Re: long antinodes?

[jhowell@yang.earlham.edu (john howell)]

On 2/25,  Ludwik Kowalski wrote:
>
SNIP
> Let me rephrase his question to show a puzzling situation. Suppose we have
> two parallel laser beams whose temporary coherence is nearly PERFECT. The
> first beam travels along the x axis passing a beam splitter positioned at
> 45 degrees. The second beam, after being reflected from a mirror, and from
> the beam splitter, is superimposed on the first one. The phase relation
> between the beams is controlled by a retarding glass plate (thickness
> between zero to lambda/4) to make sure they are out of phase. The beams are
> polarized in the same plane and an additional layer of glass is provided (to
> compensate a glass plate of the splitter, as in Michelson's interferometer).
>
> In principle, this arrangement would create two electromagnetic waves which
> are out of phase everywhere after the beam splitter. It would be a very long
> "antinode" (no energy there). Actually there would be two such antinodes,
> one pointing to the right and one pointing down.
>
> The beam splitter is controlled by nature and energy must be conserved. How
> can this be satisfied? ...
SNIP

This is a thought-provoking paradox.  I think its solution is subtle, and
is based on the fact that when light that is travelling in air reflects
(externally) off of a glass surface, the light's phase changes by 180
degrees, but no such change takes place when light that is travelling in
glass is internally reflected.

To apply this fact to the situation described in Ludwik Kowalski's note, we
must be a bit more specific about the geometry of the set-up.  Let us
assume that the coated glass beam splitter is positioned so that its
"splitting" side is "downstream" from laser 1.  When light from laser 2 is
"split" by this beam splitter, it will be undergoing an EXTERNAL
reflection, so it will undergo a 180 degree phase shift.

Ludwig's assumption is that the two beams moving along the x-axis are
exactly 180 degrees out of phase.  Let's imagine that this phase difference
is all the result of the reflection off of the beam splitter.

I've tried to sketch the experimental set-up, and if I've done it
correctly, there would actually be superposition of light along TWO
directions:  the x-axis (undeflected light from laser 1, superposed with
refclected light from laser 2) AND ALSO the y-axis (undeflected light from
laser 2, superposed with reflected light from laser 1).

The situation has an assymetry, namely: the reflection that sent light from
laser 1 along the y-axis was a reflection of light that was traveling IN
glass, reflecting OFF an air interface.   This INTERNAL reflection will not
produce any phase shift.  Consequently, the superposed light that travels
along the y axis is an "in-phase" superposition of light from the two
lasers.  The arithemetic then works out to show that this beam carries the
total energy that was incident in the two separate laser beams, so energy
IS conserved.

Details of the "arithmetic."  Note:  in the following, I'm going to pretend
that the intensity of light is equal to the square of its amplitude: a beam
with a maximum E-field of E0 will be said to have intensity E0^2.  I know
that I'm omitting a proportionality constant here; I think that constant
has a value of 1/(2 mu0*c) in vacuum, so that I should say the intensity of
such a  beam is E0^2/(2 mu0*c).  I omit the constant because:
   (a) I'm not 100% sure of its value;
   (b) the value is irrelevant here, and
   (c) omitting it makes typing an E-mail SO much easier.

The important results remain true when the constant is included -
regradless of its value!}

Suppose each laser produces light that has a max E-field of E0.  Each then
laser has an intensity of E0^2, for a total intensity output of 2*E0^2.  If
energy is conserved, we must account for all of this intensity.

A single beam would be "split" into two beams, each with intentisy E0^2/2,
so the max E-field of each beam (before we worry about superposition) is
E0/sqrt(2).

The TOTAL beam along the y axis will therefore have an "in-phase"
superposition of two beams, each with maximum E-field of E0.sqrt(2),
resulting in a total maximum E-field of 2*E0/sqrt(2) = E0*sqrt(2).  The
intensity of that beam will therefore be {E0*sqrt(2)}^2 = 2*E0^2, which is
the sum of the intensities of the two lasers.  So all the initial energy is
accounted for!}


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John Howell
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