Oh boy, a smart question I can finally answer.
The
law of conservation of energy states that energy cannot be created or lost; it can only be converted.
Thus, photons cannot be used up. A
photon is an elementary particle created at the atomic level by an electron dropping from higher orbit (due to an excited state) back to its original orbit. An electron loses energy in the form of a photon, and the higher the orbit, the more energy an electron has. (Electrons like to go back to their original orbits.) This is best represented by
phosphorescence in which the material is "charged up" and emits light as electrons drop out of their excited states.
We never see the full output of a light source because as light reflects off surfaces, photons may bounce in a way such that they do not reach our eyes. Brighter lights (with a wide spectrum of wavelengths-- white
light) increase the chances that more photons will reach our eyes. Also, light emitted from any source (or any other form of energy) tends to
spread out in multiple directions, which is why a beam gets bigger the further away it is from its target. It also explains why lasers have that "grainy" look -- since a laser outputs light of a single wavelength, the light stays in phase and does not spread out much (except in special cases, such as
shining through slits where light
light act as a wave) -- some spots are darker than others, and some are brighter, due to
constructive and destructive interference caused by a surface not being perfectly smooth (a perfectly smooth surface would reflect light of a single wavelength all in one direction, and theoretically a laser shined at it would not appear grainy). Light at the same
wavelength but out of phase by λ/2 (half a wavelength) cancel each other out, causing a dark spot in our perception of the laser beam, and light that is in phase causes a brighter spot. You can see this with a graphing calculator or
GraphCalc, a free calculator for Windows -- or Grapher, which comes free with OSX Tiger -- assuming you're using radians, try y=sin(x) vs. y=sin(x)+sin(x+pi) (destructive interference) vs. y=sin(x)+sin(x) (constructive interference).
The light emitted from a flashlight, however, is all at different wavelengths (causing us to see it as white light, depending on the ratio of different wavelenghts put out), and also explains why we see things better with white light instead of monochromatic light. With white light, the chances of light reaching us is better, because there is a larger spread of photons moving at different wavelengths, and they end up complimenting each other (not like a pat on the back!). You can see this to an extent by shining two lasers of different colors (wavelengths) at the same point and noticing that the "grainy" look is less apparent. (Try green and red!)
What we see as "colored" materials actually just absorbs light of all the other wavelengths. A green piece of paper, for instance,
absorbs the wavelengths that are not associated with "green" and reflects light of the 520-570 nm wavelengths back to your eye. Since energy cannot be lost most of this is converted into heat. (Like a microwave oven, which emits "light" at a frequency we cannot perceive with our eyes.) The light that is reflected that we do not see may get back to us in the form of ambient light. See also
black body radiation and a source that Radio mentioned to me,
http://galileo.phys.virginia.edu/classes/252/black_body_radiation.html.
We lose light output from the lens of a flashlight, as well. Light reflects off a surface when there is a change in
refractive index of the mediums light pass through -- the transition from air to glass, for instance. AR coating works by gradually changing the refractive index between air and glass through the use of multiple (hundreds?) of layers. The higher the difference in refractive index of two mediums, the more reflective a surface is.
There are also lots of particles in the air that diffract light, preventing the photons from reaching us, which is why it's harder to see things in fog (lots of
backscatter from light bouncing off water in the air) and partially why a light seems dimmer at a distance. (The other reason is because the eye is an effectively smaller target for the light to reach at a distance. Reference
inverse square law on my favorite website, Wikipedia, for more information.)
Hope this helps. This is based on what I've been learning in physics (see, school *is* useful!) and my understanding of the topic. I'd recommend taking a look at the links above, too. If I'm wrong (I really hope I'm not) please correct me (so I can do better on my physics final). Now, back to my paper.