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Reading on thin lenses

  1. Nov 8, 2008 #1
    I was reading on thin lenses, in which the lens makers' equation was proved. The proof first gave the lens (a biconvex lens) a thickness (which it would later consider negligible) so that the light would pass through one surface and out the other (so it comes into contact with 2 surfaces). The proof was based on the postulation that the image formed from contact with the 1st surface functions as the object for the 2nd surface (and vice versa). The postulation seems true for when the 1st image formed is virtual and behind the actual object, but it doesn't seem to make sense for when the 1st image formed is real and behind the surface. Can anyone clarify how the second case works?
     
  2. jcsd
  3. Nov 9, 2008 #2
    Re: Lenses

    ...o_o...
     
  4. Nov 10, 2008 #3

    Andy Resnick

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    Re: Lenses

    The derivation of the lens maker's formula which I have seen uses "transformation matrices", but those operate on rays (a ray is described as a vector (y, nu), where y is the ray height at a particular location along the optical axis and nu is the index of refraction n times the angle u), not images and objects.

    The transformation matrix of a single lens is written as:

    M[tex]\equiv R_{2}T_{12}R_{1}[/tex],

    Where [tex] R_{2} and R_{1}[/tex] are refraction matrices, and [tex]T_{12}[/tex] the translation matrix. Explicitly,

    R[tex]\equiv ((1 -P)(0 1))[/tex], where that mess in parenthesis is a poor attempt at TeXing a 2x2 matrix, and P the optical power of a surface. T[tex]\equiv ((1 0)(d/n 1))[/tex], where d/n is the 'reduced thickness'- the actual distance divided by the index of refraction.

    To get the lens maker's equation, simply let d go to 0, and for a lens in air, P = (n-1)/R, where n is the index of the lens, and R the radius of curvature of one surface.
     
  5. Nov 10, 2008 #4

    jtbell

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    Re: Lenses

    In that case, the image formed by the first surface is a virtual object for the second surface.

    Remember that with a virtual image, outgoing light rays diverge from a lens or refracting surface as if they had come from a point before the lens or surface; but the light does not actually pass through that point.

    Similarly, with a virtual object, incoming light rays converge towards a point beyond the lens or refracting surface, but never actually pass through that point because the lens or surface refracts them somewhere else. Nevertheless, you can use the distance between that point and the lens or surface as an object distance, provided that you make it negative (just as a virtual image has a negative image distance).
     
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