[3dem] RE: electron diffraction

[Philip Köck]

Thank you all very much for the answers you sent. I just picked one out to reply to.
Thank you , Nestor, for the tables.

The reason I asked is that I want to know what spatial frequencies are removed by an objective aperture
when recording high resolution (phase contrast) images.
I wasn't sure whether the objective lens settings change when one switches between imaging and diffraction.
Therefore the strange way of asking the question.
Secondly I wasn't sure whether one can use the simple formulas given that the lens is not a thin lens.
I don't know the position of the specimen with respect to the focal planes or the principal planes either.

In any case everybody seems to point me to a simple calculation.
For example for 200 kV and 1 Angstrom resolution cutoff I can take twice the angle in Nestor's table,
resulting in angle= 25 mrad.
I can also get that from Lamda = d sin(angle), if I regard the specimen as a two dimensional grating.

The next step is to use the focal length (not the distance between specimen and focal plane) as camera length,
giving a distance of the spot from the optical axis R = 72.5 micrometers for 2.9 mm focal length (f).
The formula is f * tan(angle) = R.
Most answers seem to agree about using f.

Please tell me if I got anything wrong.

Philip


-----Original Message-----
From: Wim Hagen [mailto:wim.hagen at embl.de] 
Sent: 24 August 2013 14:05
To: Philip Köck
Cc: microscopy at microscopy.com; 3dem at ncmir.ucsd.edu
Subject: Re: [3dem] electron diffraction

Hi Philip,

A simple estimation:

Angular distance in back focal plane [rad] = 2 * Bragg angle  [rad] =  Lambda (relativistic wavelength)  [m] / d (sample spacing) [m] = R (distance) [m] / F (focal length) [m]

All CM/Tecnai Twin systems have a focal length of ~2.9 mm. For 200KV, a 0.1 nm spacing gives ~25 mrad so ~72.5 micrometers distance from optical axis in back focal plane.

Put in the radius of your objective apertures for R and you can calculate where your objective apertures cut off resolution for each KV.

Best,

Wim


On Aug 23, 2013, at 4:18 PM, Philip Köck wrote:

> Hi everybody,
> 
> I'm looking for a simple estimation that I can't find in any textbook I've looked in, nor on the internet.
> I wonder if anybody can help me.
> 
> If I put a diffraction grating with spacing 1 Angstrom into the specimen holder and record an image of it.
> At what distance from the optical axis will the corresponding 
> diffraction spot (for 1 Angstrom spacing) appear in the back focal plane of the objective lens.
> 
> Assume a TEM with 200 kV and a twin lens, one that's typically used for protein structure determination.
> 
> Thanks,
> 
> Philip
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> 3dem at ncmir.ucsd.edu
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  Wim Hagen
  Senior Engineer Electron Microscopy
  European Molecular Biology Laboratory (EMBL)
  Heidelberg, Germany                    
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