[3dem] RE: electron diffraction

[Marin van Heel]

Dear Philip,

All IMAGIC CTF programs including "transfer", "ctf2d-operations" and 
"ctf2d-find" ask you for the focal distance and the objective aperture 
and they do show you the cut-off effects on the CTF by this aperture.  
In fact, when you have partial coherence in the illumination, the CTF 
cutoff is more complicated, as it is determined by the convolution of 
the source image in the back focal plane with the coherent transfer 
function multiplied by the aperture function. Almost all CTF programs in 
EM use the (incorrect) envelope approximation for dealing with partial 
coherent illumination.  People often ask me why the IMAGIC progams ask 
you for the focal distance of the objective lens: you do need it for a 
correct transfer function calculation. (See: Marin van Heel, *On the 
imaging of relatively strong objects in partially coherent illumination 
in optics and electron optics**,* /Optik/ *47* (1978) 389-408).

Cheers,

Marin

On 26-Aug-13 9:28 AM, Philip Köck wrote:
> 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


-- 
================================================================

     Marin van Heel

     Professor of Cryo-EM Data Processing

     Leiden University
     NeCEN Building Room 05.27
     Einsteinweg 55
     2333 CC Leiden
     The Netherlands
      
     Tel. NL: +31(0)715271424 // Mobile NL: +31(0)652736618
     Skype:    Marin.van.Heel
     email:  marin.vanheel(A_T)gmail.com
     and:    mvh.office(A_T)gmail.com

----------------------------------------------
     and:

     Professor of Structural Biology

     Imperial College London
     Faculty of Natural Sciences
     Biochemistry Building (Room 512)
     South Kensington Campus
     London SW7 2AZ,  UK
     email:  m.vanheel(A_T)ic.ac.uk

     Tel. UK:   +44(0)2075945316 //Mobile: +44(0)7941540625

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     Currently visiting Professor at:

     Laboratório Nacional de Nanotecnologia - LNNano
     CNPEM/ABTLuS, Campinas, Brazil
     Brazilian mobile phone  +55-19-99369051

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