Re: Experiment#1
- From: "OG" <owen@xxxxxxxxxxxxxxxxxxx>
- Date: Sun, 14 May 2006 00:54:37 +0100
"rAgAv" <ragav.payne@xxxxxxxxxxxxxx> wrote in message
news:1147494305.360220.61180@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Dear OG,
Okay, the basic purpose of this experiment is to find out what happens
when photons strike a surface, "Do they get transformed to any other
form or do they just get reflected?"
Or
"Do their wave nature get affected?" i.e. "do their frequency change?"
if yes, then it means that the energy of the light beam is either lost
or gained(according to Energy = planck's constant * frequency). The
latter is not possible according to the laws of kinematics so the
energy might be lost that means, the stationary surface which the
photons strike will gain some energy,which means the mirrors will have
to gain kinetic energy or thermal energy or any other form of
energy.(which includes energy in the form of mass, from E=mc^2)
Silver mirrors reflect about 95% of incident light, so within 2
microseconds* the laser is reduced to a million millionth of its
original
strength.
This seems kind of a confident answer. But what happens to the other
5%.
And what exactly do you mean by strength(energy or frequency or
intensity....wavelength)
Are you expecting the photons to be reflected to and fro between the two
mirrors? This may happen, but you wouldn't see any of them.
you mean their frequency gets altered so that it does'nt fall in the
visible region of 400 nm to 700nm?
Please reply.
thanx.
rAgAv.
OK
I'm not an expert on the exact formulation of reflection but I think I can
give a reasonable explanation of how the reflection is supposed to happen .
If talking about normal silvered mirrors, I think the principle that is
supposed to apply is that the Electromagnetic waves in the incident light
causes conduction band electrons in the metallic silver to oscillate and
re-transmit the incident light as a reflection. However, because the
conduction band electrons don't have complete freedom of motion and also
because of collisions between the electrons, the efficiency of this process
is not 100% so some energy is lost from the reflection and is either forward
transmitted through the silver layer into the backing material (where it
appears as heat), or contributes to the thermal energy of the conduction
band electrons themselves (i.e. also heat).
Assuming that the mirrors are not moving relative to the source / detector
the light that is reflected will have the same frequency as the original,
and the energy that is _not_ reflected ends up as heat energy in the mirror.
Since some energy is absorbed at each reflection - where we have parallel
mirrors the energy is eventually ALL absorbed (generally as heat).
What do I mean by the 'strength' of that remains? Technically I should say
the number of photons is reduced in this proportion. Certainly not the
wavelength or frequency changing.
Finally I said that if the photons are reflected to and fro 'you wouldn't
see any of them' I simply meant that if they _are_ being reflected between
the two mirrors they are NOT ending up in your eye, or camera or detector so
they quite literally aren't being 'seen'
Let me know of you have any further questions, but if you are only starting
out on this field of study I would suggest you look out some college
textbooks on how light is reflected etc.
.
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