# How does 2f-2f lens configuration work?

Say I want to image a shape which is cut out on an Aluminium sheet. If I am passing a laser through the object in a setup like the image below. What I don't understand is why is there a need of putting a beam profiler at $$f$$ from the lens since the beam is collimated. The best image I should get should be when I keep the beam profiler as close to the lens as possible to reduce divergence as much as possible. The best image I get is when I put the beam profiler at $$f$$. I don't understand why this happens. What is the reason for this? Can anyone please share some book/article on this?

• To better help you with this question I have a question for you. You say "The best image I should get should be when I keep the beam profiler as close to the lens as possible to reduce divergence as much as possible." Why do you say this? Dec 22, 2023 at 7:01
• @Jagerber48 because If the beam coming out of the second lens is collimated, it would make sense to keep the beam profiler as close to the lens as possible to reduce divergence.(because even though we say it's collimated, there is always some divergence in all practical applications) Dec 22, 2023 at 9:32
• Hmm, unfortunately I think you have some (common) misconceptions about how image formation works. What if you replace the laser + object with a candle (it's own source of radial propagating light)? Does the situation make more sense to you in that case? Dec 22, 2023 at 17:17

Your first picture does not show all of the rays needed to construct an image. From each point on the object (the square aperture), it is necessary for rays to leave that point at various angles. When those rays rejoin in the image space of a lens, then a real image will be formed. An alternate way of looking at a 4f imaging system is shown below. It's always helpful to show the rays for off-axis objects, or off-axis angles. The top picture shows the imaging condition: points on the edge of the aperture are imaged. If you move the sensor, you will see a defocused image. The bottom picture shows what is commonly called a pupil relay. A 4f system can serves as a pupil relay: a pupil relay takes collimated light through an aperture at the input and relays it to the output side. This is useful in systems where you need to relay an image or a beam over some distance: For example, if a galvo mirror scanner occupies a physical space, it can be relayed to the (internal) pupil of an objective lens. The galvo mirror would not sit close to the objective, but the pupil relay effectively relays the gavlo mirror to inside the objective lens.

If you only need to image an aperture (say, for laser ablation) you can get by with a single lens. There a lot of other considerations about the imaging conditions: effects on resolution, etc..

This is not what 4f lens systems are for, in the middle of the lenses the image is in the Fourier form, so you can filter you system in different ways (reduce noise or whatnot). As explained in Fundamentals of Photonics, Bahaa E. A. Saleh, Malvin Carl Teich: For example (From Fundamentals of Photonics, Bahaa E. A. Saleh, Malvin Carl Teich) What you're suggesting is closer to direct imaging. That's what people do with x-rays for example (because there are no practical lenses in this domain).

Edit after discussion: The question was regarding the need to Place the Beam profiler at length $$f$$. As explained in the comments the image given does not represent the underling physics. As it assumes Fresnel diffraction (Fundamentals of Photonics, Bahaa E. A. Saleh, Malvin Carl Teich), so different points on object move to a different points on the lens etc. So the need to put the Beam profiler arises because there the image is in the focal plane.

• what I'm doing is to image the object, in your case the letter F and not actually spatially filter it Dec 22, 2023 at 9:29
• I see what you mean now, ok, so this entire formalism was developed under Fresnel diffraction. This means that the rays don't move in a completely straight line. You have a cone going out from each point (like from the boundaries). This creates an image plain and a Fourier plain and that's why you need to place the Beam profiler there. In the shortest path formalism we got the lines by assuming a very short wavelength and that each point emits radially so you get a path integral. In short wavelength (like x-ray) you can do direct imaging better. Does that answer your question?
– ssm
Dec 22, 2023 at 9:53
• It is the truth, if you want you can read about it in Fundamentals of Photonics by Bahaa E. A. Saleh, Malvin Carl Teich (The pictures are from there). If you move your Beam amplifier you'll move out of the focal plane and the image will become more blurred. Same as you do with glasses. You can't have just a plane wave here, it would be inconsistent with Maxell equations to just have a square of a plane wave going through the system. you can't technically have the lines go like this. So you need to use Fresnel approximation.
– ssm
Dec 22, 2023 at 14:04
• While I love this answer because it shows Saleh and Teich which is my favorite reference on this topic, I can't upvote since it's mostly just a screenshot from the textbook. It would be improved by more prose from the answer author themselves addressing the questioners misconceptions. Dec 22, 2023 at 19:05
• @ssm I see. Thanks for the explanation Dec 22, 2023 at 20:46