Resolving Power of a Microscope

For a microscope viewing two nearby POINT objects O and O', separated by a small distance a, immersed in a medium of refractive index n, with the objective lens AB subtending a half-angle $\alpha$ at the objects (full angular aperture $2\alpha$) and the light's wavelength in that medium being $\lambda_n$ (Fig. 7.19(a)): the two objects' Airy diffraction patterns, formed effectively at infinity (on the eyepiece or a screen), have their central maxima at points I and I' respectively. Applying Rayleigh's criterion, the first dark ring of O''s pattern should coincide with I (O's central maximum), and vice versa.

Working out the geometry (Fig. 7.19(b)) for the point I where O''s first dark ring is located: since I is equidistant from the two edges A and B of the objective, the extreme ray paths $O'A$-to-I and $O'B$-to-I differ in length only over the segment near O', and this PATH DIFFERENCE works out, by straightforward trigonometry on the small triangle near O and O', to $2a\sin\alpha$.

Two distinct cases then apply, differing only in the numerical constant used for the path difference AT the first dark ring itself:

1. Non-luminous (externally illuminated) point objects -- the usual case in ordinary microscopy, where objects do not emit their own light but are lit by some external source, often through an eyepiece medium of refractive index n, so the relevant wavelength there is $\lambda_n = \lambda/n$: the path difference at the first dark ring works out, in this illuminated case, to simply $\lambda_n$ itself, giving $2a\sin\alpha = \lambda_n = \lambda/n$, i.e. $a_{min} = \dfrac{\lambda}{2n\sin\alpha} = \dfrac{\lambda}{2\,\text{N.A.}}$, where $n\sin\alpha$ is called the NUMERICAL APERTURE (N.A.) of the objective -- and correspondingly the resolving power is $R = 1/a_{min} = 2\,\text{N.A.}/\lambda$.
2. Self-luminous point objects -- applying Abbe's fuller theory of Airy-disc diffraction directly to Fraunhofer diffraction, the path difference at the true first dark ring works out instead to $1.22\lambda_n$, giving $2a\sin\alpha = 1.22\lambda/n$, i.e. $a_{min} = \dfrac{1.22\lambda}{2n\sin\alpha} = \dfrac{0.61\lambda}{\text{N.A.}}$, with resolving power $R = 1/a_{min} = \text{N.A.}/(0.61\lambda)$. …