Dear Abishek Sai,

Look the links, please.

https://books.google.az/books?id=sGcHV92A2uAC&pg=PA245&lpg=PA245&dq=Why+should+the+image+plane+be+at+fourth+focal+plane+in+a+4f+setup?&source=bl&ots=r83TsWanJe&sig=yZqZBc4924dAgbiqwR7iyBRbB74&hl=ru&sa=X&ved=0ahUKEwjHjKKPhvDMAhXIfiwKHYapDCoQ6AEIMjAD#v=onepage&q=Why%20should%20the%20image%20plane%20be%20at%20fourth%20focal%20plane%20in%20a%204f%20setup%3F&f=false

https://www.semrock.com/flatness-of-dichroic-beamsplitters-affects-focus-and-image-quality.aspx

Regards, Shafagat

Hello Abishek,

The image formed at the 4f position is indeed an image just as you would get from a single lens when the object is placed a distance 2f in front, and the image screen is 2f behind.  So with nothing but the two lenses, the output would be a collimated bean, if the input is a collimated beam.  But the advantage of a 4f system is that it allows access to the plane in the middle which can be called the Fourier plane.  At this position there is in effect a collection of collimated beams (not just one collimated beam) but importantly they propagate at a spectrum of angles.  The net effect is that the Fourier Transform of the input light field (from 2f in front of the first lens) appears at this Fourier Plane.  Blocking chosen areas in this Fourier Plane prevents their propagation to the output image (at 4f) and so allows you to filter the input (object) image - for example to pass only a range of chosen spatial frequencies to the output image, or to edge enhance the image by giving more emphasis to the higher spatial frequencies.  It also allows you to cross-correlate an image with the input image by displaying the Fourier Transform of one of the images at the middle plane, thus providing a pattern recognition function.  I hope this helps.

Regards,   Mark.

After giving yesterday’s answer to you, I had an additional thought.  Perhaps you are using the 4f system to ‘clean-up’ a collimated beam, in which case I would add the following to my previous answer.

If you apply an ideal collimated beam with uniform intensity across its area to the 4f system, it will focus to a single point at the middle plane (the 2f position).  The second lens will then deliver the best quality collimated beam at the output.  i.e. an output beam with uniform intensity across its area the same as the input beam.

But if you apply a poor quality collimated beam such as one which is not of uniform intensity across its area, for example with faint fringes across it, it will not focus to a single point at the middle plane, but spread out and the second lens would deliver a similarly poor beam to that at the input.

However, if you place an appropriately sized pin-hole in the middle at the position where the ideal beam would focus, then only the lowest spatial frequencies are able to progress to the second lens.  i.e. the light outside the pin-hole due to non-uniformity in the input beam will be blocked and not progress to the second lens.  So the second lens will then only reconstruct from the good part of your input beam.  i.e. it will deliver a nice smooth uniform collimated beam even if the input beam was of poor quality.

Regards, Mark (again).

Hi Abishek,

Your approach is true if you only consider flux collection. A collimated beam will be focused by the first lense and then re-collimated by the second.

If your approach is imaging, an object placed at the focal point of the first lense will be imaged to infinity and seen by the second lense, will be imaged at its focal point.

Regards. JM

Hi with reagrds to the above question , I have an additional doubt, suppose I am trying to image a random phase produced on an SLM (phase only) on a tissue. Is there a difference if I use 2f imaging or 4f imaging ? Is the phase produced on SLM same as the image plane neglecting the lens distortion and other effects? Or is it different in both cases ?

Jitendra

4F system (as Mark pointed out) allows accessing the Fourier plane.

By placing a “spatial frequency filter” 1f in-front of the first lens and another one 1f behind the first lens, a combined effect is produced 1f behind the second lens. This type of filtering (of individual spatial Fourier transform components of an image) allows obtaining additional information that cannot be obtained in a conventional imaging (3D structure of specimen). A single spatial frequency filter can also be used depending on the type of experiment.

If the first lens is placed 1f behind the “intermediate image plane” (formed by focusing light from a objective lens), then the light (coming out of 4F system) focuses 1f behind second lens forming an image plane. In this case filter will be placed in the middle of lens 1 and 2.

I would like to amplify the original question with my thoughts on this. Let's assume we all understand the whole Fourier transform of this thing, and we are only asking about the specific locations, and focal lengths, of the lenses.

If the object is placed anywhere to the left of lens 1 (illuminated by a plane wave), then its diffraction pattern (DP) will appear at f. So immediately, it doesn't even have to be '2f', let alone '4f'. The numerical aperture, NA, of lens 1 should be large enough to accept enough spatial frequencies to make this whole thing useful in the first place.

Placing the object at f to the left of lens 1 does cause the rays to exit the lens parallel, much like an infinity-corrected microscope. A nice image of this can be found at https://www.instructables.com/Image-Processing-Without-the-Computer/. Based on that, lens 2 can be placed anywhere and still form an image, as long as the pencils of rays from the object are intercepted to form an image to the right of lens 2. Also, if lens 2 is placed further away, it has to be larger in diameter in order to retain at least as high of an NA as lens 1, or some of the frequency components will be lost outside the lens, and the spatial filtering will be degraded.

The focal length of lens 2, call it f', can be different from f of lens 1. Putting Lens 2 in such a position that the DP between the lenses is f' away, causes the image of lens 2 to be f' away as well, so that is a " 2f' " system. But, other than achieving a magnification of 1, why do we have to have f = f' in order to do the Fourier transform imaging? My guess is that '4f' sounds catchy (maybe I am being ignorant here), and the assumption is that the total magnification is 1, but it doesn't have to be. Assuming f = f' does make it all simpler, but it could be a 2f - 2f' system and still work. Do I understand this correctly? Any thoughts or comments?