I aimed a 780 nm/1mW laser at a 1μm pinhole and took a picture of it using a lens and a CCD camera. The magnification of my imaging system is ~25 times.

I wrote some code that takes the image of the pinhole and finds the FWHM of the pinhole's diffraction pattern based on the relative pixel intensities. The FWHM is 14 pixels across. Since the pixel size on the CCD camera is 3.75 μm, this corresponds to a FWHM of 3.75 μm/pixel * 14 pixels = 52.5 μm.

Using this information, how can I determine the maximum resolution of my imaging system?

My thinking is as follows:

I took a picture of a 1μm pinhole. At 25 times magnification, the 1 μm pinhole should look like it is 25 μm across. Then I divide 52.5 μm/25 μm = 2.1 to find the ratio of actual FWHM:expected FWHM. This means that the resolution is 2.1 times the size of the 1μm pinhole, or 2.1 microns.

I don't think this is correct for several reasons:

1) it seems quite simplistic; since I am using the FWHM, my intuition tells me to consider the the Rayleigh criterion. However, I am not sure how it might fit into this scenario.

2) My units cancel out. I am trying to find the resolution in microns, but this just gives me an arbitrary '2.1'.

3) I am trying to recreate an experiment that someone did with these exact optics. The resolution they calculated was 1.4 μm but I do not know how they calculated it.


1 Answer 1


I have been performing a similar characterisation and perhaps my understanding could help out a little.

The resolution of any optical system is essentially given by its diffraction limit, which is basically limited by the physical construction of your microscope objective. In most cases, diffraction limit is given by the entrance/exit pupil of the objective. Light going through the objective undergoes diffraction at the edges of the objective, providing you a natural blurring at the edges, which results in a resolution limit. Note that the diffraction of the light by the pinhole is not what causes the resolution limit.

The idea of using a small pinhole is to provide a light source close enough to a point source. By imaging this point source, you will essentially be capturing what is known as the point spread function (PSF). The width of the PSF can be taken to be your resolution of the system.

You can calculate the PSF of your optical system as long as you know the geometry of your pupil $P(u,v)$:

$$ h(x_o, y_o) \propto \int_{-\infty}^\infty\int_{-\infty}^\infty P(u,v) \exp\left[ -i\frac{2\pi}{f\lambda} (x_ou + y_ov)\right] \, du\,dv\, , $$

This equation works only for $P(u,v)$ being the entrance pupil geometry, $f$ is the focal length of the objective. Note that you are integrating across the plane where the entrance pupil is located. (Things get more complicated if you want to work with the exit pupil)

For a circular aperture, the PSF is normally given by the Airy Pattern. Which has a width of $\sim \frac{1.22 \lambda}{NA}$, where NA is the numerical aperture of your microscope objective. The tendency is that your NA is not high enough, thus your pinhole width is actually probably about the size of your airy pattern width. Thus this is not sufficiently small to produce a light source which can be considered to be a point source.

Fret not, you could instead, assume your imaging system to have the resolution given by the PSF, and double check that this is correct by comparing what you should expect to image given by theory, and what you measure given by experiment.

According to Introduction to Fourier Optics by Goodman, the image field $U_i$ should be a convolution of the Point Spread Function (PSF) $h$ and the object field $U_o$, i.e.

$$ U_i(x_i,y_i) \propto \int_{-\infty}^\infty\int_{-\infty}^\infty h\left(\frac{x_i}{M}-\xi, \frac{y_i}{M}-\eta\right) U_o(\xi,\eta)\, d\xi d\eta\, , $$

where the proportional sign is for good measure as there are some coefficients in front and $M$ is the magnification of the system, but that does not affect the profile. The image intensity profile is given by

$$ I_i(x_i, y_i) = |U_i(x_i, y_i)|^2\,. $$

TIP: You could perform the integral computationally. I recommend setting $M=1$ in your computations, instead take $M$ into account when processing the image taken from experiment. i.e. new pixel size is given by 3.75/25 microns.


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