# How is the Rayleigh criterion connected to the Abbe limit?

I am interrested whether one can derive a formula for the point resolution (like Abbe did) of an optical system from the Rayleigh criterion (without the use of small angle approximation i.e. $$\rm{sin}(\alpha)=\rm{tan}(\alpha)$$ which is not really suitable e.g. for microscopy).

And if so whether formula is directly comparable to the Abbe limit for point (or rather line) resolution.

The Rayleigh criterion is given as: $$\theta_{min}=1.22\frac{\lambda}{D}$$ where $$\theta_{min}$$ is the smallest resolvable angle, $$\lambda$$ is the wavelength of the used lightsource and $$D$$ is the diameter of the used aperture (or of the used lens).

And the Abbe limit is given as:

$$d=\frac{\lambda}{2\,n\,\rm{sin}(\alpha)}=\frac{\lambda}{2\,\rm{NA}}$$

where $$d$$ is the smallest resolvable distance, $$n$$ is the refractive index of the medium between the object and the optical system, $$\alpha$$ is the biggest scattering angle (incident on the optical system) and $$\rm{NA}$$ is the numerical aperture.

• Obviously you are missing, at the very least, the index of refraction. Other than that, the two are virtually the same, one is basically predicting an angular resolution for an object that is infinitely far away, the other a spatial resolution for an object that is close. Since the distance to the lens is implicitly included in the numerical aperture, you could bring that over to the left hand side in the Abbe limit and get a similar formula for angular resolution (which is less useful). Today we have replaced both with point spread functions, anyway. May 24, 2016 at 20:21

Both equations are in fact structurally similar with Abbe limit given by $d= \dfrac{\lambda}{2\mathrm{NA}}$

And Rayleigh limit given by $d =1.22 \dfrac{\lambda} {2\mathrm{NA}}= 0.61\dfrac{\lambda} {\mathrm{NA}}$

where lambda is the wavelength and $\mathrm{NA}$ the numerical aperture of the light collecting lens.

The factor 1.22 comes from the definition of Bessel function of 1st kind,the fact that 1st minima of the diffraction pattern appears at 1.22 units from the central zero.

Rayleigh criterion is thus a modification of the Abbe Resolution limit. The Rayleigh criterion states that in order for 2 closely placed PSF to be resolved, the central maxima of one should lie exactly at the first minima of the second one. Since the Airy pattern is defined by the Bessel function, the minimum separation between the 2 patterns should be $1.22 \lambda/ 2\mathrm{NA}$ instead of just $\lambda/2\mathrm{NA}$ considering that the first minima will be at 1.22 times the unit from the central maxima.

• Thanks for you answer, but I think my question is still not really answered. Let me rephrase the question: How can we come form the formulation of the rayleigh criterion that I was using (θmin=1.22*λ/D) to the one you were using (d=1.22*λ/NA), without using the small angle approximation. Oct 11, 2018 at 8:43
• So would it be correct to say that Rayleigh criterion is a refined and more accurate version of the Abbe resolution limit? Dec 31, 2020 at 20:31
• Also I can also notice that one include the index of refraction and the other one doesn't. How could we go from one to the other in that case? Dec 31, 2020 at 20:55

As @Felix and @Caterina are still not satisfied I will add my 2 cents, for which I hope they are correct.

As far as I am aware Rayleigh developed his criterion on diffraction of light on slits, whereas Abbe was working on microscopy. Therefore, you have a refraction index in one and not in the other. However, you could bring the Rayleigh version close to the Abbe's. Rayleigh stated the following:

$$\begin{equation} \theta_{min} \approx 1.22 \frac{\lambda}{d}, \end{equation}$$

where $$\theta_{min}$$ represents the minimum angular radius of an Airy disk as seen from the centre of the circular aperture, $$\lambda$$ the wavelength of light and $$d$$ the diameter of circular aperture. Now this is angular separation and we have to bring it to spacial separation via $$x_{min} = R \sin\theta_{min}$$ and we get

$$\begin{equation} x_{min} \approx 1.22 \frac{\lambda R}{d}, \end{equation}$$

where $$R$$ is distance between the slit and imaging screen. Now we can convert the $$R/d$$ into the $$\sin \alpha$$ via $$\sin \alpha = \frac{d/2}{R}$$, and we get

$$\begin{equation} x_{min} \approx 1.22 \frac{\lambda}{2\sin\alpha} \approx 0.61\frac{\lambda}{\sin\alpha}, \end{equation}$$

At this point we can introduce that there is a different refractive index on one side of the slit and $$\sin\alpha$$ goes to $$n\sin\alpha$$.

$$\begin{equation} x_{min} \approx 0.61\frac{\lambda}{n \sin\alpha} \approx 0.61\frac{\lambda}{NA}. \end{equation}$$

And this is very close to Abbe's limit of

$$\begin{equation} x_{min} = 0.5 \frac{\lambda}{NA}. \end{equation}$$

So all in all it's just how you define the minimum distance at which you can still separate two sources.