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Numerical aperture and lenses

  1. Dec 30, 2011 #1

    I have two general questions regarding a lens and its numerical aperture:

    1) From this picture http://en.wikipedia.org/wiki/File:Numerical_aperture_for_a_lens.svg (from the Wiki-article http://en.wikipedia.org/wiki/Numerical_aperture) it seems to me that there is a propotionality between the NA of a lens and its focal length, i.e. long focal length = small NA. However this doesn't always seem to be true when I look at optical elements in various catalogs?

    2) I read in an old thread (https://www.physicsforums.com/showthread.php?t=304834) that only when the full aperture of a lens is used will the total NA be available. Is it possible to give a rough estimate of how much of the NA is in use?

    Best wishes,
  2. jcsd
  3. Dec 30, 2011 #2

    Andy Resnick

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    The NA is the *ratio* of aperture diameter to focal length. Keeping the lens diameter constant, then yes-varying the focal length will also vary the numerical aperture (as is the case for some zoom lenses). As a specific example, one of my microscope lenses has a NA = 1.47 and the size of the front element is 1mm or so; my 400mm camera lens has a NA of 0.18 and the front element is about 140 mm in diameter.

    By closing down the aperture stop, you decrease the NA- again, camera lenses allow you to close down the aperture stop (those funny f-numbers 2.8, 4, 5.6, etc) which decreases the NA and increases the depth of field.
  4. Dec 31, 2011 #3
    Thanks. Just to clear, what you refer to is the f/# number, right?

    On Wiki (http://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number) it says that we have f/# = 1/2NA when the lens is focused at infinity. Does this also refer to the case when I come in with a collimated beam, and focus it to a point not an infinity (I'm thinking reversal symmetry here)?
    Last edited: Dec 31, 2011
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