CONFOCALMICROSCOPY Archives

November 2004

CONFOCALMICROSCOPY@LISTS.UMN.EDU

Options: Use Monospaced Font
Show Text Part by Default
Show All Mail Headers

Message: [<< First] [< Prev] [Next >] [Last >>]
Topic: [<< First] [< Prev] [Next >] [Last >>]
Author: [<< First] [< Prev] [Next >] [Last >>]

Print Reply
Subject:
From:
Reply To:
Date:
Fri, 26 Nov 2004 10:56:03 +1100
Content-Type:
text/plain
Parts/Attachments:
text/plain (32 lines)
Search the CONFOCAL archive at
http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal

Joachim Walter wrote:

>
> What you describe would be true, if the light went through the pupil at
> all possible angles. A light bundle, which goes through the pupil at
> only one angle gets focused on one point. So I guess that in the Zeiss
> setup the beam splitter is in a conjugate plane to the pupil, and that
> the light falling on the beamsplitter is orthogonal to the reflecting
> line on the beamsplitter. Must look somehow like a parallel laser beam
> with square cross section, which gets focussed by a cylindrical lens.
> Angles in one direction, line in the orthogonal direction.
> Then the excitation light in the object forms a line, which is
> orthogonal to the line on the beamsplitter. (Line in one direction,
> angles in the other direction).
> The fluorescence light in turn has all possible angles and just a small
> part of it is reflected by the beam splitter.
> Very clever design! (if one likes slit confocals.)


For this to work the 'angles' have to be normal to the scanning slit.
So one gets a line in one direction at the image plane and in the other
direction at the pupil plane.  The 'line reflector' would be normal to
the scanning line.  Extraordinarily ingenious if they can do it.
And it has the other interesting consequence that if one worked
in reflection rather than fluorescence the 0th order would be
blocked sso that one would get a dark-field image.

                                                        Guy

ATOM RSS1 RSS2