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F-number and light intensity

  1. Jul 9, 2008 #1
    I've been searching through the internet and some of my optics books, but nowhere was I able to find the derivation of the law for a camera lens, that the intensity of light that comes on the film or chip is proportional to [tex]\frac{D^2}{f^2}=N^2[/tex], where D is the aperture diameter, f the focal length and N the f-number. Any ideas?
     
    Last edited: Jul 9, 2008
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  3. Jul 9, 2008 #2

    russ_watters

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    I've seen that equation only a few times and I'm not sure I agree with it. For one thing, it seems needlessly cumbersome since diameter is part of focal ratio. Where, exactly, did you get it?

    Typically, optics texts should tell you that the intensity of light is proportional to 1/f^2
     
  4. Jul 10, 2008 #3
    Well, it's actually in most of the books on photography, but it's never derived. They say, that when you change the aperture by a factor of sqrt(2), the intensity of light will change by factor 2.
     
  5. Jul 10, 2008 #4

    Borek

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    Something is wrong with this equation. Are you sure N is not an inverse of f-number (I will denote it by f/#)?

    First of all, by definition f/#=f/D, so it is eaxactly opposite with N=D/f (squares ignored, but they don't change anything here).

    Second, amount of light passing through the lens is not directly proportional to f/# but inversely proportional to f/# (for the sake of simplicity I am again ignoring squares). The higher the f/#, the less light passes through the lens - but the equation as listed states that there is more light.

    That's obvious - amount of the light passing through the lens is proportional to the area of the hole, this is in turn proportional to diameter squared. See the definition and it should click.
     
  6. Jul 10, 2008 #5
    ooops, you're right of course, N is the inverse of the f-number.

    Anyway, I still seem to misunderstand something. I found this 'derivation' of the equation, here: http://www.inyourfacefotos.com/fstop.htm , although I quite don't understand some of the steps there (even if it's elementary school algebra).

    Well, first of all, according to the first picture and the equations, the intensity of light at the image is proportional to f^2 (not inversely proportional). The equations are OK, so there must be something wrong with the picture, but I can't figure out what it is.

    Then I don't understand how they got the second equation in the section "aperture stops" (I_i = ...). It just doesn't make any sense to me.

    When I tried to derive it myself according to the second picture, I started with [tex]I_i = I_0 \frac{A_a}{A_i}[/tex] from which I obtained [tex]I_i = I_0 \frac{f^2}{x_i^2}[/tex], which is probably not true.

    I feel really stupid for not understanding such an elementary thing, but I'm probably missing something essential.
     
  7. Jul 10, 2008 #6

    russ_watters

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    Sorry, I read your first post too fast and didn't see that "f" was focal length, not focal ratio. That equation looks right, then.
     
  8. Jul 10, 2008 #7

    Andy Resnick

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    I'm having a tough time parsing your question, which seems to be "why is the incidance proportional to the square of the f-number?" (Incidance is the amount of light incident onto a surface). If that's the case, then the answer is most liklely because the f-number depends on the radius (or diameter) of the lens; thus the total collection area of the lens (the area) is proportional to the square of the radius, i.e. the square of the f/#.
     
  9. Jul 10, 2008 #8

    jtbell

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    The intensity of the light at the image is proportional to the area of the lens: if the lens has twice as much area, it collects twice as much light.

    For a given lens area, the intensity of the light is inversely proportional to the area of the image: if you take a given amount of light and deposit it onto half the sensor area, you get twice the intensity.

    Usually, with a camera, the object distance (from the lens) is much longer than the focal length, so the image distance very nearly equals the focal length. In this situation, the height and width of the image are very nearly proportional to the focal length. Therefore the intensity of the light at the image is very nearly inversely proportional to the square of the focal length.
     
  10. Jul 10, 2008 #9
    This is perfectly clear.

    This is the critical point. Let's say that the heigth of the object is [tex]y_0[/tex], the focal length is f, and the distance from the focal point to the film is [tex]x_i[/tex] (as in the first picture in http://www.inyourfacefotos.com/fstop.htm). Then by similarity of triangles the heigth of the image will bee [tex]y_i=\frac{y_0}{f} x_i[/tex], that is it will be inversely proportional to f and thus the intensity [tex]I \propto f^2[/tex]. The question is what is wrong in this reasoning?
     
  11. Jul 11, 2008 #10

    jtbell

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    That diagram isn't really appropriate for this question. Try the attached one instead. For an object of given size and distance from the lens, the angle [itex]\theta[/itex] is fixed. Looking at the red triangle, it should be obvious that if [itex]\theta[/itex] is fixed, then the image height [itex]h_i[/itex] is proportional to the image distance [itex]d_i[/itex]. If the object distance is much larger than the focal length, then the image distance is practically equal to the focal length. Therefore the image height is practically proportional to the focal length.
     

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  12. Jul 11, 2008 #11
    Aah, I see, the problem with my formula was that if I changed f, [tex]x_i[/tex] would change too. Your picture, jtbell, explains it very well. Thank you very much for your time.
     
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