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Lens Formula Derivation

  1. May 1, 2007 #1
    I recently took a test on lenses. One of the questions was "The image formed by a converging lens is real and is n times the size of the object. If the focal length of the lens is f, the distance from the lens to the image is?

    Now, the answer is f(n+1) but I cant seem to figure out why. Heres my work so far and what i get. (di being distance of the image, do being distance of the object, f being focal length, and n being magnification
    1/di + 1/do = 1/f so therefore 1/di = 1/f- 1/do
    n = -di/do so therefore do=-si/n

    from combining those two i come up with 1/di = 1/f - -n/di which equals 1/di = 1/f + n/di

    then i got a common denominator and cmae up with 1/di = (di + fn) / (f*di)
    then i cross multiplied and came up with f*di = di(di+f*n) so i crossed out the two di's and got f = di+fn. and then simply solved for di (what we're looking for) and got di=f-fn and from that di=f(1-n).

    Now, the answer says its supposed to be f(n+1)...where did i go wrong? I assume i messed up with a sign somewhere but i cant seem to figure out where.
     
  2. jcsd
  3. May 1, 2007 #2

    Mentz114

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    What does a negative magnification mean ? Or is one of those lengths considered negative ?
     
  4. May 1, 2007 #3
    The negative indicates that the direction of the image is oppositte of that of the object
     
  5. May 2, 2007 #4

    Mentz114

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    In that case, shouldn't this

    1/di + 1/do = 1/f

    be this
    -1/di + 1/do = 1/f ?
     
  6. May 2, 2007 #5
    erm not quite...i think you'll find its 1/f = 1/di - 1/do
     
  7. May 2, 2007 #6
    Also, a quick hint, in optics it is important to follow Cartesian sign convention, in other words take the intersection between the lens and optical axis to be zero, assume light travels left to right, any distances are measured from the lens/optical axis intersection. Therefore the object distance in this case will be negative as the object lies in front of the lens and the image distance will be positive as the image lies behind it.
    With regards to negative magnifictaion, it simply means that the image lies below the optical axis, in other words it is inverted.
     
  8. May 2, 2007 #7
    Better we take that direction as positive which is in support of direction of incident light.....and the origin is the intersection point itself.....

    here n= + di / do (in lenses it is of + sign while in mirrors -sign is there)
     
  9. May 2, 2007 #8
    I'm not saying that you can't do it that way, i'm simply saying that as an optician, that is how i was taught and how every other optician i know was taught. It will only make things harder when you confront more difficult optics problems such as multi lens systems and optical instruments if you choose not to follow Cartesian sign convention. I think Rene Descartes who devised the convention (hence the name) knew what he was talking about.
     
  10. May 3, 2007 #9
    i find this method easy ...(though it is more complex)...and that's because i can't undrestand the other conventions easily....the signs of v,u & f (if they are unknown) need not to be altered in this method and formulae can be directly applied without taking care of signs..
     
    Last edited: May 3, 2007
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