When is the image produced by a thin lens sharp?

[JulienB]

Homework Statement

Hi everybody! Being given three identical thin lenses with the same focal length, I have to determine $a$ (distance object-screen), $d_a$ (distance lens 1-lens2) and $d_b$ (distance lens 2-lens 3) so that a sharp image of the object appears on the screen regardless of the position of the optical system between the object and the screen (see picture).

Homework Equations

Lens equation: $\frac{1}{f} = \frac{1}{s_i} + \frac{1}{s_o}$ with $s_i$: distance from lens to screen and $s_0$: distance from lens to object.

The Attempt at a Solution

Well I didn't get very far because I don't really know what is the condition for an image on the screen to be sharp. Is it the case only when the lens equation is fulfilled, that is when I have a certain $s_i$ and $s_o$ so that the sum of their inverses is equal to $1/f$?

And if so, is the following thinking correct? Say a ray of light is going from $S$ to $L_1$. I want it to be sharp when meeting $L_2$, so the following equation has to hold:

$\frac{1}{x} + \frac{1}{d_a} = \frac{1}{f}$.

Is that correct? If so I can set up an equality with three equations, but I am afraid it remains dependent of $x$ then. Any clue about how to tackle such problems?Thanks a lot in advance for your answers.

 

[JulienB]

What about the matrix method for example? If I put one after the other translation and refraction matrices I would say that a ray reaching the screen has for incident angle and height:

$\begin{bmatrix} \theta_r \\ r_r \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} \theta_i \\ r_i \end{bmatrix}$
$= \underbrace{\begin{bmatrix} 1 & 0 \\ x & 1 \end{bmatrix}}_{\mbox{object} \to L_1} \underbrace{\begin{bmatrix} 1 & -1/f \\ 0 & 1 \end{bmatrix}}_{\mbox{refraction } L_1} \underbrace{\begin{bmatrix} 1 & 0 \\ d_a & 1 \end{bmatrix}}_{L_1 \to L_2} \underbrace{\begin{bmatrix} 1 & -1/f \\ 0 & 1 \end{bmatrix}}_{...} \begin{bmatrix} 1 & 0 \\ d_b & 1 \end{bmatrix} \begin{bmatrix} 1 & -1/f \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ k_3 & 1 \end{bmatrix} \begin{bmatrix} \theta_i \\ r_i \end{bmatrix}$

where $k_3$ is the distance between the third lens and the screen. Is that expression correct? And more importantly, could that bring me somewhere? I just started using the matrix notation for lenses today, so I am very unexperienced with that method and unsure about what I can and can't do with it.

I think the image will be sharp if all rays intersect at some point on the screen. I read somewhere that $C$ has to be zero for the rays to intersect at the same point independently of $\theta$, does that make sense? If so I get a crazy equation that doesn't simplify so easily, so I'd rather wait for an answer before diving into it. :)

Thank you in advance.Julien.