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I was reviewing my labnotes and got stuck on one question:

The finite dimension and size of the cantilever tip has an effect on the appearance and size of the images. Tip broadening will occur when the radius of curvature of the tip is of similar or larger size than the feature on the surface. enter image description here

The above figure shows how the height difference between 2 surface features affects the AFM lateral resolution. It can be shown that the minimum distance, d, that you can laterally resolve two sharp spikes of unequal height, $\Delta h$ on a flat surface for a given noise level of $\Delta z$, and for a hemispherical tip of radius, $R$ is: $$d=\sqrt{2R}(\sqrt{\Delta z}+\sqrt{\Delta z + \Delta h}\\$$ Derive the above expression for the minimum resolvable distance, $d$ and show that the above equation holds when $d>\sqrt{2R\Delta h}$.

I tried deriving the expression using the equation for circle: Placing 2 uniform circles with radius R on the spikes and use the centre of the first circle as my origin, and try to find the intersection point of the 2 circles and use that point to get the expression. enter image description here I get two equations: $$x^2+y^2=R^2 \\ (x-d)^2+(y+\Delta h)^2=R^2$$ However solving them lead me to nowhere: $$\begin{align} x^2-2xd+d^2+y^2+2\Delta h y+\Delta h^2 &= R^2 \\ -2xd+d^2-2\Delta h y+\Delta h^2 &= 0 \\ x &= \frac{d}{2} - \frac{\Delta h y}{d} + \frac{\Delta h^2}{2d} \\ \end{align}$$ Substitute x into $x^2+y^2=R^2$: $$\frac{d^2}{4}-\Delta h y+\frac{\Delta h^2}{2}+\frac{\Delta h^4}{4d^2} + y^2 = R^2$$ Solving y using wolfram alpha gets me $$y=\frac{\Delta h \pm \sqrt{\Delta h^2-(d^2-2\Delta h^2-\frac{\Delta h^4}{d^2}+4R^2}}{2} \\ \Delta z =y-\Delta h\\ 2\Delta z= -\Delta h \pm \sqrt{\Delta h^2-(d^2-2\Delta h^2-\frac{\Delta h^4}{d^2}+4R^2)}$$ which I think is too complicated to be the expression I needed for further derivation.


I also tried using the geometry approach:enter image description here $$A+B=d-------(1)$$ First circle :$$\begin{align} R^2 &= (R-\Delta h-\Delta z)^2+A^2 \\ &= R^2-2R(\Delta h + \Delta z)^2 +(\Delta h+\Delta z)^2 + A^2\\ 2R(\Delta h + \Delta z)&=(\Delta h+ \Delta z)^2+A^2 -------(2) \end{align}$$

Second circle:$$\begin{align} R^2&=(R-\Delta z)^2+B^2 \\ 2\Delta z R &= \Delta z^2 + B^2\\ &= \Delta z^2 + d^2 -2Ad + A^2 --------(3)\end{align}$$

Subtracting (2)-(3) and using (1): $$\begin{align}2R\Delta h &= \Delta h^2 + 2\Delta h \Delta z + A^2 - B^2 \\ &= \Delta h^2 + 2\Delta h \Delta z + d(A-B) \end{align}$$ And I don't know how to proceed from here....

Any help is really appreciated!

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You were almost there but try using the intersecting chord theorem twice to find $x$ and $y$ eliminating second order terms on the way.

The clues to using intersecting chords are the presence of all those square roots and the condition on the distance $d$.
You may have used the theorem when studying Newton's rings?

enter image description here

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  • $\begingroup$ How do I eliminate the second order terms? I get $d=\sqrt{2R(\Delta h+\Delta z)-(\Delta h + \Delta z)^2}+\sqrt{2R\Delta z - \Delta z^2}$ as my answer. $\endgroup$
    – WeiShan Ng
    Sep 2 '17 at 3:41
  • $\begingroup$ Sorry I still don't get it.....This is what I did:$$\begin{align} x^2 &=(2R-(\Delta h+ \Delta z))(\Delta h +\Delta z) \\ x &= \sqrt{2R-(\Delta h+\Delta z}\sqrt{\Delta h+\Delta z }\\y^2&=\Delta z(2R-\Delta z) \\y&=\sqrt{2R-\Delta z}\sqrt{\Delta z} \end{align}\\$$ $\endgroup$
    – WeiShan Ng
    Sep 2 '17 at 5:49
  • $\begingroup$ So we eliminate the second order terms because they are insignificant? $\endgroup$
    – WeiShan Ng
    Sep 2 '17 at 6:06

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