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Newtonian Form of Lens Equation

  1. Aug 29, 2011 #1
    Can the Newtonian Form the Lens Equation [itex]f=x'x[/itex] be derived directly from the standard form the lens equation? [itex]\frac{1}{f}=\frac{1}{d_{i}}+\frac{d}{d_{o}}[/itex]

    [itex]x[/itex] is the distance between the object and the focus on the same side of the object. [itex]x'[/itex] is the distance between the image and the focus on the other side of the lens from the first focus.
     
    Last edited: Aug 29, 2011
  2. jcsd
  3. Aug 29, 2011 #2
    Yes, overlooking the typo in the equation and the misuse of the word "focus".
     
  4. Aug 30, 2011 #3
    Oops, sorry about that. I'm not used to the LaTeX on this website with its rendering time. Can you tell me how? I derived it from a ray diagram.
     
  5. Aug 30, 2011 #4
    If you draw a picture showing the various points, it should be straightforward to show that the newton equation must be true.
     
  6. Aug 30, 2011 #5
    Isn't that just deriving it from a ray diagram? I know how to do that. I mean deriving it directly from the standard lens equation itself.
     
  7. Aug 30, 2011 #6
    I don't know what you mean, but you'll never derive the equation you typed because it is wrong. If you assume the regular lens equation is true and substitute, you quickly find that the correct alternative form of the lens equation must also be true. I personally never use it, and I use optics equations all the time, but I suppose it's useful for some specialized sorts of applications.
     
  8. Aug 30, 2011 #7
    Yes, I understand that what I have typed is incorrect. It should be [itex]\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}[/itex] and [itex]f^2=xx'[/itex]. My bad.

    Also, for your way of substitution, I have the equations [itex]|d_i-f|=x'[/itex] and [itex]|d_o-f|=x[/itex]. Substitution would involve solving for [itex]d_i[/itex] and [itex]d_o[/itex] which involves ugly quadratics. Is there a better way to do this? Or is this the only way?
     
  9. Aug 30, 2011 #8
    POOF! just got it. ah..solving at the back of the receipt from a mall. :D

    di is the same sa x', so di = x' (u can use this interchangeably)
    so as for do, do=xo

    thus,

    x'=x'+f and xo=xo+f
    substitute this to
    1/f=1/di + 1/do

    1/f = (1/x'+f) + (1/xo+f)
    1/f = {(x'+f)+(xo+f)}/{(x'+f)(xo+f)}
    do little algebra then,
    reciprocate
    f= (x'+f)(xo+f)/(x'+xo+2f)
    cross multiply
    (x'f)+(xf)+2f^2=x'xo+fxo+x'f+f^2
    simplify
    then
    2f^2=x'xo+f^2 (transpose f^2 to the other side of the equation)
    2f^2-f^2= x'xo
    POOF
    f^2=x'xo

    note: in xo , o is a subscript.

    Hope u understand :)
     
  10. Aug 30, 2011 #9
    Sorry, I don't really understand how d_i is the same as x'. x' is supposed to be the distance between the image and the focus behind the lens. Also, how can this be?
    If that were true, then we would have f=0 which is not true for all lenses (I'm not even sure if it can be true at all).
     
  11. Aug 30, 2011 #10
    Oh, the horrors! :wink: It took maybe 10 seconds writing on a paper plate that I'll later use to feed my cat, so I'm not sure what the issue is or if you've actually tried the derivation.
     
  12. Aug 30, 2011 #11
    Haha, okay. I'll try it. This is what I get:

    [itex]|d_i-f|=x [/itex]

    [itex]\implies d_i ^2-2d_i f+f^2=x'^2[/itex]

    [itex]\implies d_i^2-2d_if+f^2-x'^2=0[/itex]

    [itex]\implies d_i=\frac{2f\pm \sqrt{4f^2-4(f^2-x'^2)}}{2}[/itex]

    [itex]\implies d_i=\frac{2f\pm2x'}{2}[/itex]

    [itex]\implies d_i=f\pm x'[/itex]

    Similarly, we have [itex]d_o=f\pm x[/itex]

    Substituting, we get
    [itex]\frac{1}{f}=\frac{1}{f\pm x'}+\frac{1}{f\pm x} [/itex]

    [itex]\implies (f\pm x)(f\pm x')=f(f\pm x')+f(f\pm x) [/itex]

    [itex]{f^2}\pm{fx'}\pm{xf}\pm xx' = {f^2}\pm{fx'}+f^2\pm{fx}[/itex]

    [itex]\pm xx'=f^2[/itex]

    How do we know which sign the LHS has? Is the Trivial Inequality enough to just say that it is positive?

    Also, why doesn't the \cancel function and \begin{align*} function not work on LaTeX?

    By the way, thank you JeffKoch for forcing me to do the derivation, I was too lazy earlier.
     
  13. Aug 30, 2011 #12
    oops! sorry XD for the major mistake but i thank you for the corrections, you help me understand it too. :)
     
  14. Aug 30, 2011 #13
    The term f^2 is obviously positive if we're talking about real physical lenses, so the other side must also be positive. If you think about it, there are two ways to do this, and both make sense. :wink:
     
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