Lens design using the matrix method (three lenses)

Question

I'm trying to solve this problem:

Given are three identical thin lenses with focal length $f$. The distance between lens 1 and 2 is $d_a$, the distance between lens 2 ans 3 is $d_b$ and the distance between an object and a screen is $a$. We place the optical system in between the object and the screen. How must we choose the parameters $a$, $d_a$ and $d_b$ so that the image of the object is always sharp on the screen no matter where we place the optical system in between.

What I have done so far was to determine the system matrix of the optical system described above and I got:

$\mathcal{A} = \underbrace{\begin{bmatrix} 1 & -1/f \\ 0 & 1 \end{bmatrix}}_{\mbox{refraction } L3} \underbrace{\begin{bmatrix} 1 & 0 \\ d_b & 1 \end{bmatrix}}_{L2 \to L3} \underbrace{\begin{bmatrix} 1 & -1/f \\ 0 & 1 \end{bmatrix}}_{\mbox{refraction } L2} \underbrace{\begin{bmatrix} 1 & 0 \\ d_a & 1 \end{bmatrix}}_{L1 \to L2} \underbrace{\begin{bmatrix} 1 & -1/f \\ 0 & 1 \end{bmatrix}}_{\mbox{refraction } L1}$

$= \begin{bmatrix} 1-\frac{d_b}{f} + (\frac{-2}{f} + \frac{d_b}{f^2})d_a & \frac{-1}{f} (1 - \frac{d_b}{f}) + (1 - \frac{d_a}{f})(\frac{-2}{f} + \frac{d_b}{f^2}) \\ d_b + (1 - \frac{d_b}{f})d_a & \frac{-d_b}{f} + (1 - \frac{d_b}{f})(1 - \frac{d_a}{f}) \end{bmatrix}$

Now I am not sure how to proceed. I know that with a single lens the Gaussian lens equation goes $1/f = 1/s_o + 1/s_i$ where $s_o$ is the distance from object to lens and $s_i$ from lens to screen. I think that the (inverse of the) effective focal length of the optical system is described by cell $a_{12}$ in the system matrix, is that right? If so can I just say that the following equation must be satisfied:

$\frac{-1}{f} (1 - \frac{d_b}{f}) + (1 - \frac{d_a}{f})(\frac{-2}{f} + \frac{d_b}{f^2}) = \frac{1}{s_o} + \frac{1}{s_i} = \frac{s_o + s_i}{s_o \cdot s_i} = \frac{a - d_a - d_b}{s_o (a - d_a - d_b - s_o)}$

Unfortunately I still have a $s_o$ on the right hand side of the equation... Is that the correct way to go? Any suggestion/hint/correction?

Thank you in advance for your answers.

Julien.

Answer 1

It doesn’t seem possible to do this with any system that has a finite effective focal length $f_{eff}$. If $x$ is the distance of the object to the first focal plane and $F$ the distance between focal planes, then by Newton’s law the image plane is at a distance $x+F+\frac{f_{eff}}x$ from the object. I don’t see a way to eliminate the dependence on $x$. In physical terms, for any fixed $a$ there are at most a couple of positions of the lens system that will produce a sharp image.

Instead, consider a telescopic system. Its matrix will be of the form $\bigl(\begin{smallmatrix}A&0\\C&D\end{smallmatrix}\bigr)$. Letting $s$ be the distance to the first lens and $d=d_a+d_b$, we have $$\pmatrix{1&0\\a-d-s&1}\pmatrix{A&0\\C&D}\pmatrix{1&0\\s&1}=\pmatrix{A&0\\(D-A)s+(a-d)A+C&D}.$$ For this to be independent of $s$, we must have $A=D$ for the lens system. Together with the condition that the upper-right element of the system matrix be $0$, you now have two equations for the two unknowns $d_a$ and $d_b$, so it should be fairly straightforward, though perhaps tedious, to solve for them. The symmetry of the system matrix and of the physical system suggest that $d_a=d_b$, so you might try making that substitution first. Expressing distances as multiples of the individual lens focal length $f$ might also make the calculations simpler.

To find $a$, observe that the offset of a ray at the image plane must be independent of its incident angle, which translates into the condition $(a-d)A+C=0$, that is, $a=d_a+d_b-C/A$.

Answer 2

Yes, the effective focal length of the optical system is $ F = -{1 / a_{12}} $. The image of the object is always sharp on the screen no matter where we place the optical system in between means that the distance from the object to the lens $s_0 $ is infinite. This requires $a >> F$. And, the screen needs to be at the focal point of the optical system. The distance from the secreen (focal point of the system) to the rear lens is $s_{2f} = -{a_{11} / a_{12}}$ (See: The Light Fantastic, A Modern Introduction to Classical and Quantum Optics, by I. R. Kenyon, Appendix B).

Using the Gaussian lens equation for the system $ {1 \over F} = {1 \over s_o} + {1 \over s_i}$, we get $ {1 \over F} = {1 \over s_{2f}} = {-a_{12} \over a_{11}} $ and $ {1 \over F} = {1 \over Fa_{11}} $. Thus, $ a_{11}=1 $. \[ 1 - {{d_b} \over {f}} + ({-{2} \over {f}} + {-{d_b} \over {f^2}})d_a = 1 \] For instance $ {d_b}=3f $ and $ {d_a}=3f $ or $ {d_b}=4f $ and $ {d_a}=2f $ satisfies this equality.