Heisenberg's microscope used to demonstrate the Observer Effect/Uncertainty Principle begins by stating that the resolving power of any microscope is $$\Delta x=\frac{\lambda}{\sin \epsilon}$$ Being $\Delta x$ the minimum distance resolvable by the microscope, $\lambda$ the wavelength of light used, and $\epsilon$ the angle formed by the cone of light coming off the object which can pass through the microscope lens.

Now, since only the position in one dimension is required, I think it is reasonable to assume that instead of a circular lens being used, a prismatic lens can be used, which focuses all parallel rays of light to a line rather than a point, which should be enough to determine the $x$-position of a particle.

I know the cause of this limit is diffraction, but I can't quite get how to derive it. My first thought was to imagine two points in the lens's focal plane, the light coming from which is made into parallel rays of light by the lens, and then with those parallel rays calculate the angles of the fist diffraction minima using the simple ray diagrams usually used to demonstrate diffraction. Then find the distance between two points in which one of the the first minima of one pattern overlaps the other's maximum.

I have some problems with this method. First, I am not sure how to calculate the diffraction angles if the light is coming in at an angle, because there wil be a phase difference in the light emanating from different points in the slit. Second, I think this is too specific a case to apply to any microscope used to resolve the $x$-position of a particle.

Please help.


1 Answer 1


This problem comes down to the fact that, in general, the diffraction integral cannot be solved. I think some of the important features of the problem can be found without solving it.

Imagine the following 2D physical system, $x$ and $z$. Two point sources, with cylindrical waves emanating from them in phase are located a distance $z_0$ from left of a slit screen. The two points are separated by a distance $x = \Delta x$. The slit screen has an aperture in it of width $d$ centered at $x=0$. Finally, there is a second imaging screen at a distance $z_1$ to the right of the slit screen. The electric field at the slit screen is given by

$$ E(x',z_0) = E_1(x',z_0) + E_2(x',z_0) = E_0\frac{e^{i k \sqrt{x'^2 + z_0^2}}}{\sqrt{x'^2 + z_0^2}} + E_0\frac{e^{i k \sqrt{(x'-\Delta x)^2 + z_0^2}}}{\sqrt{(x' - \Delta x)^2 + z_0^2}} $$

where $k=2 \pi/\lambda$. The electric field at the imaging screen is given by the diffraction integral. Examine the diffraction integral for the second term in the above electric field (see the above link for coordinate notation) to find:

$$ E(x,z_1) = \frac{1}{i \lambda} \int_{-d/2}^{d/2} \frac{e^{i k \sqrt{(x'-\Delta x)^2 + z_0^2}}}{\sqrt{(x' - \Delta x)^2 + z_0^2}} \frac{e^{i k \sqrt{(x - x')^2 + z_1^2}}}{\sqrt{(x - x')^2 + z_1^2}} dx' $$

if I first make the substitution $u = x' - \Delta x$ and then follow it with the substitution $\chi = x + \Delta x$, I find that this term becomes

$$ E(x,z_1) = \int_{-d/2}^{d/2} E_1(u,z_0) \frac{e^{i k r}}{r} du + \int_{-d/2- \Delta x}^{-d/2} E_1(u,z_0) \frac{e^{i k r}}{r} du - \int_{d/2 - \Delta x}^{d/2} E_1(u,z_0) \frac{e^{i k r}}{r} du $$

where I have defined

$$ \frac{e^{i k r}}{r} = \frac{e^{i k \sqrt{(\chi - x')^2 + z_1^2}}}{\sqrt{(\chi - x')^2 + z_1^2}}. $$

So, the field at the imaging screen due to the point offset from the horizontal axis by $\Delta x$ consists of 3 terms, a term identical to the field at the imaging screen due to the on axis point, but offset by an amount $\Delta x$ and two extra integrals.

I can (approximately) ignore the two extra integrals if I assume that the separation between the two sources, $\Delta x$ is much smaller than the diameter of the aperture, $d$. Given this, the second integral is approximately equal to the third integral and they cancel.

At this point I have found that the electric field at the imaging screen is given by the sum of two terms which are identical integrals but one is offset from the other by an amount $\Delta X$, but the integral is still, in general, impossible to integrate.

To make the problem tractable, I specialize to the case of Fraunhofer diffraction. In which case I can write down the result of the integrals as

$$ E(x,z_1) = E_0 d \; \mathrm{sinc} \left( \frac{\pi d x}{\lambda z_1} \right) + E_0 d \; \mathrm{sinc} \left( \frac{\pi d (x - \Delta x)}{\lambda z_1} \right) $$

The peak of the second distribution overlaps with the first null of the first distribution when $\pi d x / (\lambda z_1) = \pi$ (this is the Rayleigh criterion) or, in other words, diffraction doesn't limit my resolution when $\Delta x \ge \frac{\lambda z_1}{d}.$

Now, on to optics. In order to form an image on the imaging screen, I need to put a cylindrical lens in to the slit (I assume a thin lens for simplicity, here). The imaging condition for this lens is given by geometric optics and is

$$ \frac{1}{z_0} + \frac{1}{z_1} = \frac{1}{f} $$

where $f$ is the focal length of the thin lens. The magnification of the system is $m = -z_0/z_1$. I won't use the negative sign from now on, but its there. This also modifies the argument of the sinc functions above with $x \rightarrow x/m$, it is changing the size, but not the shape of the image on the imaging screen. The diffraction limit is now given by $\Delta x \ge \frac{m \lambda z_1}{d} = \frac{\lambda z_0}{d}.$ Notice that I can't improve on the diffraction limit with magnification.

Finally, geometrically, the most divergent ray that can leave the point source on axis is defined by the problem statement to be $\sin(\epsilon) = d/(2 z_0)$ and I have at last

$$ \Delta x \ge \frac{\lambda}{2 \sin(\epsilon)}. $$


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