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A SEM operates at $30keV$ and uses an objective lens with spherical aberration constant, $C_s=2mm$.

  1. Calculate the optimal probe forming angle for this system.
  2. Calculate the resolution of this SEM.

This is a tutorial question for my Master's Imaging course. In the solutions, first the de Broglie wavelength is computed (which is fine), then:

  1. An equation is introduced for the optimal probe forming angle which I can't find anywhere (in my notes or online) and so don't understand its origin, this is: $$ \alpha_{opt}=1.1\left(\frac{\lambda}{C_s}\right)^{\frac{1}{4}} $$
  2. An equation (or actually 2) is introduced for the resolution which, again, I can't seem to find anywhere and so, again, fail to understand, these are: $$ resolution=\frac{d_{min}}{2}=\frac{1.3\times\lambda^{\frac{3}{4}}\times C_s^{\frac{1}{4}}}{2} $$

If anyone could shed some light as to where these are coming from it would be very much appreciated.

Cheers

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1 Answer 1

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(I ended up figuring it out)

Turns out the total probe size is given by combining the effects of diffraction and spherical aberration, i.e. ${d_{tot}}^2={d_{C_s}}^2+{d_{diff}}^2$ and. $$ d_{diff}=\frac{1.22\lambda}{\alpha}, $$$$ d_{C_s}=\frac{\alpha^3C_s}{2}. $$

Differentiating ${d_{tot}}^2$ w.r.t. $\alpha$ and setting this equal to zero gives a minimum. This can be re-arranged for $\alpha$ to give the optimisation angle, $\alpha_{opt}$, in terms of $\lambda$ and $C_s$ (which is the 1st expression I asked about in the question).

Then setting $\alpha =\alpha_{opt}$ in the original expression, $d_{min}$ can be found in terms of $\lambda$ and $C_s$ (which was the numerator of the 2nd expression I asked about in the question).

Resolution is then just defined to be $\delta_{min}=\frac{d_{min}}{2}$ which concludes all my original problems.

(I just answered it in case anyone else finds it useful)

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