Does the substage condenser numerical aperture cap total resolution of an imaging system?

Question

Amongst formulas to estimate the theoretical resolution of a transmitted light microscope, I often see:

$$ \delta = \frac{1.22\lambda}{NA_{obj} + NA_{cond}} $$

Where $NA_{cond}$ is the numerical aperture of the condenser lens, and $NA_{obj}$ is the numerical aperture of the objective lens. $\delta$ is the resolution estimate.

This formula, however, implies that when $NA_{cond}$ is larger than $NA_{obj}$ , it will lead to increase in the resolution, despite the objective lens not being able to accept the light cone with a larger half-angle than what it was designed for.

My intuition here is that the actual formula should be

$$ \delta = \frac{1.22\lambda}{2 \times min(\{NA_{obj}, NA_{cond}\})} $$

So that the largest half angle will not benefit the resolution, because it will be limited by the least half angle of the lenses involved.

What am I missing?

Answer 1

This expression is only valid for "properly configured" microscopes where $NA_{\text{obj}} \approx NA_{\text{cond}}$. For any system that strays from this you would have to calculate the Point Spread Function of the optical system. More often the Modulation Transfer Function is calculated and used. The resolution for an optical system is usually given by the first zero of the PSF. For circular apertures of radius $r$ this gives a PSF of

$PSF \propto J_1(ka\sin(\theta))$ The first zero is when $$r = \frac{3.83}{\pi}\frac{\lambda}{n\sin\theta} = 1.22\frac{\lambda}{2n\sin\theta} = \frac{1.22\lambda}{2NA} \approx \frac{1.22\lambda}{NA_{\text{obj}}+NA_{\text{cond}}}$$

Which we define to be our resolution limit.