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Posted By: Dogrock Light in a room - 12/27/05 02:37 AM
If sitting in a room with the light on I've found to hard to understand how the light reflecting from the walls enters my eyes to give such clear definition. Why doesn't it cancel out as it all reflects from the four walls, before reaching my eyes.
Posted By: Anonymous Re: Light in a room - 12/27/05 05:53 AM
You will find pretty good answers in any elementary level Physics Book.
"Things" can not cancel out exactly because many things are involved in making such a cancellation happen.Namely Wavelength , Angle of Relfelction , Frequency ,Geometry and Number of Photons.And then you have a limitation of your eyes itself to measure the distinctness.
Posted By: RM Re: Light in a room - 12/27/05 01:53 PM
how many times can a wave of light reflect or refract off a surface before being converted into heat or something else? Or does it just 'bounce' for ever?

I remeber I once asked my physics teacher what would happen if you were to light a peice of magnesium in a closed box where the edges were all mirrors and then opened it long after the magnesium had burned out. She said you would not see light because it would have been converted into heat.
Posted By: Dogrock Re: Light in a room - 12/28/05 02:57 AM
We could paint all the walls, floors and ceiling the same colour, and have all the same distance from the light source. Assuming every photon must come to our eyes in a straight line it's strange that some parts of the walls don't seem much brighter than others. I wondered if we see mostly the light from the bulb reflecting directly off the wall to our eyes, or is there a play of light from all the surfaces. I'd like to be able to sit in a room and explain to someone exactly how they're seeing.
Posted By: Anonymous Re: Light in a room - 12/28/05 08:45 AM
A highly polished surface is not easy to obtain..
A mirror has this property.
A painted suface doesnt give you this option..
The surface irregularities gets replaced by varying depth of Elemental Paint.
Coherence is still not achieved to show you the interference.
Any doubts?
Posted By: Anonymous Re: Light in a room - 12/29/05 07:35 AM
A laser is coherent light, pulses at the same wavelength and frequency. It ought to be possible to interfere with laser light and "cancel" it out. A defense for the star wars offense.
Posted By: Anonymous Re: Light in a room - 12/29/05 08:31 AM
hahaha ..
was it a doubt or an answer?
The Hologram projection uses principles of Interference in a room.But it is a different kind.
The question was different I suppose.
Posted By: Uncle Al Re: Light in a room - 12/29/05 07:27 PM
A lens is a Fourier transform element. If you expose a sheet of photographic film to ambient light all you get is perceptibly unstructured grey. Interpose a Fourier filter and you get a different information domain that is more useful. Both intensity and phase matter in conveying information. Naked film ordinarily only captures intensity.

It needn't be a net positive lens as such. Any such Fourier filter element will adequately do - pinhole, zone plate, binary optics, Fresnel optics, concave mirror, holographic plate of an optical element... No matter what the modality, it must process both intensity and phase.

If phase information were retained in the naked film case, as by holography, you would capture a visible image. In fact, it would be a full 3-D image. But that requires self-correlated light - coherent in both space and time - plus physical depth of recording medium.
Posted By: Dogrock Re: Light in a room - 12/31/05 02:57 PM
Obviously I need to know more about the eye than the photons to give any useful explanation. Answers you don't expect are less satisfying but always more informative. Just took a wrong turn from reading too much about how light moves. It said it took six photons to activate a cell in your eye, of course I didn't think of all the extra stuff an eye does before you can see anything.
Posted By: Uncle Al Re: Light in a room - 12/31/05 06:11 PM
A human retinal cell is a single photon receptor. All you need is one photon for detection. Pre-exposure reduces sensitivity.

Silver emulsion requires at least three photons/grain, and oxygen will reverse the exposure if the second and third ones are slow in arriving. One then achieves maximum sensitivity for low light levels and very long exposures (astronomy) by

1) Pre-exposing the film to average one or two photons/grain. But first...

2) Bake in hydrogen gas or add formate to create a reducing environment without oxygen. Lastly,

3) Expose the hypered film at low temps to turn off everything but the photochemistry.

The simple answer is a cold CCD detector. X-ray crystallography structure determination went from months using film to maybe a week using single point counters to hours using a planar CCD array. 5% error, data vs. modeled fit, used to be astoundingly good. Modern hardware routinely gives 3% error and two more decimal places in the results.

The beauty of science (and computer programming) is that you need only be clever once. After that, follow instructions and the universe will always cooperate. Yesterday's Nobel prize is tomorrow's homwork.
Posted By: FourFourTwo Re: Light in a room - 01/27/06 04:12 PM
Quote:
Originally posted by Rob:
how many times can a wave of light reflect or refract off a surface before being converted into heat or something else? Or does it just 'bounce' for ever?

I remeber I once asked my physics teacher what would happen if you were to light a peice of magnesium in a closed box where the edges were all mirrors and then opened it long after the magnesium had burned out. She said you would not see light because it would have been converted into heat.
I too have been wondering about this. It sounds a bit dodgy to me.
Does the act of bouncing off a mirror automatically convert part of a photons energy into heat? If so, how many reflections does it take to convert a photon 100% into heat? If the light has lost energy to heat, wouldn't it then appear red-shifted?
Is there any difference in the light you see directly from a source 10 feet away as opposed to light reflected in a mirror 5 feet away from the source and 5 feet away from the viewer?
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