| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
16 Nov 2005 05:04:28 AM |
| Object: |
A question about lens |
I am not a physicist. I love the answers I get asking questions to this
community.
Could light bend at a rate so high that it "goes back" in the same
direction of incidence? In other words, can a refractive enough
material (or distribution of materials) form a lens which starts
behaving like a mirror?
(I know if you bend an optical fiber, you get this behavior through
geometry. But I am not asking that.)
Is reflection nothing but this "over refraction"?
Are there objects dense enough to reflect all the light passing
through? It would be impossible to see through such objects. Unlike
black holes absoring energy, such objects would reflect energy.
Thanks in advance,
-Bhushit
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| User: "Andy Resnick" |
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| Title: Re: A question about lens |
16 Nov 2005 07:48:35 AM |
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wrote:
I am not a physicist. I love the answers I get asking questions to this
community.
Could light bend at a rate so high that it "goes back" in the same
direction of incidence? In other words, can a refractive enough
material (or distribution of materials) form a lens which starts
behaving like a mirror?
Maybe you are thinking of "total internal reflection". That requires
light exiting a region of high refractive index and entering a region of
low refractive index.
Or maybe you mean something mroe exotic, something called a "phase
conjugate mirror"- unlike normal reflection, the light is sent back in
the same direction it was incident on the mirror. This is a non-linear
optical phenomenon (4-wave mixing).
(I know if you bend an optical fiber, you get this behavior through
geometry. But I am not asking that.)
Is reflection nothing but this "over refraction"?
Both are simple descriptions for the more general case of propogation in
a non-uniform media. In general, no- reflection is not "over refraction".
Are there objects dense enough to reflect all the light passing
through? It would be impossible to see through such objects. Unlike
black holes absoring energy, such objects would reflect energy.
I'm not sure it has to do with density, per se. I don't see any reason
why something could not be completely reflective, or a perfect
scatterer: the only constraint is that a + r + t = 1, where a is the
coefficient of absorption, r the coefficient of reflection, and t the
coefficient of transmission. This is just conservation of energy.
Invent a material with a = t = 0, and then it's automatically completely
reflective. Doubtful that such a material exists, maybe some sort of
aerosol.
--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
Case Western Reserve University
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| User: "Richard Herring" |
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| Title: Re: A question about lens |
16 Nov 2005 08:28:23 AM |
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In message <dlfdbv$pki$1@eeyore.INS.cwru.edu>, Andy Resnick
<andy.resnick@op.case.edu> writes
I'm not sure it has to do with density, per se. I don't see any reason
why something could not be completely reflective, or a perfect
scatterer: the only constraint is that a + r + t = 1, where a is the
coefficient of absorption, r the coefficient of reflection, and t the
coefficient of transmission. This is just conservation of energy.
Invent a material with a = t = 0, and then it's automatically
completely reflective. Doubtful that such a material exists, maybe
some sort of aerosol.
I don't think you can achieve that at all wavelengths, even in
principle, thanks to Kramers and Kronig.
--
Richard Herring
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| User: "PD" |
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| Title: Re: A question about lens |
16 Nov 2005 09:40:34 AM |
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wrote:
I am not a physicist. I love the answers I get asking questions to this
community.
Could light bend at a rate so high that it "goes back" in the same
direction of incidence? In other words, can a refractive enough
material (or distribution of materials) form a lens which starts
behaving like a mirror?
Yes indeed. First things first: Visualize the boundary between high
index of refraction and low index of refraction, and an axis
perpendicular to that boundary. Recall that the law of refraction says
that the ray's angle with respect to that axis is steeper in the
*lower* index of refraction. This means that the limit you are
referring to occurs for light *leaving* a lens. This phenomenon is
called "total internal reflection".
It also occurs in swimming pools, if you look upward from underneath.
You will see the whole vista above the surface, horizon to horizon,
contained in a circle. Outside that circle you will see only the bottom
of the pool, reflected in the surface of the water.
(I know if you bend an optical fiber, you get this behavior through
geometry. But I am not asking that.)
Actually, it is exactly total internal reflection that makes a fiber
work. There is no "mirrored" surface on the inside of a fiber casing.
Is reflection nothing but this "over refraction"?
Not typically, no. Mirrored surfaces reflect for a different reason.
Typically, it is because they are conductors, not because they are
dense.
Are there objects dense enough to reflect all the light passing
through? It would be impossible to see through such objects. Unlike
black holes absoring energy, such objects would reflect energy.
Thanks in advance,
-Bhushit
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| User: "tadchem" |
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| Title: Re: A question about lens |
16 Nov 2005 06:56:08 AM |
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wrote:
I am not a physicist. I love the answers I get asking questions to this
community.
<there's no accounting for taste ;-)>
Could light bend at a rate so high that it "goes back" in the same
direction of incidence? In other words, can a refractive enough
material (or distribution of materials) form a lens which starts
behaving like a mirror?
Possibilities:
fiber-optic U-shaped light conduit
cube corner reflectors
retroreflective beads (cat's eye reflectors)
http://www.bikexprt.com/bicycle/reflectors/reflwrk.htm
Practically, each of these is only effective over a limited waveband.
(I know if you bend an optical fiber, you get this behavior through
geometry. But I am not asking that.)
That is the U-shaped light conduit
Is reflection nothing but this "over refraction"?
No. Refraction requires that the energy enter a second medium.
Reflection requires the energy be *prevented* from entring a second
medium. The 'cat's eye' reflectors provide 100% reflection for
wavelengths at which the index of refraction is a particular value
(2.000 IIRC). At certain angles of reflection (depending on the index
of refraction) the reflection is 100%:
http://scienceworld.wolfram.com/physics/TotalInternalReflection.html
Are there objects dense enough to reflect all the light passing
through?
If you are including all electromagnetic energy from radio waves to
cosmic rays, then the answer is no. Ordinary matter as we know it
cannot do that.
It would be impossible to see through such objects. Unlike
black holes absoring energy, such objects would reflect energy.
They would be perfect mirrors. One could see a reflection of
*everything* around them like looking at one of those big spherical
'surveillance' mirrors mounted high on the wall in stores.
Tom Davidson
Richmond, VA
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| User: "" |
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| Title: Re: A question about lens |
16 Nov 2005 05:11:39 AM |
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I have passed your question on to GOD (The General Overall Director) to
be answered during his next press conference.
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| User: "Gregory L. Hansen" |
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| Title: Re: A question about lens |
16 Nov 2005 01:15:54 PM |
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In article <1132139068.316293.167750@g44g2000cwa.googlegroups.com>,
<joshipura@gmail.com> wrote:
I am not a physicist. I love the answers I get asking questions to this
community.
Could light bend at a rate so high that it "goes back" in the same
direction of incidence? In other words, can a refractive enough
material (or distribution of materials) form a lens which starts
behaving like a mirror?
(I know if you bend an optical fiber, you get this behavior through
geometry. But I am not asking that.)
Is reflection nothing but this "over refraction"?
Are there objects dense enough to reflect all the light passing
through? It would be impossible to see through such objects. Unlike
black holes absoring energy, such objects would reflect energy.
In a way, yes. But maybe not in the way you're thinking.
Snell's law says
n1 sin(a1) = n2 sin(a2)
where n1 and n2 are the indices of refraction of two materials, and a1 and
a2 are the angles perpendicular to the surface. Usually air is one of the
materials, with n=1 (or close enough). Looking at the relative size of
the angles,
n1/n2 = sin(a2)/sin(a1)
Going from the lower to higher n, the light ray bends towards the
perpendicular, or straightens out. That's sort of the opposite of
reflecting. But going from the higher n to the lower, the light bends
more severely. When the angle of refraction is 90 degrees (parallel to
the surface), sin(90 deg)=1, and call n2=1 for air (a2=90),
n1 = 1/sin(a1)
When the condition is met you get total internal reflection, which is the
case for fiber optics. It's still reflected at an angle, not directly
back. To reflect directly back (angle of incidence = 0, sin(a1)=0) the
index of refraction n1 has to go to infinity.
There's also partial reflection at an interface. For light hitting
perpendicular, the amount reflected is
R = (n-1)/(n+1)
assuming air or vacuum at one interface. When n gets large, R goes to 1,
total reflection at the interface. n doesn't get that large. But with
multiple interfaces, the reflection can be increased. More so if the
thicknesses of each layer are designed so that reflected light interferes
constructively. Then it's a dialectric mirror, which can be more
reflective than any shiny metal. Similar technology can make
anti-reflective surfaces, often used on lenses. In principle you can make
mirrors or anti-reflective coatings only for a limited range of
wavelengths, but in practice only visible light counts, and you can do
quite well in that 800 or so nanometer range of wavelengths. With
computer-based modeling and modern methods of controlling surface
thicknesses, you can make reflection versus wavelength curves with
virtually any shape you like, and participants at optical conferences have
competed in making reflection curves that most closely match arbitrary
silhouettes like the Taj Mahal.
--
"A few months in the laboratory will save a few hours in the library."
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| User: "Matthew Lybanon" |
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| Title: Re: A question about lens |
17 Nov 2005 10:12:26 AM |
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Look up "total internal reflection" (sometimes you may find it without the
"internal").
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| User: "" |
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| Title: Re: A question about lens |
18 Nov 2005 04:08:15 AM |
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Thanks for replies I got so far.
I am not talking about "total internal reflection" where in the light
comes from denser side and bounces back into denser medium - like the
swimming pool phenomenon or fiber glass phenomenon someone describes.
I am also not talking about gradient in refractive index or U bend or
some other geometric technique as someone describes.
I am talking about light coming from air on a "glass pane", bent so
much that it comes back in the air. It is much like "perfect mirror"
someone describes.
I wonder whether light coming from air can even go parallel to the pane
surface on incidence. That would mean the light just "disappears" -
neither passed through, nor thrown back and the object remains
invisible.
I hope this clarifies.
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| User: "PD" |
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| Title: Re: A question about lens |
18 Nov 2005 10:11:45 AM |
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wrote:
Thanks for replies I got so far.
I am not talking about "total internal reflection" where in the light
comes from denser side and bounces back into denser medium - like the
swimming pool phenomenon or fiber glass phenomenon someone describes.
I am also not talking about gradient in refractive index or U bend or
some other geometric technique as someone describes.
I am talking about light coming from air on a "glass pane", bent so
much that it comes back in the air. It is much like "perfect mirror"
someone describes.
I wonder whether light coming from air can even go parallel to the pane
surface on incidence. That would mean the light just "disappears" -
neither passed through, nor thrown back and the object remains
invisible.
No, and a glance at Snell's law reveals why. In passing a boundary from
low to high index of refraction (say, air to glass), light is bent to
be more *toward* the perpendicular to the boundary, not *away* from the
perpendicular to the boundary. So there is no approach angle from the
air side that will produce an angle in the glass that is so steep that
the light would never come out.
PD
I hope this clarifies.
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| User: "" |
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| Title: Re: A question about lens |
24 Nov 2005 12:45:06 AM |
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I understand "towards the medium". It means as the medium gets denser,
the light goes farther from air-glass border.
Think of a pencil half in air and half behind a glass block. This glass
has a nob attached to control its refraction. As I turn the nob, the
refraction changes.
This means the refraction angle should change from
the a side of perpendicular line (pencil appears broken) to
the perpendicular line (pencil appears perpendicular to the surface of
incidence) to
the other side of perpendicular line (pencil broken the other way) to
"reverse direction of total internal reflection", the other side right
angle to perpendicular line (pencil disappears) to
obtuse angle from the perpendicular line from the other side (pencil's
remaining parts are seen in the air)
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