Numerical aperture of a Keplerian telescope

[Michele Conni]

Homework Statement

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Design an afocal Keplerian telescope to imagine an object of $L = 5\, mm$ with a resolution of $R = 2\, \mu m$ and a magnification of $M=-2$; assume that the wavelength is $\lambda = 500\, nm$.
Don't use lenses faster than $F/1$.
Using the optical invariant, what is the numerical aperture $NA$ of the objective lens?
Calculating $NA$ directly from resolution, what should the $NA$ be?

Homework Equations

A definition of the Lagrange (or optical) invariant can be found in Wikipedia (https://en.wikipedia.org/wiki/Lagrange_invariant); to report it here, it can be derived in every position of the optical axis of a system from the marginal and the chief ray (which are defined by the aperture and the field stop positions respectively):
$$H=n(u\bar{y}-\bar{u}y)$$
Here, $u$ and $\bar{u}$ indicate the angles of the marginal and the chief ray, respectively, while $y$ and $\bar{y}$ indicate their distances from the optical axis; $n$is the index of refraction of the material of propagation.

At the object $y=0$ and $\bar{u} = 0$, therefore (considering the paraxial case and assuming that the aperture of the system is circular and the object with the center in the optical axis):
$$H=nu\bar{y}=NA\frac{L}{2}=0.6\frac{\lambda}{r_0}\frac{L}{2}$$

Finally, the numerical aperture is defined as $NA=n\sin{\theta}$; in air and under the paraxial approximation (i.e. small angles and big distances between the optical elements) $NA\simeq \theta$.

The Attempt at a Solution

The Keplerian telescope uses two convex lenses to dilate a bundle of optical rays (http://electron9.phys.utk.edu/optics421/modules/m3/telescopes.htm, first picture), and it can be easily derived that the focal lengths of the two lenses are related by the equation $f_2 = \abs{M}f_1$ (in this case $\abs{M}=2$).
Since it is not specified, I suppose that the object is at infinite distance form the first lens, and I place it with its center on the optical axis, so that (if the optical axis is defined on the axis $z$) it goes from $x=-2.5\, mm$ to $x=+2.5\, mm$.
If I use the fastest lens possible, i.e. $F/1$, the relationship between diameter and focal length of the first lens is $f_1=D_1$, and $D_2=f_2=2f_1=2D_1$.

I solved the second question: if I force the resolution of the system to be equal to the diameter of the Airy disk of a circular aperture (https://en.wikipedia.org/wiki/Airy_disk), I get:
$$R=2r_0=2(0.6 \frac{\lambda}{NA})$$
from which:
$$NA=1.2\frac{\lambda}{R}=0.3$$
and the answer is right.

However, I am not able to solve the first question: there is nothing that forces you to choose a different f-number for your lenses, and if you choose them to be equal, both of them can be considered as aperture stops (since the marginal rays are parallel to the optical axis, because they come from $-\infty$. Nonetheless, if you decide that the first lens is the aperture stop, it can be easily derived that the Lagrange invariant becomes $H=\frac{L}{6}$, while if the second one is chosen it changes to $H=\frac{L}{12}$. From this, the numerical aperture should be derived from $H=NA \frac{L}{2}$, but, since the solution is $NA=0.25$, none of these are right.

How do I solve this? Where am I wrong?

 

[Swamesh]

Hey, did you get the solution for first question? I am stuck on that problem.If you did get, kindly help me.