Calculating the resolution limit using Fermat's principle

[albertrichardf]

Hi,
I read the Feynman Lectures Volume 1, Chapter 27, section 27-7, which can be here. In the lecture he describes the fundamental limits of resolution and provides a criterion.

Here is the diagram I am referring to, figure 27.-9:

There are two light sources, $P$ and $P'$ There is an optical system, which is just a line $SR$ and it gets the light from sources $P$ and $P'$ to converge. Suppose that the image of $P$ is formed at $T$, which is where the lines $PST$ and $PRT$ meet. Now draw the lines $P'RT$ and $P'ST$. If the two points $P$ and $P'$ are to be resolved, Feynman gives the formula 27.17:
$$t_2 - t_1 < \frac {1}{\nu} $$

where $t_2$ and $t_1$ are the times for the paths $P'RT$ and $P'ST$, respectively and $\nu$ is the frequency of the light.

He then goes on to state that this equation is exactly equivalent to the following:
$$ D = \frac {\lambda}{n sin (\theta)} $$
where $D$ is the distance $PP'$, $\lambda$ is the wavelength of the light, $n$ is the refractive index of the light at point $P$ and $\theta$ is the opening angle of the lens at $SR$.

I have tried to derive this equation, but I ended up with a factor of 2 in the denominator. That is, I got:
$$ D = \frac {\lambda}{2n sin (\theta)} $$

To get that I dropped perpendiculars between $P'S$ and $PS$, and $P'R$ and $PR$. Then I calculated the difference in distances and summed the differences. The sum of the differences gives me $P'R - P'S$, since $PR = PS$. Then I end up with:
$$ P'R - P'S = 2 D sin\theta $$

I then multiplied throughout by $\frac nc$ to get the times on the left hand side. Then I used the inequality that was given to obtain the result I did.

What did I do wrong in this derivation?

Thank you for answering.

 

[Charles Link]

If $\theta$ is the half-angle of the lens, then I agree with your result, which I derived by a Taylor series expansion (with $D$ as the variable), to compute the difference in $t_2-t_1$. $\\$ The derivation is relatively straightforward, but slightly complex, involving the Pythagorean theorem, and having the projections of $t_2$ and $t_1$ onto the lens be equal to $w$ and $w-2D$, with $SR=2w-2D$. $\\$ If $\theta$ is the full angle of the lens, then the approximation $2 \sin(\theta) \approx \sin(2 \theta)$ is needed to get Feynman's result. $\\$ Anyway, that's what I got, but I didn't double-check my calculus and algebra very carefully.

 

[albertrichardf]

If $\theta$ is the half-angle of the lens, then I agree with your result, which I derived by a Taylor series expansion (with $D$ as the variable), to compute the difference in $t_2-t_1$. $\\$ The derivation is relatively straightforward, but slightly complex, involving the Pythagorean theorem, and having the projections of $t_2$ and $t_1$ onto the lens be equal to $w$ and $w-2D$, with $SR=2w-2D$. $\\$ If $\theta$ is the full angle of the lens, then the approximation $2 \sin(\theta) \approx \sin(2 \theta)$ is needed to get Feynman's result. $\\$ Anyway, that's what I got, but I didn't double-check my calculus and algebra very carefully.

Thank you for replying.
He refers to the angle $\theta$ as the "opening angle of the lens", which seems to mean that $\theta$ is actually the full angle of the lens. However, he states that $D > \frac {\lambda}{n sin \theta}$ is exactly equivalent to $t_2 - t_1 > \frac 1 \nu$ and he does not give any indication that this formula applies only for small angles.

 

[Charles Link]

Thank you for replying.
He refers to the angle $\theta$ as the "opening angle of the lens", which seems to mean that $\theta$ is actually the full angle of the lens. However, he states that $D > \frac {\lambda}{n sin \theta}$ is exactly equivalent to $t_2 - t_1 > \frac 1 \nu$ and he does not give any indication that this formula applies only for small angles.

Using $\theta$ as the half angle, I got the result that $2 n D \sin(\theta)=\lambda$, and the only assumption there is that $D$ is small. $\\$ Let $s=PR=PS$, I got that the distance corresponding to $t_2$, which I will call $\bar{t}_2=[(s \cos(\theta)^2+w^2]^{1/2}$, and $\bar{t}_1=[(s \cos(\theta))^2+(w-2D)^2]^{1/2}$. I then expanded $\bar{t}_1$ in a Taylor series, with $D$ as the variable.

 

[albertrichardf]

Using $\theta$ as the half angle, I got the result that $2 n D \sin(\theta)=\lambda$, and the only assumption there is that $D$ is small. $\\$ Let $s=PR=PS$, I got that the distance corresponding to $t_2$, which I will call $\bar{t}_2=[(s \cos(\theta)^2+w^2]^{1/2}$, and $\bar{t}_1=[(s \cos(\theta))^2+(w-2D)^2]^{1/2}$. I then expanded $\bar{t}_1$ in a Taylor series.

I also got the same result, although I did not assume that D is small.
From geometry,
$$s - P'S = D sin \theta $$

$$ P'R - s = D sin \theta $$

And then the sum of the two equations gives $P'R - P'S = 2 D sin \theta$
Somehow he gets rid of the factor of 2.

 

[Charles Link]

I do think you made the assumption somewhere that the location is such to have $\theta$ be what it is, which is essentially that $D$ is small. The reason is that in the limit of $D$ being very large, $\bar{t}_2-\bar{t}_1=RS$. The difference being proportional to $D$ is a first order result for small $D$. As $D \rightarrow +\infty$, the difference does not go to $+\infty$. $\\$ Additional note: These relations regarding resolution of images are , in general, somewhat imprecise, so they really do not need to be highly precise in an algebraic sense to be worth mentioning, as Feynman did here.

 

[albertrichardf]

I see, thank you for answering. If they are imprecise, then I suppose that the small angle approximation of $sin 2\theta = 2 sin \theta$ would also work, even if it was not mentioned. That would explain how he obtained the end result, provided that $\theta$ is the angle between $PR$ and $PS$.