# Feynman Lectures Vol. 1: error in the formula for resolving power?

In chapter 27 of Feynman Lectures on Physics Vol. 1, section 7 on resolving power, Feynman states the rule for optical resolution:

two different point sources can be resolved only if one source is focused at such a point that the times for the maximal rays from the other source to reach that point, as compared with its own true image point, differ by more than one period.

So the rule as stated by Feynman is

$$t_2 - t_1 > 1/f$$

where $$f$$ is the frequency of light.

Feynman then continues:

If the distance of separation of the two points is called $$D$$, and if the opening angle of the lens is called $$θ$$, then one can demonstrate that $$\, t_2 - t_1 > 1/f \,$$ is exactly equivalent to the statement that $$D$$ must exceed $$λ/(n \ \text{sin} \theta)$$, where $$n$$ is the index of refraction at $$P$$ and $$λ$$ is the wavelength.

However, what I get is $$D> \frac{λ}{2 n \ \text{sin} \theta}$$ i.e. an extra factor of $$2$$ in the denominator. This holds when $$D$$ is sufficiently small (as is clear from the picture). If we define $$t$$ as the time to propagate from $$P$$ to $$S$$ or $$R$$ then by small angle trigonometry (not small in $$\theta$$!) we get that (omitting $$c$$ and $$n$$) $$\ t_1 = t - D\ \text{sin} \theta \$$ and $$\ t_2 = t + D\ \text{sin} \theta \$$, hence my result.

Am I making some trivial mistake? If not, how come this has not been spotted before?

N.B.: Feynman writes that $$\theta$$ is the opening angle of the lens, whereas in the picture it is depicted as the half-angle. Taking $$\theta$$ to mean the full angle $$\angle SPR$$ the resulting denominator should be $$2 n \ \text{sin} \frac{\theta}{2}$$.

• Is this the same as the Airy disk formula, see wikipedia. Is an aperture implied by angle theda? It could be an error in his notes, it's possible. Apr 11, 2019 at 13:38