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Suppose I have the following simple optical setup consisting of a coherent (laser) collimated beam travelling through focusing lens 1. The beam focuses to a diffraction limited spot with a $\frac{1}{e^2}$ intensity radius of approximately $w = \frac{\lambda}{2NA}$, where $\lambda$ is wavelength and $NA$ is the numerical aperture of lens 1.

A common method to measure the spot size at the focus of lens 1 is to collect the beam with a second lens (lens 2) and to direct the beam onto a single pixel detector which integrates the entire optical signal. One can then sweep a sharp blade (knife edge) across the beam profile at the lens 1 focus and integrate the signal at the detector for each knife edge position. This would enable the measurement of the beam radius (simply fit the measured curve to an error function to extract the Gaussian beam radius). Ideally the measured spot agrees with the physical waist size.

Case 1 optical setup

enter image description here

Now consider a slightly altered setup, where an iris is placed right before lens 2. Suppose that all the light is still captured by lens 2 and directed onto the detector. Now run the knife edge measurement. Is the measured spot size the same in this case as case 1?

enter image description here

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    $\begingroup$ Just wondering: if the spot width is in the order of the wavelength, then your beams before and after lens 2 appear overoptimistic to me. You‘ll have diffraction there from the iris‘ diameter, won‘t you? Hence, results on size measurement can‘t be identical. It will probably indicate a larger spot. $\endgroup$
    – MS-SPO
    May 10, 2023 at 1:15
  • $\begingroup$ Yes this is true, the diameter of the beam at the detector in my image should be larger. However the knife-edge measurement occurs at the focused spot location right after lens 1, so the knife edge measurement (error function) at the detector will correspond the beam size at the plane of the knife edge. There is no imaging occurring in this setup, just signal integration! The fundamental question is: does restricting the NA on the collection lens mean that the spot size measured (via knife edge) at the focus after lens 1 is larger than if there is no restriction (i.e., case 2 vs case 1). $\endgroup$ May 12, 2023 at 18:34
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    $\begingroup$ Let me put it that way: what counts, is being a wave, not names, intended functions, wishes. In this arena adding usually means worsening. However, I will accept convincing data, e.g. from experiment, simulation etc. $\endgroup$
    – MS-SPO
    May 12, 2023 at 19:45

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In the first case as you move the knife edge down you first block the higher order spatial frequencies decreasing the overall power of the beam. As you continue blocking the beam you block more and more lower order frequencies until you start blocking negative higher order spatial frequencies.

In the second case as you move the knife edge down you block the higher order spatial frequencies but, these are already getting partially block by the iris so you will not register the same change in power.

So to answer your question simply, no. It will depend on the radius of the iris and also its location.

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  • $\begingroup$ There is one paradox if this is true. At lens 1 focus, the wave fronts interfere to create a small spot $w$ according to the input $NA$. Physically this is the size of the beam at the focus. Now with the iris in place, we measure a larger spot size since we're capturing a portion of spatial frequencies. The paradox is that during the knife edge scan, you are interacting with the beam at the focus of lens 1, where there exists a small spot $w$. The knife edge interacts with this beam size. How can the detector then measure a bigger spot? Is this some type of observer effect? $\endgroup$ May 15, 2023 at 23:07
  • $\begingroup$ The detector may measure a bigger spot size, but the actual spot is still $w$, right? In which case, scanning the knife edge through the focus will not scatter any light except for when the blade interacts with the region contained in $w$. $\endgroup$ May 15, 2023 at 23:09
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    $\begingroup$ Recall the method that the knife edge test uses to measure beam size. You are using the single pixel detector to measure intensity. This intensity is given by $I(x) = \int_{-\infty}^{\infty}\int_{-\infty}^{x} e^{-\frac{x^2+y^2}{w^2}} dx dy$ The position of the blade is given by x. This gives you the error function you fit to find the beam size. With the iris in place you cannot say that you are integrating over infinity anymore since you are chopping some of the spatial frequencies off. The detector only detects total intensity and you are blocking some of that intensity with the pinhole. $\endgroup$
    – Bryan
    May 18, 2023 at 5:21

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