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


>At 09:26 PM 4/1/2003 -0500, Mark Holm wrote:
>>
>>
>>Say I have a perfectly flat diagonal, split neatly into two parallel halves,
and
>>   that the split is in the plane that includes the center of the image
plane,
>>the center of the diagonal and the center of the primary.  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.  Because it is on a 45 degree angle, the 100 nanometer
step
>>causes a  282.8 nanometer path length difference, but the path is also offset
by
>>141.4 nanometers.
>>
>>The offset is the troublesome part.  I haven't yet figured out how to allow
for
>>it in the path length calculation.  In calculating the path lengths to a
>>particular point on the image plane, I need to allow for the fact that the
>>offset paths are coming from a slightly different place on the primary.
Because
>>the primary is curved, this will change the path length.  The primary curve
is
>>shallow, so the path length difference due to the offset at the primary
should
>>be considerably less than 141.4 nanometers.  Therefore, the resulting path
>>length difference should be less than, but probably not too much less than
282.8
>>nanometers.
>>
>>My poor brain isn't up to the calculation just now.  I'll have to let it wait
>>for a time when I am more alert.  Perhaps someone has already been down this
>>path and can quote the answer from memory.
>>
>>If my analysis above isn't too far off the mark, wave front errors resulting
>>from diagonal mirror errors should be multiplied by slightly less than the
>>square root of 2 (1.414 etc.) as well as by the usual factor of two applied
to
>>reflections.
>>
>>Mark Holm
>>mdholm{at}telerama.com
>>

Mark -- I addressed the question of the path length difference for a
surface error in a diagonal in an earlier version of this thread  (Subject:
Re: ATM Flats  Date: Fri, 14 Mar 2003 15:01:52 -0400).  To save space I had
simply stated the following as a fact without any explanation:

"...3. The optical path difference at 45 degrees incidence is sqrt(2)
times the normal separation (not a factor of 2 as in normal
incidence)..."

In other words a further multiplication by the usual factor of 2 is incorrect.

Perhaps an explanation is now called for.  I'll try my best without a blackboard.
Set up an x-y coordinate system.  Draw a square of side 1 with its corners
at (0,0), (1,0), (1,1), (0,1).  The figure 1 is arbitrary and the units
don't matter. Draw the diagonal from (0,1) to (1,0).  The diagonal line
represents the surface of the error-free diagonal.  The left edge of the
square represents an incoming wave front; the bottom edge represents an
outgoing wave front.  The path length of any ray from the incoming to the
outgoing wavefronts is exactly 1.0; i.e., it equals the LENGTH OF THE SIDE
OF THE SQUARE.

Now suppose there is a region on the surface offset from the error-free
part by some amount E (your "100 nanometer step error" for
example).  To be definite let's say the region is offset to cause an
INCREASE in path length for rays that strike it.  This region can now be
represented by another diagonal line running from (0,1+E*1.41) to
(1+E*1.41, 0).  The new diagonal line is the diagonal of a square of side 
1+E*1.41, and hence the difference in path lengths is just E*1.41.  We are
finshed with the computation.  There is no need to increase the path length
difference by another factor, such as 2.

If any of the above is unclear please let me know.

Martial Thiebaux
Rawdon Hills, Nova Scotia

--- BBBS/NT v4.00 MP