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echo: atm
to: ATM
from: burrjaw{at}earthlink.net
date: 2003-04-02 20:30:18
subject: Re: ATM How Good Does a Diagonal Need to Be?

To: atm{at}shore.net
From: Jim Burrows 
Reply-To: Jim Burrows 


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At 21:26 2003-04-01 -0500, Mark Holm wrote:

>The wave front error caused by a diagonal mirror error is affected by the
>45 degree angle.
>
>This is not an easy thing for me to visualize.  I have chosen the
>physicist's usual route of imagining the simplest possible geometry.
>
>Now if I slide one half of the diagonal perpendicular to the diagonal
>surface by say 100 nanometers, I will have the same effect as a diagonal
>with a 100 nanometer step error on its surface.  If this were a
>perpendicular folding mirror instead of a 45 degree diagonal, the 100
>nanometer step would cause a 200 nanometer path length difference.

Sorry, this requires an HTML picture; I just can't hack an ASCII drawing.
MikeHolm.jpg
Ray ABC (wavefront AA') is bouncing off the 45 deg diagonal BB' to form
wavefront CC'.  Note that the optical path length ABC = A'B'C'.  Now
displace this part of the diagonal down by 100 nm so that ray AB now goes
ADC'.  The path length difference is only *one* side (BD) of the square
with diagonal 200 nm, or 141 nm.

         -- Jim Burrows
         -- mailto://burrjaw{at}earthlink.net
         -- http://home.earthlink.net/~burrjaw
         -- Seattle N47.4723 W122.3662 (WGS84)
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At 21:26 2003-04-01 -0500, Mark Holm wrote:
The wave front error caused by a
diagonal mirror error is affected by the 45 degree
angle. This is not an easy thing for me to
visualize.  I have chosen the physicist's usual route of imagining
the simplest possible geometry.
Now if I slide one half of the diagonal perpendicular to the diagonal
surface by say 100 nanometers, I will have the same effect as a diagonal
with a 100 nanometer step error on its surface.  If this were a
perpendicular folding mirror instead of a 45 degree diagonal, the 100
nanometer step would cause a 200 nanometer path length
difference.
Sorry, this requires an HTML picture; I just can't hack an ASCII drawing.
MikeHolm.jpg

Ray ABC (wavefront AA') is bouncing off the 45 deg diagonal BB' to form
wavefront CC'.  Note that the optical path length ABC =
A'B'C'.  Now displace this part of the diagonal down by 100 nm so
that ray AB now goes ADC'.  The path length difference is only
*one* side (BD) of the square with diagonal 200 nm, or 141 nm.

        --
Jim
Burrows
        --
mailto:%2F%2Fburrjaw{at}earthlink.net"
eudora="autourl">mailto://burrjaw{at}earthlink.net;
        --
http://home.earthlink.net/~burrjaw"
eudora="autourl">http://home.earthlink.net/~burrjaw;
        --
Seattle N47.4723 W122.3662 (WGS84)


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--=====================_7234618==.REL--

--- BBBS/NT v4.00 MP
* Origin: Email Gate (1:379/1.100)
SEEN-BY: 633/267 270
@PATH: 379/1 633/267

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