# What equation should I use to determine retinal image size generated by 10 micron DIA input beam?

If I plot the effective focal spot diameter ($$S$$) achieved by using the popular expression: $$S = \frac{4L}{\pi}\frac{F}{D}$$, where L is input beam wavelength, $$F$$ is the focal length of the lens, and $$D$$ is the input beam diameter, then as $$D$$ decreases you reach a point where $$D = S$$. If you continue to decrease $$D$$, the converging lens seems to behave as a diverging lens.

For example, if I have a 633 nm collimated beam, and a lens with $$F =$$ 10 mm (on the order of the typical human eye lens to retina distance), and I incrementally decrease $$D$$, then at $$D$$'s lower than 90 microns the resulting retinal spot size (image) would be larger than 90 microns.

Is this actually what occurs or does the above equation break down at this scale? If the latter, what equation should I use to determine spot size for very small input beams through the eye lens focused onto the retina (i.e. ~10 microns or so)?

You hit upon an important concept. Indeed as the aperture, $$D$$, becomes small enough (even though $$D \gg L$$) diffraction from the aperture dominates whatever simple quadratic phase curvature the lens imparts. To lowest-order the diffracted half-angle, $$\theta$$, can be found from the product of $$D$$ and $$\theta$$ and put equal to $$L$$. So $$\theta \approx L/D$$.