Ask the Physicist!

With the recent publication of PHYSICS IS... there are now three Ask the Physicist books! Click on the book images below for information on the content of the books and for information on ordering.

QUESTION:
Why we cannot feel the momentum or the force of the sunlight?

ANSWER:
The most intense light from the sun is about λ=550 nm=5.5x10^-7 m (yellow); the corresponding frequency is f=c/λ=3x10⁸/5.5x10^-7=5.45x10¹⁴ Hz. The momentum of one photon is p=hf/c=6.6x10^-34x5.45x10¹⁴/3x10⁸=1.2x10^-27 kg·m/s. The photon flux (photons per second per square meter) for photons of this wavelength is about Φ=3x10²¹ s^-1·m^-2. So if I estimate your area to be about 1 m², there are about 3x10²¹ photons hitting you every second. The total momentum imparted to you over that second (assuming that you absorb the photons) is about Δp=3x10²¹x1.2x10^-27=3.6x10^-6 kg·m/s. The average force you feel is about F=Δp/Δt=3.6x10^-6N; and this force is spread uniformly over your area. This is really small; for example, if we take the typical weight of a one dust particle to be about 10^-8 N, the weight of about 360 dust particles spread over your whole body is about what you "feel" from the sunlight.

QUESTION:
Assume there are 2 black bodies, 1 with a mass double of the other. If both black bodies reach a temperature of 5000K, do both emit blue light as black body radiation, with the heavier body emitting a brighter light. Or does the heavier body emit white light while the lighter body emits blue light? Why?

ANSWER:
A black body does not radiate one color but rather a spectrum of colors. The spectrum for 5000 K is shown above. The most intense color of this spectrum is at 580 nm which is the green color shown to the right. The intensity of the light emitted does not depend on the mass, it depends on the surface area of the black body; if you double surface area, you double the amount of energy being radiated.

QUESTION:
The demonstration wherein one pours water from a clear vessel and one shines a laser pointer (a red laser pointer in my case) through the water and it appears the laser beam bends with the water pouring out of the clear vessel; What causes the beam to bend with the water? Would this same type of effect work with other fluids than water, such as air or glass?

ANSWER:
It is caused by total internal reflection. If the angle which the light hits the surface between the water and the air is glancing enough, the light will be reflected back into the water rather than be refracted into the air. Total internal reflection can happen whenever light strikes the interface with a material of smaller index of refraction; the minimum angle for which it can occur depends on the relative indices of refraction.

QUESTION:
If matter above absolute zero emits thermal radiation—and thermal radiation is actually electromagnetic radiation—what temperature is required for matter to emit visible light?

ANSWER:
Technically, visible light is emitted by all objects because the spectrum of emitted radiation is continuous and includes all wavelengths. However, you can say at what temperature the visible part of the spectrum is more intense than any other emitted radiation, about 5000 K. Interestingly, that is about the temperature of the surface of the sun.

QUESTION:
Which main part of the electromagnetic spectrum emitted from the sun, is reflected by mirrors on to a water boiler. For generating renewable energy?

ANSWER:
It depends on two things, the spectrum emitted by the sun and the reflectivity of the mirror itself. The sun has the most energy emitted at wavelengths in the visible region but a significant amount is also in the infrared region. If the reflectivity is fairly flat over this region for the metal you choose as a mirror, the spectrum reflected will be similar to the spectrum of the sun. Both aluminum and silver are at about 90% reflectivity over th range of most of the spectrum and therefore the energy reflected will be very similar to the energy which strikes the mirror. The other thing to consider is that whatever absorbs the reflected radiation should absorb as much of the incident radiation on it as possible, i.e. be "black".

QUESTION:
Why does light not bend through glass from air if the angle of incidence is at 90degrees to the boundary?

ANSWER:
I can think of several reasons:

Snell's law requires it: n₁sin

θ₁=n₂sinθ₂ implies that if θ₁=0 then θ₂=0.

If θ₁≠0, the ray defines a unique plane which is normal to the surface on which the refracted ray would travel. If θ₁=0 and θ₂≠0, how would you know which direction the refracted wave travels?

Huygen's principle would have plane waves striking parallel to the surface generating plane waves also parallel to the surface inside the second medium.

QUESTION:
if i put a point sized light source inside a pond then depending upon the refractive index of water the some of the light rays will come out.I have figured it out that a circle will be formed at the pond surface.my question is whether THE LIGHT WILL COME FROM INSIDE THE CIRCLE OR OUTSIDE.(basically i'm asking TIR takes place at angle greater than critical or less than critical)

ANSWER:

Inside.

QUESTION:
Approximately how many photons of sun light are in a square inch over one second? Assuming the sun is directly over head, clear day, standard earth atmosphere at sea level. How many photons in one square inch over one second from a one ~~foot~~ candle source at one foot and would it then be a simple matter of multiplication if I doubled the intensity of the source? The light is coming from a 575 w, 120 v, 3600 kelvin halogen lamp for the second question. I teach the art of lighting design for theater and dance.

COMMENT:

Reading below, you will see that there is an error in the questioner's units since the ft-candle is a measure of intensity, not of power. The source would be characterized by its power, namely a candle or a lumen (which is a fraction of a candle). So I have changed "foot candle" above to just candle.

ANSWER:

I have come to appreciate how convoluted measurement of light is! Furthermore, since the questioner asked for photons/in²/s, one runs into the problem that photons of different color (frequency or wavelength) have different energies. The energy e per photon of a particular wavelength λ is e=hc/λ where c=3x10⁸ m/s is the speed of light and h=6.6x10^-34 J∙s is Planck's constant. The Joule (J) is the SI unit for energy and is most familiar to nonphysicists as the amount of energy delivered per second by a power source of 1 W. So 100 photons of red light delivers much less energy than 100 photons of blue light because blue has a much shorter wavelength. Light intensity is energy/second/area, so equal photons are not equal intensities. Almost any source of light will contain a whole spectrum of light. The spectra of both the sun and the lamp in the second part of the question are well-approximated as black bodies. The curve to the left labelled 5000 K is what the radiation from the sun looks like as a function of wavelength. It peaks at about about λ=555 nm (0.555 μm), yellow. One can find the maximum intensity wavelength (in meters) for any absolute temperature T black-body spectrum using Wien's law, λ_max=2.9x10^-3/T. So the peak for the halogen lamp in the second part of the question is λ_max=2.9x10^-3/3600=810 nm, in the infrared. The problem is that all the "photometric" units (the candle and everything derived from it) are defined in a way which incorporates the sensitivity of the human eye; "radiometric" units are more what a physicist would prefer, the raw data as it were. It is very difficult to unfold one to the other. I will do my best!

To estimate the photon flux I will assume that all the sun's light is at 550 nm. Now, I found that the average intensity of the sun at the distance the earth is from the sun is about I=1367 W/m². Using the expression above, the energy of a yellow photon is e=3.6x10^-19 J/photon. Therefore the flux of photons is Φ=I/e=3.8x10²¹ (photons/s/m²)(1 m²/1550 in²)=2.45x10¹⁸ photons/s/in²). To get a more accurate estimate of the photon flux would involve very complicated mathematics to get the net flux from the continuum of wavelengths, but this calculation certainly gives a reasonable order-of-magnitude estimate.

Another way to approach this is to express the intensity in lux, I=110,000 lux for bright sunlight. Using an online converter to SI units, I=0.16 W/m². Why is this so different from the value of 1367 W/m² above? I believe it is because the lux includes only the visible light which is a very tiny slice of the whole spectrum. So this is probably a much more useful way for the questioner to approach this problem. The photon flux then (maybe we should call it "visible photon flux") is approximated as Φ_visible=I/e=4.4x10¹⁷ (photons/s/m²)(1 m²/1550 in²)=2.9x10¹⁴ photons/s/in².

The second part of the question is trickier because of the units used. The "candle" is a unit measuring power (like a Watt), energy/s. It was historically based on the brightness of a single candle. However, as you get farther away from a source, its brightness gets smaller, so we introduce a quantity usually called intensity or flux which is indicative of energy at some distance passing through some area in one second. The usual unit is the lumen (lm), 1 lm=1 candle∙1 ft²/(4πr²) where r is the distance from the source to where the intensity is measured. Knowing that 4πr² is the area of the sphere with the source at the center, a lumen is seen to be the fraction of the total rate at which energy passes through the 1 ft² at the distance R. [This can be more elegantly expressed by defining a sterradian (sr), the three dimensional analogue of angle called a solid angle, Ω=A/r² (see picture at the right). So you can write 1 lm=1 candle∙Ω/(4π).] So the lumen is also a measure of power but only a well-defined fraction of the total power. The foot-candle is defined as 1 ft-candle=1 lm/ft²; the metric equivalent is the lux, 1 lux=1 lm/m². The ft-candle and lux are both measures of intensity, power/area.

Now to the second part of the question. For a distance of one foot from a 1 candle source, the intensity is 1 ft-candle=10.76 lux. Now, as above, convert this to Watts/m² using the online calculator, I=1.6x10^-6W/m². And so, Φ_visible=I/e=4.4x10¹² (photons/s/m²)(1 m²/1550 in²)=2.9x10⁹ photons/s/in². In this case, I have again used the 555 nm conversion which might not be appropriate for an 810 nm max lamp, but I suspect it may be ok for you since you are interested in visible light.

Regarding your question about doubling intensity, if you double the intensity of the entire spectrum uniformly, then there would be just a doubling of the number of photons. If, for example, you change intensity by increasing the temperature of the lamp, the distrubution would change so that all bets are off.

This may be a long, rambling answer, but I had to learn a lot as I went along. Keep in mind that the photon fluxes are order-of-magnitude, not exact.

QUESTION:
I want to know is there any expression for relativistic force as we have for momentum?

ANSWER:
In classical mechanics, force F only only has meaning in terms of the acceleration a it causes some mass m to have, F=ma. This can be more generally stated as the force is the time rate of change of linear momentum, F=dp/dt. Using the relativistic expression for p, this defines force in special relativity. Usually, though, it is much more useful, both in classical mechanics and relativistic mechanics, to solve problems using the potential energy V(r) associated with a force, V(r)=-∫F∙dr.

QUESTION:
I am very curious - do different electromagnetic waves/frequencies affect each other in any way? If one photon hits another or if they pass each other, do they affect each other in any way? For instance, if I have a flashlight with a stream of light, and another flashlight with a stream of light shining perpendicularly through the first one, I know that the streams do not seem to affect each other in any way - but do they? In ANY way?

ANSWER:
Indeed, photons interact with each other. However, for all practical purposes, two flashlight beams are not sufficiently intense for there to be an observable rate of interaction. Physicists do study the interaction between two photons, though. One well-known example is the interaction of a high-energy photon with the electric field of a nucleus (and therefore a photon) to create an electron-positron pair.

QUESTION:
Is higher frequency really more energy? It is thought that higher frequency has more energy. However, the higher the frequency the smaller the wavelength. With that said, though the higher frequency is delivering more energy to a a given location, it is giving less total energy in a given area. A lower frequency is giving less energy to a given location, but more energy in an area. So the TOTAL energy of the wave would seem to be equal. Greater area(wavelength), lower frequency. Higher frequency, lower area(wavelength).

ANSWER:
I am afraid your thinking is muddled. Electromagnetic waves of a given frequency are a collection of particles called photons; the smallest amount of electromagnetic wave you may have is a single photon which will correspond to a particular frequency. The energy E of a single photon is proportional to its frequency f, E=hf where the proportionality constant h is Planck's constant. So if 1000 red photons hit a wall and 1000 blue photons hit a wall, the blue photons will bring the most energy.

QUESTION:
If an object approached a plane mirror at 3/4 the speed of light, would it be approaching its image at 1.5 times the speed of light?

ANSWER:
No, because the image does not exist. Light leaves you and travels to the mirror at the speed of light and then the reflected light comes back to you at the speed of light. The light you emit and the light you receive would both be doppler shifted toward the blue so what you saw would not be true color.

QUESTION:
Why is it that hot objects such as lightbulb filaments emit light while cold objects such as ourselves emit no light at all?

ANSWER:
Well, let's first define "light" as any electromagnetic radiation, not just the visible spectrum. All objects radiate light and the wavelengths they predominantly radiate depends on temperature. A human body has a temperature around 300 K (80⁰F) and a tungsten filament has a temperature of around 3000 K (5000⁰F). The picture to the right shows the radiation for both of these temperatures; also note the visible spectrum indicated by the colored vertical bands near 0.7 microns. At 3000 K, the radiation is most intense in the region of visible light; at 300 K there is almost no intensity of visible light and the spectrum is most intense around 10 microns which is in the "invisible" infrared spectrum. Night vision goggles are sensitive to infrared radiation and enable you to see "cold" objects in dark situations.

QUESTION:
I am baffled by the fact that if you have a light source and two (or more) mirrors, you multiply the total amount of light that illuminates a room by the number of reflecting surfaces. Does this mean that we increase the total amount of light energy without having to add energy to the light source? Or does each reflection carry less energy than the incident beam from the original light source?

ANSWER:
Yes, you can brighten a room with mirrors; but there is no problem of energy conservation because you are simply using light which would otherwise have been absorbed by the walls. But, not 100% of the light is reflected because if it were, you could turn off the light and the room would not go dark; and the room would just get brighter and brighter if you left the light on. I can do a rough calculation to give you an idea of how long light would bounce back and forth. Suppose that the mirrors were 99% reflective (much more than actual mirrors are) and you had two parallel mirrors separated by a distance of 3 m. Since the speed of light is 3x10⁸ m/s, the time between reflections is 3/3x10⁸=10^-8 s. There are therefore 10⁸ reflections per second. At each reflection the intensity decreases by a factor of 0.99, so after n reflections the intensity has been reduced by a factor of 0.99ⁿ. Suppose we look at how much light is left after 0.1 ms=10^-4 s, 10,000 reflections: 0.99¹⁰⁰⁰⁰≈2x10^-44 or 2x10^-42%. I think we can agree that it is almost instantaneously gone!

QUESTION:
How do CDs create rainbows when we shine white light on them?

ANSWER:
Because they have many closely spaced lines which act like a diffraction grating.

QUESTION:
why are soap bubbles colorful?what colors are observed when a soap bubble is illuminated by monochromatic light?

ANSWER:
The reason is that a soap film has two surfaces from which reflection of light can occur and they are very close to each other. Therefore, light of certain colors will interfere constructively while light of other colors will interfere destructively. A pretty clear explanation as well as a calculator may be found on the hyperphysics site. If you illuminate the bubble with monochromatic light, you will only see that color or no light reflected at all.

QUESTION:
In a 20 X 20 room you have a movie projector on the east side with a screen on the west wall. On the south side you have another projector with a screen on the north wall. What happens to the pictures on the screens when the two projector beams cross at a 90 degree angle. Do the pictures remain cohesive or are they distorted? What is happening in the convergence zone where the beams are crossing? In this instance is light acting as a wave or particle?

ANSWER:
Light waves obey an equation which is linear; what that means is that when two waves pass the same place the net amplitude of light is equal the the sum of the two individual amplitudes. In your case, what this means is that the light from one projector passes right through the light from the other unaffected. Wave or particle? It depends on what you are looking for. The light, just like all electromagnetic radiation, is both waves and particles. (This does not mean a mixture.)

QUESTION:
Is there a Doppler shift between the centre and outside of a wheel. Say for example a standard merry go round you find at a sideshow, if I stand in the middle and look at something on the outer does my image Doppler shift. The reason it is confusing to me is because your angular rotation is the same for both people but not your linear speed.

ANSWER:
If you were asking about sound rather than light, there would be no Doppler shift because for normal waves there is no effect if there is no component toward or away between the source and the observer. For light, however, which is what you ask about, there is a Doppler shift. If your question had asked for a source and and observer moving on straight parallel paths with different speeds (sort of what you asked), then you could relatively compactly write the Doppler shift equation relative to the observer. In the figure to the right, the observer will see the source moving with a speed v and light comes at an angle θ as shown. In this case, f'=f/[γ(1+(vcosθ/c))] where γ=1/√[1-(v²/c²)] and f and f' are the frequencies at the source and observer, respectively. If θ=90⁰, f'=f/γ; this is called the transverse relativistic Doppler effect. But, in your scenario, each will see the other to be at rest. But both are actually accelerating relative to an inertial frame, so the situation is not so simple. You can see a discussion of this scenario on Wikepedia. Suppose that the source is at a distance r₁ from the center, the observer is at a distance r₂, and the angular velocity is ω. Then assuming that the distance traveled by the observer in the time it takes the light to travel between the two is very small, I find that f'≈f√[(1-(r₁ω/c)²)/(1-(r₂ω/c)²)]. I just realized that you actually asked an easier question, namely that r₂=0 and r₁=R where R is the radius of the merry-go-round, so, f'=f√(1-(Rω/c)²). This is exactly the same as the transverse Doppler effect above, f'=f/γ. [Sorry that the answer is much longer than needed. Whenever I misread a question and do a lot of interesting and more general work, I am loathe to delete it! The concise answer for you is that f'=f/γ.]

QUESTION:
if both electric and magnetic components are absent in light what will happen

ANSWER:
That is like saying "if there were no atoms in my pencil, what would it be?" Light is electric and magnetic fields and if either were to disappear, it would not be light.

QUESTION:
I know that visible light has much more energy than microwaves but how is it that microwaves can heat food while visible light can't? Not even gamma rays (which we wouldn't want to use for our food anyway) can heat thing like microwaves.

ANSWER:
First, it is not energy that matters, rather power, the rate at which energy is delivered to what you are trying to heat. While it is true that a visible-light photon has more energy than a microwave photon (E=hf), you cannot make a statement about total energy without knowing how many photons of each you have. So, if you concentrate light from the sun to a spot you can burn paper. But, there must also be something going on beyond power, something about how the radiation interacts with what you want to heat. For example, even very intense light will not greatly heat up a sheet of glass because almost none of the light will be absorbed, thereby giving its energy to the glass. Microwave radiation is just the right frequency to excite vibrations of water molecules, so many of the impinging photons are absorbed and give their energy to the food—heating it.

QUESTION:
the speed of light is constant,yet we know gravity pulls on light,wouldn't light aimed directly towards a high source of gravity,such as a black hole move faster ?

ANSWER:
As light falls into a black hole, it gains energy, but not by speeding up. The frequency increases as it falls meaning the energy of each photon increases, but the speed stays just the same.

QUESTION:
Light behaves like a particle and a wave(in that it collides with matter and it has no real mass). Right? SO, why does light travel from objects straight into our eyes (and everywhere else) why doesn't it collide with ALL of the other light in between said object and our eyes and then when the light reaches our eyes it would not have traveled directly from the object but from a multitude of angles, thus distorting the image of the object(plausibly too much to even see anything).

ANSWER:
Light does not always travel straight from an object to your eyes. The reason you see a blue sky and not a black sky is because some light from the sun strikes a molecule or a dust particle in the atmosphere, scatters from that point to your eye. Sometimes light from some source finds two paths to your eye and the light from those two paths can interfere with each other either being brighter or darker than you would expect; this is called diffraction but usually we do not see this because the wavelength of light is so short. When you are looking at a leaf outside, you are really looking at light which originally came from the sun.

QUESTION:
We know there are EM waves. All other waves need a medium to propagate. How does the EM wave propagate? There must be an ether.

ANSWER:
Here is what we know for certain: the speed of light in a vacuum is independent of the velocity of the source or the velocity of the observer. There is therefore not a medium with respect to which it moves. It is no puzzle how light propogates, Maxwell's equations clearly predict electromagnetic waves which have a speed c in vacuum. If you want to believe in the жther, be my guest. But it is sort of like religion, you need faith and will not get proof. It is not needed in electromagnetic theory, and if it is not needed, why add it? Did you ever hear of Occum's razor?

QUESTION:
My question is about the doppler effect. It is understood that a moving objects sound waves are bunched up in front of it, and elongated behind it, which creates the sound effect that we hear. What is throwing me off is the bunching up of sound waves in front of the object. Why would the sound waves bunch up in front of the object, wouldn't they just blend together and harmonize? I keep thinking in my mind that the sound difference we here on a moving object is the harmonizing of the sound waves together as the waves get closer to your ear and not that they are just bunched up together.

ANSWER:
What does that mean, "blend together and harmonize"? Harmonize usually means when two or more sounds blend together. But the source of sound we are talking about here is a single sound, so harmonizing seems to be inappropriate. Although most sounds are complex, it is easiest to understand by first assuming that the sound you are listening to is a single wave with a single wavelength λ and frequency f which travels with a speed c in still air. Think of the source sending out the peak of the wave once every 1/f seconds. So, a peak goes out and then, 1/f seconds later, the next peak goes out. But, if the source has a speed v, it has moved a distance x=v/f while the first peak has moved a distance c/f=λ. Therefore, the distance between peaks, which is the new wavelength λ', is given by λ'=λ±x=c/f' where you choose -(+) sign if the source is moving toward (away from) you and f' is the frequency you hear. So, the waves either "bunch up" if the source is coming toward you and "spread out" if the source is going away from you. The Doppler effect is usually expressed in terms of the frequency, f'=f/[1±(v/c)]. [Incidentally, no generality is lost if you consider a more complex wave form than a simple single-frequency wave, because any complex wave form may be represented by a superposition (add up a bunch) of simple waves of single frequencies. Since the new frequency is proportional to the old frequency, you get exactly the same shape of complex wave but just "bunched up" or "spread out". This is called Fourier analysis.]

QUESTION:
My question relates to electromagnetic radiation waves like light. I don't understand what the wave looks like in three dimensions. Usually I am asked to picture it like a sea wave or an undulating piece of paper but then how could a radio wave be received by many receivers at the same time in many locations out of the plane of the wave? The other question on waves is why there is a wave at all. What cause the peaks and troughs or the energy to change direction?

ANSWER:
If you have a point source on a water surface, you generate wavefronts that propogate like circles outward (2-dimensional waves). But, in three dimensions, the wavefronts are spheres and they propogate outward as shown in the little animation to the right. So really, 2D waves are not such a good representation of 3D waves. There is "a wave at all" because the radio antenna creates waves which propogate outward. Most antennas are directional, that is they do not send out spherical waves but wave fronts which are more intense horizontally so that energy is not wasted sending waves into the ground or into space; maybe the waves are more cylindrical than spherical. The animation to the left shows a cylindrical wave but for a radio antenna, the cylinder axis would be vertical instead of horizontal.

FOLLOWUP QUESTION:
I now understand the the wave is promulgated as a sphere (or maybe a cylinder) but I cannot get my head around how that translates into the generally depicted view of the light wave being like an undulating snake. With the sphere analogy surely I would just see arcs of a circle as an observer?

ANSWER:
The pictures above show only the wave fronts, which usually represent the maximum amplitude of the outgoing waves. But the spaces between the surfaces shown also have the "undulating" waves in them. If you were to look out along a line and measure the electric field at each point in space at one time, you would see your "undulating snake". The graph to the left shows what the results of your electric field measurements along a radial line would be. The high spots are where you would be crossing the wave-front surfaces in the 3D pictures above. As time goes on, the whole wave would move to the right. The reason that the amplitude decreases as you get farther away from the source is that the waves carry energy and, as you get farther from the source, that energy gets spread out over a larger area but the total energy of the wave has to remain constant. (The energy is proportional to the square of the electric field.)

QUESTION:
Why are the ripples formed on the surface of water alwalys circular no matter which type of source has produced it .. i mean if v drop a square or a triangular stone then too v end up getting only circular ripples ??

ANSWER:
They are not always circular. Drop an 8' 2x4 in the water and nice straight waves will come from the long side. But on the corners the waves will come out rounded where the corners were because of diffraction. Any feature which is not large compared to the distance between water waves will exhibit diffraction. So, if you drop a square pebble in the water it will have approximately circular waves almost immediately because it is small compared to the distance between water waves. The feature of diffraction can be understood using Huygen's principle which says the next wave may be generated by assuming each point on the previous wave behaved like a point source (making perfect little circles) and then drawing the envelope of all the little circles. The picture below illustrates that for the the 2x4. I have only drawn a few of the Huygen's wavelets at the left end to illustrate how the corners get rounded. As the wave (red) spreads farther the whole rectangle gets rounder and rounder until, finally when you are much farther than 8' from the 2x4, it becomes almost perfectly round

QUESTION:
I was working on a ship and was down in one of the holds cleaning out the bilges. The hatch cover was closed. No sunlight was entering the hold except for that coming through eight 1.5 inch diameter holes spaced out on the hatch cover. The eight holes cast eight bright circles of light on the deck 70 feet below the hatch cover, creating eight perfectly focused images of the clouds passing overhead. I get that part - pinhole cameras. However, after a couple hours when my eyes were well adjusted to the dark, I noticed another optical effect while sitting down there taking a break. About 25 feet above the hatch cover was parked the boom of a crane, running diagonally across the cover. When my eyes adjusted, I saw a perfect shadow of this crane on the deck of hold, a dimmer version of the shadow that would be seen if the hatch cover were to be opened. In no way was there a regular lay-out of the eight holes in relation to the crane boom. There were not eight different shadows - only one that somehow must have been a composite. Is there a name for this effect?

ANSWER:
Here is what I think is happening. Presumably, since you talk about shadows, it was a sunny day. The eight bright spots you see on the floor were eight separate images of the sun formed by your eight "pinholes"; the clouds you saw were clouds passing directly in front of the sun. As a check, I would expect that the diameter of these images would be about 7" since the angle subtended by the sun in the sky is about 0.5⁰, so d≈70'(12"/1')x0.5⁰x(π/180⁰)=7.3". But, the sun is only a tiny amount of the whole image formed by each pinhole; the whole rest of the sky is there but at a much lower intensity. The crane is part of that bigger image for each pinhole and I suspect there are really eight images which are pretty much overlapping because the spacing between pinholes is probably small compared to the size of either the crane or its shadow. I also think we should call the shadow an image instead.

QUESTION:
I have a little device that illustrates some optics principle or other, but I don't understand how it operates. It's a small spool (once held thread), one end of which has its opening covered with Al foil (so as to keep out light). A sharp pin was used to create a single, small hole in the foil. The other spool end is open, but a small, iron nail has been poked thru the end (so it's perpendicular to the long axis of the spool). When the spool is held close to the eye so that the nail end is closest to the eye and the other end directed toward a source of sufficient light, and then the nail moved so that it crosses the line of vision, the nail shadow appears to enter the field from the oppostie direction from which it actually is moving. So, my question is: What's going on here? I assume there's a camera obscura connection, but that doesn't help me.

QUERY:
Sorry, I am not getting the picture. How close is the nail to the eye. What do you mean by the shadow?

FOLLOWUP:
One would usually hold the device within a couple of cm from the eye, so as not to confuse the issue; nevertheless, the phenomenon remains observable on out to maybe 15 cm. But as you move the device farther from the eye, you must know to allow the eye to focus farther out, at the spools farther end light source (pinhole), as opposed to its closer end where the nail is moved back and forth across the opening. At several cm out, it takes a young person's young eyes to focus on the actual nail. And by shadow, I mean a black image (as opposed to the silver-colored nail) that forms on the retina.

ANSWER:
This is an interesting device. I made one so that I could be sure I knew what was happening (I used a pin instead of a nail). This is essentially a pinhole camera except we are not using it to make an image in the usual way. Now, all the light coming through the pinhole is coming radially away from the hole and the nail, if it is below the center line, for example, blocks light which came from above the center line when entering the hole. The brain now interprets this lack of light as meaning that a dark object was placed above the center line and that is what you see. This is the same reason that the image, if we use this as a pinhole camera, is inverted—light from above goes through the hole to below, from the left to the right, etc. I found it works best if the nail end is almost against the eye; and you do not want to focus on the nail, the "image" will appear to be far away. I put image in quotes because it is not really an image. It is very much like a virtual object formimg a real image (for optics afficionados).

QUESTION:
I was wondering if: Someone was shinning a flashlight toward you. And someone else, say at 90 degrees, emitted a light beam or a beam of a certain frequency. Could the light from the flashlight be blocked? Can you cancel out a light frequency, with another frequency?

ANSWER:
For starters, the flashlight has a continuum of frequencies (white light) so each frequency would have to be dealt with individually, clearly impractical. In addition, the flashlight light is incoherent, that is the light is a complete hodgepodge of light waves having no particular phase relationship among themselves; to get interference effects you need coherence (waves all in lockstep with each other). Finally, if you were able to do this, it would certainly not be from 90⁰ for the following reason. Suppose you had a flashlight with one single wavelength and the light was coherent (you would call that a laser). In principle, if you shone an identical wave in the opposite direction you would establish a standing wave where you would have a dark spot every half wavelength of the light. But, since the wavelength of visible light is like 600 nm=6x10^-7 m, you would not be able to observe these dark spots without very special instrumentation. Note that you can never make the light go away completely, you can never cancel it all out.

QUESTION:
Why does the wavelength of light change but the frequency doesn't when it moves into a different medium?

ANSWER:
Frequency is a property given the electromagnetic radiation at its source. If you have an electron oscillating with a particular frequency, then that will be the frequency of the radiated wave. What the wavelength will be depends on what the speed of the wave is. Take a simple example: the frequency is 5 cycles per second and the velocity is 10 meters per second. Then the wavelength will be the distance the wave travels in one cycle (called the period) which is, in this case, 1/5 of a second. Hence the wavelength will be 2 meter. If the the speed were instead 5 m/s, the wavelength would be 1 meter. When light enters another medium, what changes is the speed and, since the frequency is characteristic of the source rather than the medium, it does not change. Hence, the wavelength must change.

QUESTION:
Why do electron microscopes get better pictures than regular light microscopes?

ANSWER:
The problem with optical microscopes is that you cannot look at anything which is comparable to or smaller than the wavelength of the light. Visible light has wavelengths of a few hundred nanometers, ~6x10^-7 m, so what you are looking at has to be bigger than that, a few microns maybe. The reason is diffraction, light does not just travel in nice straight lines when obstructions or aperatures get to be on the order of the wavelength, e.g. light can be bent around a corner. If light does not go in nice straight lines, geometrical optics doesn't work. On the other hand, we know that particles like electrons can behave like waves and their wavelengths are inversely proportional to their momentum. So, if we make an electron go fast enough it will have a wavelength much less than light and therefore let us "see" much smaller things.

QUESTION:
I learned the basic premise that things moving fast can have negative effects on other things when I was in kindergarden. In elementary school I came up with a question. If light travels so fast, why doesn't it completely obliterate anything it hits, by smashing it to pieces? I had my sister ask her physics teacher. The answer he sent back to me was that light was too unfocused, and that objects with zero rest mass do not have that kind of effects on things which do have mass. (Great answer to give to a kid in elementary school.) This question remained with me as I got older, and inquiring my own physics teachers I only learned enough to become further confused about the subject. I know when something with positive rest mass, like a bullet or a baseball hits something it transfers momentum and kinetic energy. I was told that the effect light has on objects heating them etc, is an effect of kinetic energy. But I am still having trouble really understanding the concept. With learning the difference between energy, and momentum only helping slightly. If I throw a baseball with high velocity and it hits something, like a block of metal the impact of the ball will cause the block to move. If I shine a laser at the block and just keeping adding more and more power would it ever be possible to accelerate the block with the laser? (Assuming it could avoid being melted)

ANSWER:
You are interested in force so you need to ask about linear momentum (which we usually think of as p=mv, but I will modify this a little below) because, in a collision, each object experiences a force which is proportional to the change in its momentum. When a ball strikes another ball, during the time they are in contact each exerts a force on the other (which are equal and opposite, Newton's third law). The struck ball, initially at rest, rebounds forward because of this force and the striking ball slows down because of the equal/opposite force it feels. If the striking ball moves faster, it has more momentum and therefore exerts a larger average force during the collision. In fact, the force is precisely equal to the time rate of change of momentum; this is Newton's second law. Now, to your question. Light can exert a force but not because it goes fast but because it carries momentum. Even though photons have no mass, they still do have momentum. So, when light strikes something, it does exert a force on it; that force, for everyday situations, is just too small to have much of an effect because normal intensities of light just do not have all that much momentum. I will give you a couple of examples of so called light pressure. GPS satellites, to work accurately, must have their orbits calculated extremely accurately and corrections for radiation pressure from light from the sun must be applied. It has been proposed that future spaceships in the solar system could have gigantic reflecting sails and use radiation pressure as a means to be accelerated away from the sun; each photon comes in with a momentum p=E/c where E is the energy of the photon and c is the speed of light. Since the photons are reflected, their change in momentum is 2E/c which is also the momentum transferred to the spaceship.

QUESTION:
So if light from a flashlight is a form of energy, can it be felt? Is it possible for a blind person to know if he is under a street light? How bright does a light have to be for me to feel it?

ANSWER:
Indeed, light does exert a force, normally referred to as radiation pressure. When light strikes you it is either absorbed or reflected. Light which is reflected exerts twice as much force as light which is absorbed because it does not lose any energy. But, the force for any reasonable intensity is extremly small and for the force to be big enough to feel would require an intensity where you would be burned up by the heat from the light you absorbed. To give you an idea of the magnitude, a laser pointer exerts a force on the order of 3x10^-12 N; to exert a force of about 1 oz you would need about 100 billion laser pointers.

QUESTION:
How can laser light cast a pattern resembling the double slit experiment when shone on the tip of a needle, a structure that does not resemble a double slit?

ANSWER:
The only way this "resembles" a double slit pattern is the appearance of fringes. Diffraction can occur any time you have more than one path light can take to reach the screen. Imagine light coming from right from the two sides of the needle. Then, these two rays could interfere either destructively or constructively depending on the viewing angle, so the pattern could look very much like a double slit pattern.

QUESTION:
I am a photographer and general tinkerer. I have started building pinhole cameras and during the course of building them I came across some equations on obtaining the optimal pinhole size. I find I understand the equations well, but I can not find a solid explanation for the 2.44 constant used in the calculation for the Airy Disk (2.44 x light wave length x focal length). Everything I have read is very vague and I want an explanation. Why do we use this number? I read that it is like pie, it just is, but is to what, waves? Is it a speed? What does it represent? I want to understand.

ANSWER:
This verges on being too technical for this web site, but I always want to encourage folks to "have an enquiring mind", so I will take a stab at it. This question requires that you understand a little about diffraction, interference of light waves. If we shine light through a slit and the slit is very large compared to the wavelength of the light we are looking at, we get an "image" of the slit on a screen. If the rays of light hitting the screen come from very far away compared to the size of the slit, the image of the slit will be the same size as the slit itself. The sun coming in through a window creates an image of the window, the same size, on the wall. So, I now want to make images of smaller and smaller slits; what happens is that we come to a point where as we make the slit smaller, the image starts getting bigger. The figure above shows what the "image" of a very narrow slit looks like (below) and a graph of the intensity (above). It is fairly straightforward to show (see any elementary physics textbook) that asinθ=λ gives the angle of the first dark spot in the pattern for wavelength λ; the geometry showing a and θ are shown to the right. Now, a pinhole is not a slit, but it is very similar and you would expect its pattern to be very similar. Indeed, instead of getting stripes you get a bullseye as shown to the left. Doing the analysis similar to that which leads to the angle of the first minimum gives a slightly different result, asinθ=1.22λ where a is the diameter of the hole. In essence, this is where your factor of 2.44 comes from; where does it come from? It gets pretty technical here! When you solve the pinhole diffraction problem, you work in cylindrical coordinates because the problem has cylindrical symmetry. As is often the case with problems with this symmetry, the solution (for the intensity in this case) involves a Bessel function, a special mathematical function which whole books have been written about. The intensity is given by I=I₀[J₁(Ѕkasinθ)/(Ѕkasinθ)]², where J₁ is the first order Bessel function and k=2π/λ. Now, we are interested in when the intensity is first zero; the first zero of J₁(x) is for x=3.832, so Ѕkasinθ=(πa/λ)sinθ=3.832, or asinθ=1.22λ. Next, we convert the angle to lengths by approximating (see diagram above) sinθ≈x/R so x=1.22Rλ/a or 2x=2.44Rλ/a; 2x is the diameter of the smallest spot to which a collimated beam of light can be focused. Since you say that you "understand the equations well" except where 2.44 comes from, I guess my task is complete. Apparently (from the little research I did) the optimal size is if the minimum spot size equals the hole size, i.e. 2x=a_opt, so a_opt=√(2.44Rλ). There is an alternative form of optimum size which is based on the Rayleigh criterion for resolving two spots (which stipulates that the central maximum of one image is on the first minimum of the second) which has a_opt=√(3.66Rλ). (A similar problem is the structure of the compound eye of insects, elegantly discussed in Feynman's Lectures on Physics, Vol. 1.)

QUESTION:
Why does light travel in a srtaright line and sound don't?

ANSWER:
Light does not travel in a straight line necessarily any more than sound does. The simplest example is a mirror which clearly changes the direction of light. Light is also bent by entering a different medium which is how lenses work. Light is also bent (ever so slightly) by gravity. Light is also bend by diffraction (for example, you can never make a perfectly sharp shadow of something because light bends as it passes the edges.

QUESTION:
I am a photographer. I will tell you the facts. I take pictures in a studio with a strobe that is off camera. When I move the strobe closer to the subject (person I'm photographing) the exposure increases and I can use a smaller aperture in my camrera (i.e. the subject reflects more light into my camera, so I can 'stop down the lens', which closes the diameter of the lens, so less light enters). However, if I leave the strobe in place and move only the camera closer to the subject the exposure remains the same. In fact, even if I move the camera 100 feet away from the subject the exposure of the subject (although he appears smaller in the viewfinder) remains the same. Now why is that so? The light travels the same distance so shouldn't the exposure remain the same?

ANSWER:
First, you need to understand what is called, in physics, the intensity of the light. Intensity is defined as energy per second per square meter carried by the light. In your case, since you use a strobe, I will refer (incorrectly, but it serves our purpose) to intensity as energy per square meter since the light all comes in a very short time. The important thing to understand about intensity is that it falls off like 1/d² where d is the distance you are from your subject. The reason for this is fairly easy to understand. The light goes away from your subject as a spherical pulse and the total energy in the pulse must stay the same and the area of the sphere when it reaches you is 4p d²; ; so to keep to total energy constant it must be spread out more thinly. Thus, if you are 50 m from your subject, the light intensity will be 4 times greater than if you are 100 m from your subject.

If you move the strobe closer to the subject the light illuminating the subject will be greater so the light coming from the subject will be greater so you will have to stop down your camera. This, as you note, is not surprising. However, if you move the camera farther away, it does seem, at first analysis, that you should have to open the camera up because as you get farther away the intensity gets smaller. However, if you think about it a moment, you will realize that although the intensity decreases like 1/d², the size of the image of your subject also decreases like 1/d². Since the size of the image is smaller it takes less light to get a proper exposure of the subject by exactly the same amount as the intensity has decreased. If you use a telephoto lens to make the image the same size you would have to open your lens up to get a proper exposure. (Note added: I have been informed that a telephoto lens, by its construction, lets in more light for the same f-stop. So, I should say that using a device which enlarges the image without increasing the amount of light available will result in underexposure of the film.)

QUESTION:
Is there a color filter that can absorb color but transmit shades of gray?

ANSWER:
If you mean is there a simple piece-of-glass kind-of-thing which turns what would have been a color image into a black and white image, I believe that the answer is no. All light is characterized by its wavelength and lights of different wavelengths correspond to different colors. (You might want to read the answer to the question below for more detail.) "Shades of gray" are not colors and therefore do not actually exist in the light which forms the image.

On the other hand, there are certainly ways to create a black and white (B&W) image from what would have been a color image if viewed by your eye. The simplest to understand, of course, is B&W film in a camera. Here we have a chemical which undergoes some chemical reaction (turning to a different chemical) if light strikes it. The brighter the light is, regardless of its color, the more of the chemical reaction takes place so that you are actually forming an image based on brightness of the light rather than color of the light. This is an oversimplification because photographic film is not "linear", that is the chemistry is not equally sensitive to all colors; for example, most B&W films are relatively insensitive to red light which is why someone's (bright) red shirt often looks black on a B&W picture even though you might have thought it should look lighter. This is also the reason dark room lights are red. You can also make B&W images electronically; digital cameras often have an option of taking B&W pictures. Scanners can also create a B&W image from a colored object.

So, the bottom line is that a color image is one where the image is formed by using wavelength information whereas a B&W image uses brightness information, and no "passive" device can create the brightness information from the color information. Filters are passive devices which block out certain colors and transmit others.

QUESTION:
I am a student at the University and am writing an article on alternative therapies. This topic, while interesting, is very difficult to scientifically ground. I have some questions that I think you may be able to answer. It has been so long since I've taken a physics class, that I just want to be absolutely sure that what I am writing is factual. I would appreciate your time to look over and reply to these questions:

Does color literally vibrate? Is it just our eyes that vibrate when we look at a color, or does that color actually vibrate? If so, do different colors vibrate at different intensities?
Does light vibrate?
Are light and color both considered energy, or are they thought of as the same thing? Are they physically interchangable?
Are our bodies comprised of energy, including light and color?

ANSWER:
First, what is light? Light is a very narrow region of what is called the electromagnetic (EM) spectrum. It is that type of EM radiation to which our eyes are sensitive. Other examples of EM radiation are radio waves (far "above" what we can see), radiant heat, also called infrared (just "above" what we can see), x-rays (somewhat "below" what we can see), ultraviolet radiation (just "below" what we can see), and nuclear gamma radiation (far "below" what we can see).

Now what do I mean by "above" or "below"? EM radiation is a type of wave and all EM waves move through space with the same speed (186,000 miles per second). Being waves, they pass us or hit us with a characteristic frequency and have a characteristic wavelength; think of a water wave: if you are standing still on the shore looking at waves coming in, frequency is, for example, the number of waves per minute which hit the beach and wavelength is the distance between two adjacent waves. If you think about it (this is subtle!), if all waves travel with the same speed which you know, like light waves do, then if you know the wavelength, you also know the frequency (or vice versa). Thus waves can be characterized by either wavelength or frequency; I think wavelength is a little easier to visualize, so in the examples above, "above" means a longer wavelength than our eyes are sensitive to and "below" means a shorter wavelength than our eyes are sensitive to. The shortest wavelength which we can see is about 4000 angstroms (an angstrom is 1/10,000,000,000 meters, so you see that an angstrom is a very short distance) and the longest is about 7000 angstroms.

So what is different between 4000 and 7000 angstrom light? If 4000 strikes your eye you perceive violet; if 7000 strikes your eye, you perceive red; If 10000 or 1000 strikes your eye you perceive nothing (black or darkness). Thus, you see that what color is is simply a qualitative description of the wavelength of the light. Incidentally, another wave with which you are familiar is sound; the pitch of sound is determined by the wavelength of the sound waves. So pitch is to sound as color is to light.

Now, I haven't told you what is doing the "waving" in EM radiation. For water waves, the water "wiggles"; for sound waves, the air "wiggles", both vibrating. For EM radiation, it is more complicated and subtle (light can propagate through totally empty space where nothing is available to "wiggle") and I think you get the gist of the general idea without going into details. If you want more details, let me know.

With that background, let's answer your questions (not necessarily in the order you asked them):

"Does light vibrate?" Sure, just like the waves hitting the beach with some frequency, the light hits your eye (or anything else it hits) with some frequency. But you never perceive this because the frequency is amazingly large (for example 400,000,000,000,000 times per second for red light). Nevertheless, your eye/brain evidently have a way of measuring the color because you perceive it.
"Does color literally vibrate? Is it just our eyes that vibrate when we look at a color, or does that color actually vibrate?" Color is just a qualitative measure of a length of the wave, as stated above, so it does not vibrate. I am not an expert in vision, but I know that the eye does not "vibrate" in response to the incident light but absorbs the energy of the light in a time averaged way.
"If so, do different colors vibrate at different intensities?" Well, 'not so', but you have raised another issue, intensity. Intensity has nothing to do with color; rather it is a measure of another property of waves which I did not discuss above--how big are they? E.g., water waves an inch high have a different "intensity" than water waves a foot high. (Actually, intensity is defined to depend on the square of the height, so in the preceding example the intensities would differ by a factor or 12x12=144.) Intensity is a measure of how bright light is, so bright and dim green lights would have the same color but different intensities.
"Are light and color both considered energy, or are they thought of as the same thing? Are they physically interchangeable?" Light carries energy with it as it moves along. This is easily seen by feeling the warming effect of light that strikes you; you probably, when you were a kid, used a magnifying glass to focus sunlight on something and burn it--energy. Again, be careful to realize that color is only a way of characterizing the light, so of course light and color are not interchangeable. What color does is determine how much energy the light can carry. Purple light carries more energy than a comparable amount of Red light because the shorter the wavelength, the more energy. That is why x-rays can penetrate into your body but visible light does not.
"Are our bodies comprised of energy, including light and color?" Einstein showed that all matter is a form of energy. In that sense, our bodies are comprised of energy. Just because light carries energy does not mean that you can conclude that our bodies are comprised also of light.