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Optics - focused vs. defocused image

  1. Jul 26, 2015 #1
    Hello,

    (I am new here, so hopefully this is the right location to ask my question. please let me know if there is more relevant forum)

    I am trying to understand 2 situations:

    A) I have an object, lens and a detector. The detector is at the focal plane of the lens. I am reducing the aperture of the lens.

    B) Same as before, (section A above), but now the detector is NOT at the focal plane of the lens, and again I am reducing the aperture of the lens.

    I am trying to understand the difference between the 2 states.

    Please let me know if I was clear enough.


    Thank you,
    Mark
     
  2. jcsd
  3. Jul 26, 2015 #2

    sophiecentaur

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    You would need to specify the situation in a bit more detail, for a really good answer to your question but I could say that focussing a small image onto a large detector may not be the best approach - if you just want a good signal output. Stopping down a lens will give greater latitude (depth) of focus.
    It's normally a matter of choosing optics to suit the detector size / redsolution, if you want best value for money.
    Give us some more idea of what you actually want to know / do.
     
  4. Jul 26, 2015 #3
    Thank you Sophiecentaur.

    I will try to elaborate.

    I have a laser beam, pointing at some surface. The surface is NOT at the focal plane of my camera. The image is defocused. I receive an image of speckles. If I reduce the camera aperture, my spot of speckles simply reduced in size (diameter).

    I don't fully understand what happens when I do the same as above, but now the surface is AT the focal plane of the camera.

    I mean, I am wondering what's the fine difference of reducing the camera aperture when your image is in focus or not in focus?

    I am not sure, but does Fourier Optics could help me get some intuition on that? What does a lens/camera do?
     
  5. Jul 26, 2015 #4

    sophiecentaur

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    Sorry but that is confusing me. Where is the camera? Is it being used to photograph the image on the "surface"? A diagram might help.
     
  6. Jul 26, 2015 #5
    the camera is "looking" at the spot that is being illuminated by the laser on the surface (I've added a diagram).

    one time the surface is in FOCUS of the camera, and one time the surface is in DEFOCUS of the camera.

    In both cases I start "closing" (or reducing) the camera aperture.

    Now I am trying to understand what is happening in each case, and what's exactly the difference. Also, how does reducing the aperture affects the "depth of focus"?



    Hopefully it's more clear now.


    Thank you!!!
     

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  7. Jul 26, 2015 #6

    sophiecentaur

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    OK Thanks for the clarification.

    First: conventional camera imaging - say of a point source image. The reason that reducing aperture makes focussing less critical is that the 'cone' of rays, coming from the whole of the area of the lens is tighter / narrower. The criterion for judging whether or not an image is 'in focus' is how wide a projected 'spot' appears. A small variation in image plane position will cause the spot to be larger than when you have perfect focus. If the cone is more acute, there is more room for error of position. If you're being fussy, you can make a distinction between Depth of Field and Depth of Focus - which are terms relating to the acceptable error (blur) due to variation of object distance or 'film plane' position. They amount to the same thing in many contexts. http://www.edmundoptics.co.uk/technical-resources-center/imaging/depth-of-field-and-depth-of-focus/ [Broken]
    I am never sure of the reasons why laser light has the effects it does. The speckle is due to diffraction of the coherent light by random irregularities of the surface it hits - fairly understandable. But when you have a very intense spot from a laser, projected on a screen, you may find that your sensor is driven into saturation and could be giving the impression of a wider spot size when the spot size is more like the effect of blurring. If you could attenuate the beam first or reduce the exposure time drastically, you might find that you get a different result from your aperture /distance combinations.
    Alternatively, could it be that, with such a coherent beam, you are getting a more specular reflection, which could present the camera with an object that is at infinity, rather than at the surface in question. I haven't thought that through completely so it could just be BS. Perhaps someone could put me /us right on that????
     
    Last edited by a moderator: May 7, 2017
  8. Aug 3, 2015 #7
    Thank you very much for your thorough answer.
     
  9. Aug 3, 2015 #8

    sophiecentaur

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    I should be interested to hear if you ever get results with drastic under-exposure. A dense neutral filter in front of the camera lens could also have the same effect. BTW, how many pixels does the camera image of the spot occupy? Sometimes, the camera characteristic can account for odd optical effects; worth considering, perhaps.
     
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