. It shows the direction and magnitude of the two vectors over time. It also documents the fixed inter-wavefront distance, a.k.a. wavelength (λ). So if Light is a wave, what is it displacing? The fields themselves, which is why it is the only wave which doesn’t need a medium (i.e.: it can travel in a vacuum).
The vector we use to represent the orientation of a light wave as it moves through space is the Electric Field vector. In the image at above right, the Electric Field Vector happened to point straight up. However, the Electric Field can point in any direction that is perpendicular to the direction of propagation. In the diagram below, the light beam (collection of different light waves) is coming in from the right. Initially, there are multiple orientations. A polarizer is a screen with an array of parallel metal wires. By passing the light through a plane polarizer, only the component vectors of the electric field vectors that are parallel to the wires pass through. At this point, the light wave is polarized linearly.
By passing linearly polarized light through a quarter wave plate, you can turn it into circularly polarized light. That is light where the electric field vector constantly spins around the axis of propagation, as in the image at right. This is a lower-yield topic than linear polarization, which is used in stereochemistry classification. The clinical application of circular polarization is birefringent microscopy, used to identify gouty crystals and amyloid proteins in certain protein-deposition diseases.
The Electromagnetic Spectrum is the model we have to unify all the different types of light (EM radiation). As long as the Electric and Magnetic fields oscillate as described earlier, we have light. The frequency of the oscillation determines the “color” of light. We humans can see only a teensy fraction of all the possible frequencies (and thus wavelengths) of light (~700 nm to ~400 nm). We are effectively “colorblind” to the universe. The visible light spectrum is known by the mnemonic “ROYGBIV,” from low to high energy, and fits into the whole EM spectrum as shown below:
Type of EM Radiation
Tip to learn position in spectrum
When created (not on MCAT, for context only)
The most powerful light in the known Universe.
Created by the Big Bang only, once in all of spacetime.
Made the “Incredible HULK”, a VERY high energy human.
Produced as a result of the interconversion of matter and energy (Nuclear phenomenon). Hence the name γ decay (other types being α and β ).
Strong enough to punch through a human body to make an image (X-ray).
The types of light rays produced by the excitation and relaxation of inner-shell electrons of heavy metals.
Name is “Above Violet”
Name is “Below Red”
Sun (or re-radiated by asphalt in summer)
Name means “smaller wavelength than radio”
Gigantic. Human→ Building sized
(λ ~1m to 100m)
Radio towers on Buildings.
The reason we evolved to see such a narrow spectrum has to do with the photoreactivity of biological pigments. According to the photoelectric effect, radio waves are too low-energy to excite electrons in most metals, and UV will kick those same metal electrons straight off the atom, ionizing it. If the electron were flung off into space, its energy would be lost to the plant. A “Goldilocks” level of invested light energy leads to the capture of the electron and its associated potential energy. Magnesium is the metal ion in chlorophyll, and it absorbs red or blue light as shown in the diagram at right. Since what is not absorbed is reflected, this is the reason that plants, and thus our Earthly environment, is green.
In the autumn, as the leaves senesce, different pigments are expressed which reflect different parts (red and yellow) of the visible spectrum. Likely, the ability to capture these energies of light is why the “visible spectrum” is visible. Xenobiologists make a good argument for alien species seeing these frequencies for the same reasons we do.
The word “photon” is the name given to light when we treat it like a particle. However, we use this interchangeably with the phrase “light wave.” Max Planck figured out that for a given frequency of light, there is a specific amount of energy associated with wavelength. The formula that relates the two is:
Energy of a single wave of Light = h× frequency in Hz
The constant h is called Planck’s constant, and equals 6.6 x 10-34 J∙s. Looking at the units, if:
f = # cycles/sec
then Planck’s constant must have the units as follows:
h=(Energy/cycle) × time
Which means that it if we were to draw a photon out as the function of the Electric Field vector over time, we would see a sinusoidal graph like the ones below for different colors of visible light.
Planck’s constant would then be the yellow area, the area under one oscillation (aka one wavelength, one sine curve). It would represent the “packet” or quantum of energy delivered by one cycle of oscillation of light. The faster a given light wave oscillates, the more frequently those packets of Energy are delivered to the detector (your eye). As you can see, in the same time (width of the picture), the violet light delivers 10 quanta, but the red wave delivers just over 2.
So, in practical terms, higher frequency light is more energetic, because it delivers equally-sized packets of energy to your eye more frequently.
Note about Power of Light beams: E=hf tells us about the Energy contained in one whole infinitely long wave, or the photon equivalent. Planck’s constant only tells us about the size of the packet of energy delivered by one oscillation. Use E=hf to talk about the energy of photons as countable items. Thus, to calculate the Power of a light source (rate of Energy delivery), we would need calculate the rate of photon delivery, which would also equal the rate of photon generation. You can’t really answer questions about noncoherent (not laser) light without calculus, so don’t worry about it. Remember, though, if the question does talk about a laser, all the energy delivery will add constructively. Use units as a guide, and you will usually get the right answer based on that alone.
When a wave passes through a gap, like the rocks at right, it radiates out in circles from the gap. Clearly, diffraction happens to waves other than light. In fact, diffraction, reflection, dispersion, and refraction happen to all waves. Notice that there are actually two slits in the rocks in the picture at right, causing an interference pattern between the two diffracted waves. This is akin to Young’s double-slit experiment, which showed that light acts like a wave (the answer the MCAT wants for what Young’s experiment tells us).
Later versions of the experiment showed that recording the results and monitoring what was passing through the slits actually caused light to act like a particle, violating the earlier results. Weird, and fun to contemplate, but not as relevant as the vocab: bright fringes and dark fringes are what the regions of constructive and destructive interference are called, respectively. They are shown as they would appear on the projection screen below at left. The picture to the right gives a bird’s eye view of how the light waves that are diffracting out of the two slits interfere to make the bright and dark fringes.
When a light ray hits a reflective surface, it bounces back at an angle away from the normal line that is determined by the Law of Reflection:
θincident = θreflected
θi = θr
Both θi and θr are defined as the angle the respective ray makes with the line that is normal to the surface.
This law applies to both planar mirrors and curved mirrors (see the section on Geometric Optics below).
When light is in vacuum, it travels without banging into anything, so its speed is constant, and is the fastest thing in the Universe:
c = 3.0 x 108 m/s
In dense but translucent materials (glass, plastic, water), there are atoms. Light gets absorbed and re-emitted by their electron clouds as it passes through the medium. This process is not instantaneous. The denser the material, the slower the light travels. We define the index of the refraction by the following formula:
↑ Density → ↑ index of refraction (n)
Thus, by definition, vacuum has an nvacuum = 1. Air is mostly empty space, so nair = 1.0003, which is effectively 1, and should be treated as such for the MCAT. The densest medium is diamond ndiamond = 2.42.
Snell’s Law quantifies refraction. The subscripts 1 and 2 in the above diagram are arbitrary. When solving a problem, either the initial medium or the second medium could be denser. In the image above, the light ray is bending toward the normal, so it must be going from less dense to more dense. If you back-traced ray Q, and sent a light ray along that line but in the other direction, it would be leaving the dense material, and would bend away from the normal along ray P. The equation is:
n1sin θ1 = n2sin θ2
The above diagram shows what happens as you increase θincident from 0∘ when going from a dense to a less dense medium. Initially, you refract away. As θincident increases, eventually, θrefracted = 90∘. At this point, θincident = θcritical, the incident angle at which light rays will refract until they are parallel to the interface, never leaving the medium. As θincident increases further, θrefracted eventually reaches > 90∘. At this point, the refracted ray actually bounces back inside the medium. This is a phenomenon called total internal reflection.
Clinically, this is the basis of the fiber optic cables which allow endoscopes to transmit an image from within the body effectively undiminished in intensity to an eyepiece or camera located outside the body.
When any wave enters a slower medium at a right angle to the surface of the new medium, it passes straight through unmolested. Plug the values into Snell’s Law for yourself (hint: sin 90∘ = 1) to test it out.
If the wave enters a planar refractive optic (a pane of glass) at an angle, it refracts towards the normal on entry, and refracts away an equal amount upon leaving. This leads to no image distortion, but does create an offset of where the image appears to be compared to where the original object (light source) actually is. Try looking out of your window at an angle, then raising the window sash to run this experiment yourself.
The higher energy the light wave, the more it refracts. This is the phenomenon called dispersion, seen in the prism at right.
Mnemonic: Prisms Point Purple Preferentially
Example Problem: Diamond has an index of refraction of 2.42. What is the critical angle?
n1sin θ1 = n2sin θ2
n1sin θc = n2sin 90o = n2
θc = sin-1 n2/n1
θc = sin-1 nair/ndiamond
θc = sin-1 1/2.42 = sin-1 0.413
θc = 24.4o
Thus, since diamond has the highest index of refraction (2.42), it can reflect internally even when the θincident of the light wave trying to leave is pretty small. The “brilliant” cut of a diamond ensures that there are many surfaces that can intersect with exiting light waves so that θincident > 25o, “trapping” the light inside and forcing it to bounce around a few times before finally escaping. This long path length allows enough time for visibly appreciable dispersion to occur. This is the reason for diamonds being so sparkly!
De-emphasized on the MCAT, but you should be able to follow them. Some really clear ones are found at HyperPhysics.
, all variables are in units of distance (usu. cm)
Meaning of positive value
Meaning of negative value
(there is a place in the real world for parallel incident light rays to focus)
(no place to focus in real world)
(Projectable onto a screen in the real world)
(The incident light rays are originating from an object in the real world, and thus the object is on the same side as the light rays going in to the optic)
(Object on other side from light rays going in. ONLY seen in complex optics, where image of one lens serves as object of next)
TIP: ALL real images coming out of a single optic are inverted.
ALL virtual images coming out of a single optic are upright.
Ray diagrams are useless on the MCAT unless you are checking whether individual optics splay or focus rays in a complex system. The focal length equation will get you quantitative answers. It is slow and chunky when it comes to answering conceptual problems, with which the test is replete. I have seen several questions just asking what the definition of focal point is: “The place where parallel incident light rays converge.” Focal length is the distance between the center of the optic and the focal point. A source that is close will emit rays that splay out from the source, putting an angle between the rays. The sun does this, too, but the distance of the radius of the Earth’s orbit is so great that the sun’s rays come in effectively parallel to one another. So, on the MCAT: Distant light source = Distant object = parallel incident rays.
The best algorithm for quick solution and understanding of optics is two-stage. Figure out what kind of system you are dealing with, then look at the relationship between object distance (o) and focal length, (f).
Convergent lenses and mirrors act the same, regardless of whether they are lenses or mirrors: they bring the rays closer together. Thus, statements about one type of simple convergent optic is true for all convergent systems (concave mirror or convex lens).
Divergent systems all act the same: they cause the rays to splay apart.
Both convergent and divergent systems are spherical optics, with a focal point at ½ of the radius of the circle they describe. The center of the circle, therefore is a distance of 2f from the center of the optic.
It is best to learn about optics by examining your real-life experience with some common simple optics. The following example optics were chosen for accessibility. Map what you learn below onto your experience with these objects.
Glasses for nearsightedness
Lady’s Makeup Mirror
Hallway corner mirror
Since f is negative for divergent systems, any real object (positive o) will lead to a virtual image:
. Sign analysis is going to tell us about the sign of i.
. Therefore i will be negative, and the image virtual.
Just remember, the best way to understand this is simply the divergent mirror (hallway corner mirror): For divergent systems, the image will be upright, virtual, and shrunken.
This holds true for divergent lenses, too - like lenses for myopia.
For convergent systems, just use the magnifying glass as your example optic. Everything about the convergent mirror (makeup mirror) is exemplified by real- life truths about the magnifying glass. The image below is the simplest way to understand everything about convergent systems. For the OCD in you, see below for the validation of the diagram using the focal length and magnification equations. The arrows show what happens when the object distance falls in the ranges relative to f and and 2f.
Summary: This flowchart outlines the simple approach to optics. Get used to using it, and learn how to draw the above diagram for convergent system quickly from memory.
Optics are generally not perfect. Two classes of image aberrations are testable on the MCAT.
Technically, for parallel light rays that are some distance from the optical axis to converge to form a perfect image at the focal point, the optic shouldn’t be spherical. It should be parabolic, as in the image below. In practice, milling mirrors into parabolas is much more expensive than milling them into spherical shapes.
Thus, only certain applications mandate parabolic mirrors. A light source at the focal point of a parabolic car headlamp will produce parallel outgoing light rays. This results in an even spread of light on the roadway for increased safety.
Because different colors of light refract to different degrees (Prisms Point Purple Preferentially), lenses will not focus all the different colors of light perfectly. This effect increases the farther away the light ray passes through the optic from the optical axis (see image below).
You can check this out with any glasses for myopia. The effect is noticeable if you look through the edge of the lens more than the center. In the image above, there is a red and blue aberration around the temple of the glasses.
. Sign analysis is going to tell us about the sign of i again in each case
, so by assigning common denominators on the left:
, by combining terms: , and by flipping both sides:
, so let’s plug in to the magnification equation:
, so the magnification equation can be rewritten as:
for all types of optics.
So, let’s use the above to verify the summary in the image above.
if o = f,
, so , so . No image will form
if o < f,
, and plugging in to the focal length equation:
. Therefore i will be negative, and the image virtual.
f, since 0 < o < f
, in other words, the image will be upright and magnified
if o > f,
, and plugging in to the focal length equation:
. Therefore i will be positive, and the image real.
SUB-CASES of o > f:
if f < o < 2f,
, since f < o < 2f
, in other words, the image will be inverted, magnified, and real.
if o = 2f
, in other words, the image will be inverted, equal in size to the object, and real.
if o > 2f
, since o > 2f
, in other words, the image will be inverted, shrunken, and real.