Does a mirrored ceiling increase photon count in a room with a candle?

  • Context: Undergrad 
  • Thread starter Thread starter spectragal
  • Start date Start date
  • Tags Tags
    Mirrors Photons
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
15 replies · 4K views
spectragal
Messages
10
Reaction score
0
Hypothetical question:
Someone is in a 6' x 6' x 6' lightly ventilated space with a mirrored ceiling and a lit candle on a small table. Is there more photons in the room due to the mirror on the ceiling? Is there twice as much light as there would have been if there were no mirror in the room? I am trying to wrap my mind around mirror matter and many such subjects.
 

Discussion

Physics news on Phys.org
spectragal said:
Is there more photons in the room due to the mirror on the ceiling?
Yes. Some of the photons which would have been absorbed by the ceiling are reflected back into the room.

spectragal said:
Is there twice as much light as there would have been if there were no mirror in the room?
No. Not all of the photons emitted from the candle go to the mirror, also, not all of the photons that reach the mirror are reflected.
 
Thank you for that.
I have one more question. I understand that the ceiling material without the mirror, and some of the walls, would have absorbed the majority of the photons.
Now using the scenerio with the mirror, the mirror would absorb some of the photons but not as many, correct? On one hand I know this makes sense but on the other hand the mirror is also a material that can absorb photons, it is made of atoms too it just happens to be reflective.
I am not sure what my question is other than they are both supposedly solid objects, but in fact are constructed of atoms that will absorb the electromagnetic force. It is amazing that just because the mirror is shiny and reflective that photons do not get abosrbed into the material in the same way.
Care to elaborate?
 
I guess what I am getting at is, pretend that the photon is blind - I know how strange this sounds.
Both the ceiling and the mirror of both spaces is made of glass. One ceiling is painted with a flat chalky grey substance and the other is painted with a slick shiny silver/grey substance; therefore, the second ceiling is reflective.
The blind photon does not distinguish between either as it is blind. Both the substances would 'feel' nearly identicle in regards to penetration and density. They only look different.
What is it about reflectivity that would make a photon bounce back into the space or be less absorbed by a reflective surface than a non-reflective one.
 
The photon is blind. It is the surface that emits light due to stimulation from the photon. You have two different surfaces.

It is like asking why two radios tuned to a different station are playing different songs even though they are in the same room.
 
Reflection occurs because of a change in the material properties. For EM, in particular, conductivity. Mirrors reflect because the silver backing is highly conductive and glass is not.
 
I was under the assumption that photons do not " bounce " per se, but are actually given off from electrons they have been " absorbed " by, minus some quantity of energy.

Is that incorrect ?

*edit* http://en.wikipedia.org/wiki/Photoelectric_effect
 
Last edited:
I have a related question: Upon reflection, how does the re-admitted photon "know" how to precisely cause the angle of relection to equal the angle of incidence? If the photon simply bounced off the mirror like a ball, I can easily picture how this would work. Much harder is for me to picture how the incoming photon can "tell" the re-admitted one which way to go. Duhh...?
 
Isaacsname said:
I was under the assumption that photons do not " bounce " per se, but are actually given off from electrons they have been " absorbed " by, minus some quantity of energy.
That part doesn't happen. If it did, then mirrors would make things redder. The energy loss in reflection is a reduction in the number of photons, not a reduction in the amount of energy per photon.
 
spectragal said:
I guess what I am getting at is, pretend that the photon is blind - I know how strange this sounds.
Both the ceiling and the mirror of both spaces is made of glass. One ceiling is painted with a flat chalky grey substance and the other is painted with a slick shiny silver/grey substance; therefore, the second ceiling is reflective.
The blind photon does not distinguish between either as it is blind. Both the substances would 'feel' nearly identicle in regards to penetration and density. They only look different.
What is it about reflectivity that would make a photon bounce back into the space or be less absorbed by a reflective surface than a non-reflective one.

Actually the difference in something be reflective and non reflective can be as simple as the texture. Like how you can grind or polish a rough surface down to a smooth shiny one.

The paint in your example would cause a drastic difference in how it "feels" to a photon. At the scale of a photon, your paint adds a huge amount of material to go through.
 
DaleSpam said:
That part doesn't happen. If it did, then mirrors would make things redder. The energy loss in reflection is a reduction in the number of photons, not a reduction in the amount of energy per photon.

Ahhh, sorry, my mistaken understanding there. Thanks. I recall now, having watched a video explanation of why light travels through glass just last week .
 
We know electrons are very small. How photons are reflected by electrons on the solid metal surface? Which factor influence on the reflexibility of electromagnetic wave?
 
daumphys said:
Which factor influence on the reflexibility of electromagnetic wave?
I already said, conductivity.
 
DaleSpam said:
I already said, conductivity.

Does every wavelength electromagnetic wave have the same reflexibility?
 
daumphys said:
Does every wavelength electromagnetic wave have the same reflexibility?
No. A material's conductivity varies with frequency.