How Everything Works Home Page 10 Most Recent Questions and Answers (out of 1595) Printer Friendly Version

1595. If a penny fell from the Empire State Building, could it actually punch a hole in the sidewalk?
A famous urban legend states that a penny dropped from the top of the Empire State Building will punch a hole in the sidewalk below. Given the height of the building and the hardness of the penny, that seems like a reasonable possibility. Whether it's true or not is a matter that can be determined scientifically. Before we do that, though, let's get some background.

Falling rocks can be dangerous and, the farther they fall, the more dangerous they become. Falling raindrops, snowflakes, and leaves, however, are harmless no matter how far they fall. The distinction between those two possibilities has nothing to do with gravity, which causes all falling objects to accelerate downward at the same rate. The difference is entirely due to air resistance.

Air resistance—technically known as drag—is the downwind force an object experiences as air moves passed it. Whenever an object moves through the air, the two invariably push on one another and they exchange momentum. The object acts to drag the air along with it and the air acts to drag the object along with it, action and reaction. Those two aerodynamic forces affect the motions of the object and air, and are what distinguish falling snowflakes from falling rocks.

Two types of drag force affect falling objects: viscous drag and pressure drag. Viscous drag is the friction-like effect of having the air rub across the surface of the object. Though important to smoke and dust particles in the air, viscous drag is too weak to affect larger objects significantly.

In contrast, pressure drag is strongly affects most large objects moving through the air. It occurs when airflow traveling around the object breaks away from the object's surface before reaching the back of the object. That separated airflow leaves a turbulent wake behind the object—a pocket of air that the object is effectively dragging along with it. The wider this turbulent wake, the more air the object is dragging and the more severe the pressure drag force.

The airflow separation occurs as the airflow is attempting to travel from the sides of the object to the back of the object. At the sides, the pressure in the airflow is especially low due as it bends to arc around the sides. Bernoulli's equation is frequently invoked to help explain the low air pressure near the sides of the object. As this low-pressure air continues toward the back of the object, where the pressure is much greater, the airflow is moving into rising pressure and is pushed backward. It is decelerating.

Because of inertia, the airflow could be expected to reach the back of the object anyway. However, the air nearest the object's surface—boundary layer air—rubs on that surface and slows down. This boundary layer doesn't quite make it to the back of the object. Instead, it stops moving and consequently forms a wedge that shaves much of the airflow off of the back of the object. A turbulent wake forms and the object begins to drag that wake along with it. The airflow and object are then pushing on one another with the forces of pressure drag.

Those pressure drag forces depend on the amount of air in the wake and the speed at which the object is dragging the wake through the passing air. In general, the drag force on the object is proportional to the cross sectional area of its wake and the square of its speed through the air. The broader its wake and the faster it moves, the bigger the drag force it experiences.

We're ready to drop the penny. When we first release it at the top of the Empire State Building, it begins to accelerate downward at 9.8 meters-per-second2—the acceleration due to gravity—and starts to move downward. If no other force appeared, the penny would move according to the equations of motion for constant downward acceleration, taught in most introductory physics classes. It would continue to accelerate downward at 9.8 meters-per-second2, meaning that its downward velocity would increase steadily until the moment it hit sidewalk. At that point, it would be traveling downward at approximately 209 mph (336 km/h) and it would do some damage to the sidewalk.

That analysis, however, ignores pressure drag. Once the penny is moving downward through the air, it experiences an upward pressure drag force that affects its motion. Instead of accelerating downward in response to its weight alone, the penny now accelerates in response to the sum of two force: its downward weight and the upward drag force. The faster the penny descends through the air, the stronger the drag force becomes and the more that upward force cancels the penny's downward weight. At a certain downward velocity, the upward drag force on the penny exactly cancels the penny's weight and the penny no longer accelerates. Instead, it descends steadily at a constant velocity, its terminal velocity, no matter how much farther drops.

The penny's terminal velocity depends primarily on two things: its weight and the cross sectional area of its wake. A heavy object that leaves a narrow wake will have a large terminal velocity, while a light object that leaves a broad wake will have a small terminal velocity. Big rocks are in the first category; raindrops, snowflakes, and leaves are in the second. Where does a penny belong?

It turns out that a penny is more like a leaf than a rock. The penny tumbles as it falls and produces a broad turbulent wake. For its weight, it drags an awful lot of air behind it. As a result, it reaches terminal velocity at only about 25 mph (40 km/h). To prove that, I studied pennies fluttering about in a small vertical wind tunnel.

Whether the penny descends through stationary air or the penny hovers in rising air, the physics is the same. Of course, it's much more convenient in the laboratory to observe the hovering penny interacting with rising air. Using a fan and plastic pipe, I created a rising stream of air and inserted a penny into that airflow.

At low air speeds, the penny experiences too little upward drag force to cancel its weight. The penny therefore accelerated downward and dropped to the bottom of the wind tunnel. At high air speeds, the penny experienced such a strong upward drag force that it blew out of the wind tunnel. When the air speed was just right, the penny hovered in the wind tunnel. The air speed was then approximately 25 mph (40 km/h). That is the terminal velocity of a penny.

The penny tumbles in the rising air. It is aerodynamically unstable, meaning that it cannot maintain a fixed orientation in the passing airstream. Because the aerodynamic forces act mostly on the upstream side of the penny, they tend to twist that side of the penny downstream. Whichever side of the penny is upstream at one moment soon becomes the downstream side, and the penny tumbles. As a result of this tumbling, the penny disturbs a wide swath of air and leaves a broad turbulent wake. It experiences severe pressure drag and has a low terminal velocity.

The penny is an example of an aerodynamically blunt object—one in which the low-pressure air arcing around its sides runs into the rapidly increasing pressure behind it and separates catastrophically to form a vast wake. The opposite possibility is an aerodynamically streamlined object—one in which the increasing pressure beyond the object's sides is so gradual that the airflow never separates and no turbulent wake forms. A penny isn't streamlined, but a ballpoint pen could be.

Almost any ballpoint pen is less blunt than a penny and some pens are approximately streamlined. Moreover, pens weigh more than pennies and that fact alone favors a higher terminal velocity. With a larger downward force (weight) and a smaller upward force (drag), the pen accelerates to a much greater terminal velocity than the penny. If it is so streamlined that it leaves virtually no wake, like the aerofoil shapes typical of airplane components, it will have an extraordinarily large terminal velocity—perhaps several hundred miles per hour.

Some pens tumble, however, and that spoils their ability to slice through the air. To avoid tumbling, a pen must "weathervane"—it must experience most of its aerodynamic forces on its downstream side, behind its center of mass. Arrows and small rockets have fletching or fins to ensure that they travel point first through the air. A ballpoint pen can achieve that same point-first flight if its shape and center of mass are properly arranged.

Almost any ballpoint pen dropped into my wind tunnel plummeted to the bottom. I was unable to make the air rise fast enough to observe hovering behavior in those pens. Whether they would tend to tumble in the open air was difficult to determine because of the tunnel's narrowness. Nonetheless, it's clear that a heavy, streamlined, and properly weighted pen dropped from the Empire State Building would still be accelerating downward when it reached the sidewalk. Its speed would be close to 209 mph at that point and it would indeed damage the sidewalk.

As a final test of the penny's low terminal velocity, I built a radio-controlled penny dropper and floated it several hundred feet in the air with a helium-filled weather balloon. On command, the dropper released penny after penny and I tried to catch them as they fluttered to the ground. Alas, I never managed to catch one properly in my hands. It was a somewhat windy day and the ground at the local park was uneven, but that's hardly an excuse—I'm simply not good at catching things in my hands. Several of the pennies did bounce off my hands and one even bounced off my head. It was fun and I was more in danger of twisting my ankle than of getting pierced by a penny. The pennies descended so slowly that they didn't hurt at all. Tourist below the Empire State Building have nothing fear from falling pennies. Watch out, however, for some of the more streamlined objects that might make that descent.

1594. Are smart meters bad for people’s health? Is this actually not knowable at this time? -- ED
If by smart meters you mean the devices that monitor power usage and possibly adjust power consumption to periodically, then I don't see how they can affect health. Their communications with the smart grid are of no consequence to human health and having the power adjusted on household devices is unlikely to be a health issue (unless they cut off your power during a blizzard or a deadly heat wave).

The radiated power from all of these wireless communications devices is so small that we have yet to find mechanisms whereby they could cause significant or lasting injury to human tissue. If there is any such mechanism, the effects are so weak that the risk associated with it are dwarfed by much more significant risks of wireless communication: the damage to traditional community, the decline of ordinary human interaction, and the surge in distracted driving.

1593. Why can't the Japanese stop the chain reaction in the Fukushima Daiichi nuclear reactors? — FE
The Japanese did stop the chain reactions in the Fukushima Daiichi reactors, even before the tsunami struck the plant. The problem that they're having now is not the continued fissioning of uranium, but rather the intense radioactivity of the uranium daughter nuclei that were created while the chain reactions were underway. Those radioactive fission fragments are spontaneously decaying now and there is nothing that can stop that natural decay now. All they can do now is to try to contain those radioactive nuclei, keep them from overheating, and wait for them to decay into stable pieces.

The uranium atom has the largest naturally occurring nucleus in nature. It contains 92 protons, each of which is positively charged, and those 92 like charges repel one another ferociously. Although the nuclear force acts to bind protons together when they touch, the repulsion of 92 protons alone would be too much for the nuclear force—the protons would fly apart in almost no time.

To dilute the electrostatic repulsion of those protons, each uranium nucleus contains a large number of uncharged neutrons. Like protons, neutrons experience the attractive nuclear force. But unlike protons, neutrons don't experience the repulsive electrostatic force. Two neutron-rich combinations of protons and neutrons form extremely long-lived uranium nuclei: uranium-235 (92 protons, 143 neutrons) and uranium-238 (92 protons, 146 neutrons). Each uranium nucleus attracts an entourage of 92 electrons to form a stable atom and, since the electrons are responsible for the chemistry of an atom, uranium-235 and uranium-238 are chemically indistinguishable.

When the thermal fission reactors of the Fukushima Daiichi plant were in operation, fission chain reactions were shattering the uranium-235 nuclei into fragments. Uranium-238 is more difficult to shatter and doesn't participate much in the reactor's operation. On occasion, however, a uranium-238 nucleus captures a neutron in the reactor and transforms sequentially into neptunium-239 and then plutonium-239. The presence of plutonium-239 in the used fuel rods is one of the problems following the accident.

The main problem, however, is that the shattered fission fragment nuclei in the used reactor fuel are overly neutron-rich, a feature inherited from the neutron-rich uranium-235 nuclei themselves. Midsize nuclei, such as iodine (with 53 protons), cesium (with 55 protons), and strontium (with 38 protons), don't need as many neutrons to dilute out the repulsions between their protons. While fission of uranium-235 can produce daughter nuclei with 53 protons, 55 protons, or 38 protons, those fission-fragment versions of iodine, cesium, and strontium nuclei have too many neutrons and are therefore unstable—they undergo radioactive decay. Their eventual decay has nothing to do with chain reactions and it cannot be prevented.

How quickly these radioactive fission fragment nuclei decay depends on exactly how many protons and neutrons they have. Three of the most common and dangerous nuclei present in the used fuel rods are iodine-131 (8 days half-life), cesium-137 (30 year half-life), and strontium-90 (29 year half-life). Plutonium-239 (24,200 year half-life) is also present in those rods. When these radioactive nuclei are absorbed into the body and then undergo spontaneous radioactive decay, they damage molecules and therefore pose a cancer risk. Our bodies can't distinguish the radioactive versions of these chemical elements from the nonradioactive ones, so all we can do to minimize our risk is to avoid exposure to them or to encourage our bodies to excrete them by saturating our bodies with stable versions.

1592. Suppose you have a metal sphere in vacuum and you begin putting electric charge on that sphere. Neglecting possible discharges, how much charge can the sphere store? An unlimited amount? -- BC
By asking me to "neglect possible discharges," you're asking me to neglect what actually happens. There will be a discharge, specifically a phenomenon known as "field emission". Neglect that discharge, then yes, the sphere can in principle store an unlimited amount of charge. But on route to infinity, I will have had to ignore several other exotic discharges and then the formation of a black hole.

What will really happen is a field emission discharge. The repulsion between like charges will eventually become so strong that those charges will push one another out of the metal and into the vacuum, so that charges will begin to stream outward from the metal sphere.

Another way to describe that growing repulsion between like charges involves fields. An electric charge is surrounded by a structure in space known as an electric field. An electric field exerts forces on electric charges, so one electric charge pushes on other electric charges by way of its electric field.

As more and more like charges accumulate on the sphere, their electric fields overlap and add so that the overall electric field around the sphere becomes stronger and stronger. The charges on the sphere feel that electric field, but they are bound to the metal sphere by chemical forces and it takes energy to pluck one of them away from the metal.

Eventually, the electric field becomes so strong that it can provide the energy needed to detach a charge from the metal surface. The work done by the field as it pushes the charge away from sphere supplies the necessary energy and the charge leaves the sphere and heads out into the vacuum. The actually detachment process involves a quantum physics phenomenon known as tunneling, but that's another story.

The amount of charge the sphere can store before field emission begins depends on the radius of the sphere and on whether the charge is positive or negative. The smaller that radius, the faster the electric field increases and the sooner field emission starts. It's also easier to field-emit negative charges (as electrons) than it is to field-emit positive charges (as ions), so a given sphere will be able to hold more positive charge than negative charge.

1591. I have an active but paraplegic friend who is building an electric off-road scooter using DC motors. Those motors will have to reverse directions frequently while under load. Will they tolerate immediate reversals, or must there be a delay? — JO, Valley Springs, California
Modern brushless DC motors are amazing devices that can handle torque reversals instantly. In fact, they can even generate electricity during those reversals!

Instant reversals of direction, however, aren't physically possible (because of inertia) and aren't actually what your friend wants anyway. I'll say more about the distinction between torque reversals and direction reversals in a minute.

In general, a motor has a spinning component called the rotor that is surrounded by a stationary component called the stator. The simplest brushless DC motor has a rotor that contains permanent magnets and a stator that consists of electromagnets. The magnetic poles on the stator and rotor can attract or repel one another, depending on whether they like or opposite poles—like poles repel; opposite poles attract.

Since the electronics powering the stator's electromagnets can choose which of the stator's poles are north and which are south, those electronics determine the forces acting on the rotor's poles and therefore the direction of torque on the rotor. To twist the rotor forward, the electronics make sure that the stator's poles are always acting to pull or push the rotor's poles in the forward direction so that the rotor experiences forward torque. To twist the rotor backward, the electronics reverses all those forces.

Just because you reverse the direction of torque on the rotor doesn't mean that the rotor will instantly reverse its direction of rotation. The rotor (along with the rider of the scooter) has inertia and it takes time for the rotor to slow to a stop and then pick up speed in the opposite direction. More specifically, a torque causes angular acceleration; it doesn't cause angular velocity. During that reversal process, the rotor is turning in one direction while it is being twisted in the other direction. The rotor is slowing down and it is losing energy, so where is that energy going? It's actually going into the electronics which can use this electricity to recharge the batteries. The "motor" is acting as a "generator" during the slowing half of the reversal!

That brushless DC motors are actually motor/generators makes them fabulous for electric vehicles of all types. They consume electric power while they are making a vehicle speed up, but they generate electric power while they are slowing a vehicle down. That's the principle behind regenerative braking—the vehicle's kinetic energy is used to recharge the batteries during braking.

With suitable electronics, your friend's electric scooter can take advantage of the elegant interplay between electric power and mechanical power that brushless DC motors make possible. Those motors can handle torque reversals easily and they can even save energy in the process. There are limits, however, to the suddenness of some of the processes because huge flows of energy necessitate large voltages and powers in the motor/generators and their electronics. The peak power and voltage ratings of all the devices come into play during the most abrupt and strenuous changes in the motion of the scooter. If your friend wants to be able to go from 0 to 60 or from 60 to 0 in the blink of eye, the motor/generators and their electronics will have to handle big voltages and powers.

1590. Does light speed up as it gets further from the sun? The force of gravity has an effect on it, right? — JF, Ireland
Although that sounds like a simple question, it has a complicated answer. Gravity does affect light, but it doesn't affect light's speed. In empty space, light is always observed to travel at "The Speed of Light." But that remark hides a remarkable result: although two different observers will agree on how fast light is traveling, they may disagree in their perceptions of space and time.

When those observers are in motion relative to one another, they'll certainly disagree about the time and distance separating two events (say, two firecrackers exploding at separate locations). For modest relative velocities, their disagreement will be too small to notice. But as their relative motion increases, that disagreement will become substantial. That is one of the key insights of Einstein's special theory of relativity.

But even when two observers are not moving relative to one another, gravity can cause them to disagree about the time and distance separating two events. When those observers are in different gravitational circumstances, they'll perceive space and time differently. That effect is one of the key insights of Einstein's general theory of relativity.

Here is a case in point: suppose two observers are in separate spacecraft, hovering motionless relative to the sun, and one observer is much closer to the sun than the other. The closer observer has a laser pointer that emits a green beam toward the farther observer. Both observers will see the light pass by and measure its speed. They'll agree that the light is traveling at "The Speed of Light". But they will not agree on the exact frequency of the light. The farther observer will see the light as slightly lower in frequency (redder) than the closer observer. Similarly, if the farther observer sends a laser pointer beam toward the closer observer, the closer observer will see the light as slightly higher in frequency (bluer) than the farther observer.

How can these two observers agree on the speed of the beams but disagree on their frequencies (and colors)? They perceive space and time differently! Time is actually passing more slowly for the closer observer than for the farther observer. If they look carefully at each others' watches, the farther observer will see the closer observer's watch running slow and the closer observer will see the farther observer's watch running fast. The closer observer is actually aging slightly more slowly than the farther observer.

These effects are usually very subtle and difficult to measure, but they're real. The global positioning system relies on ultra-precise clocks that are carried around the earth in satellites. Those satellites move very fast relative to us and they are farther from the earth's center and its gravity than we are. Both difference affect how time passes for those satellites and the engineers who designed and operate the global positioning system have to make corrections for the time-space effects of special and general relativity.

1589. I understand that water changes to its gaseous state when boiled at 212 degrees F. But why does that same water, in the form of falling rain, change to its gaseous state at a much lower temperature after it strikes a hot surface, such as a paved road? — JA, North Carolina
Liquid water can evaporate to form gaseous water (i.e., steam) at any temperature, not just at its boiling temperature of 212 F. The difference between normal evaporation and boiling is that, below water's boiling temperature, evaporation occurs primarily at the surface of the liquid water whereas at or above water's boiling temperature, bubbles of pure steam become stable within the liquid and water can evaporate especially rapidly into those bubbles. So boiling is a just a rapid form of evaporation.

What you are actually seeing when raindrops land on warm surfaces is tiny water droplets in the air, a mist of condensation. Those droplets happen in a couple of steps. First, the surface warms a raindrop and speeds up its evaporation. Second, a small portion of warm, especially moist air rises upward from the evaporating raindrop. Third, that portion of warm moist air cools as it encounters air well above the warmed surface. The sudden drop in temperature causes the moist air to become supersaturated with moisture—it now contains more water vapor than it can retain at equilibrium. The excess moisture condenses to form tiny water droplets that you see as a mist.

This effect is particularly noticeable when it's raining because the humidity in the air is already very near 100%. The extra humidity added when the warmed raindrops evaporate is able to remain gaseous only in warmed air. Once that air cools back to the ambient temperature, the moisture must condense back out of it, producing the mist.

1588. Why does ice float on water? — Y, Laguna
Solid ice is less dense than liquid water, meaning that a liter of ice has less mass (and weighs less) than a liter of water. Any object that is less dense than water will float at the surface of water, so ice floats.

That lower-density objects float on water is a consequence of Archimedes' principle: when an object displaces a fluid, it experiences an upward buoyant force equal in amount to the weight of the displaced fluid. If you submerge a piece of ice completely in water, that piece of ice will experience an upward buoyant force that exceeds the ice's weight because the water it displaces weighs more than the ice itself. The ice then experiences two forces: its downward weight and the upward buoyant force from the water. Since the upward force is stronger than the downward force, the ice accelerates upward. It rises to the surface of the water, bobs up and down a couple of times, and then settles at equilibrium.

At that equilibrium, the ice is displacing a mixture of water and air. Amazingly enough, that mixture weighs exactly as much as the ice itself, so the ice now experiences zero net force. That's why its at equilibrium and why it can remain stationary. It has settled at just the right height to displace its weight in water and air.

As for why ice is less dense than water, that has to do with the crystal structure of solid ice and the more complicated structure of liquid water. Ice's crystal structure is unusually spacious and it gives the ice crystals their surprisingly low density. Water's structure is more compact and dense. This arrangement, with solid water less dense than liquid water, is almost unique in nature. Most solids are denser than their liquids, so that they sink in their liquids.

1587. When electricity comes out of the wall and through a lamp, where does the circuit loop complete? Does the circuit go all the way back to the power plant? — J, Florida
The electric circuit that powers your lamp extends only as far as a nearby transformer. That transformer is located somewhere near your house, probably as a cylindrical object on a telephone pole down the street or as a green box on a side lawn a few houses away.

A transformer conveys electric power from one electric circuit to another. It performs this feat using several electromagnetic effects associated with changing electric currents—changes present in the alternating current of our power grid. In this case, the transformer is moving power from a high-voltage neighborhood circuit to a low-voltage household circuit.

For safety, household electric power uses relatively low voltages, typically 120 volt in the US. But to deliver significant amounts of power at such low voltages, you need large currents. It's analogous to delivering hydraulic power at low pressures; low pressures are nice and safe, but you need large amounts of hydraulic fluid to carry much power. There is a problem, however, with sending low voltage electric power long distances: it's inefficient because wires waste power as heat in proportion to the square of the electric current they carry. Using our analogy again, sending hydraulic power long distances as a large flow of hydraulic fluid at low pressure is wasteful; the fluid will rub against the pipes and waste power as heat.

To send electric power long distances, you do better to use high voltages and small currents (think high pressure and small flows of hydraulic fluid). That requires being careful with the wires because high voltages are dangerous, but it is exactly how electric power travels cross-country in the power grid: very high voltages on transmission lines that are safely out of reach.

Finally, to move power from the long-distance high-voltage transmission wires to the short-distance low-voltage household wires, they use transformers. The long-distance circuit that carries power to your neighborhood closes on one side of the transformer and the short-distance circuit that carries power to your lamp closes on the other side of the transformer. No electric charges pass between those two circuits; they are electrically insulated from one another inside the transformer. The electric charges that are flowing through your lamp go round and round that little local circuit, shuttling from the transformer to your lamp and back again.

1586. I have a 70 to 300 mm lens with f-5.6. But I can manually take it up to f-22. What does that mean and how does it work? Also why can't I bring it down to say f2.8? — AR, Pakistan
The f-number of a lens measures the brightness of the image that lens casts onto the camera's image sensor. Smaller f-numbers produce brighter images, but they also yield smaller depths of focus.

The f-number is actually the ratio of the lens' focal length to its effective diameter (the diameter of the light beam it collects and uses for its image). Your zoom lens has a focal length that can vary from 70 to 300 mm and a minimum f-number of 5.6. That means the when it is acting as a 300 mm telephoto lens, its effective light gathering surface is about 53 mm in diameter (300 mm divided by 5.6 gives a diameter of 53 mm).

If you examine the lens, I think that you'll find that the front optical element is about 53 mm in diameter; the lens is using that entire surface to collect light when it is acting as a 300 mm lens at f-5.6. But when you zoom to lower focal lengths (less extreme telephoto), the lens uses less of the light entering its front surface. Similarly, when you dial a higher f-number, you are closing a mechanical diaphragm that is strategically located inside the lens and causing the lens to use less light. It's easy for the lens to increase its f-number by throwing away light arriving near the edges of its front optical element, but the lens can't decrease its f-number below 5.6; it can't create additional light gathering surface. Very low f-number lenses, particularly telephoto lenses with their long focal lengths, need very large diameter front optical elements. They tend to be big, expensive, and heavy.

Smaller f-numbers produce brighter images, but there is a cost to that brightness. With more light rays entering the lens and focusing onto the image sensor, the need for careful focusing becomes greater. The lower the f-number, the more different directions those rays travel and the harder it is to get them all to converge properly on the image sensor. At low f-numbers, only rays from a specific distance converge to sharp focus on the image sensor; rays from objects that are too close or too far from the lens don't form sharp images and appear blurry.

If you want to take a photograph in which everything, near and far, is essentially in perfect focus, you need to use a large f-number. The lens will form a dim image and you'll need to take a relatively long exposure, but you'll get a uniformly sharp picture. But if you're taking a portrait of a person and you want to blur the background so that it doesn't detract from the person's face, you'll want a small f-number. The preferred portrait lenses are moderately telephoto—they allow you to back up enough that the person's face doesn't bulge out at you in the photograph—and they have very low f-numbers—their large front optical elements gather lots of light and yield a very shallow depth of focus.

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