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Lens power/focal length question

  1. Mar 8, 2005 #1
    A friend and I are getting conflicting answers for a focal length/power question. He says he gets 22.5 diopters, whereas I get 5.1 diopters, so I was wondering who's screwing up. Here is the question and my answer:

    Q: If a +20 diopter lens is a symmetric double convex lens of index 1.50, what is its focal length and power when placed in a tank of water with index 1.33?


    A:

    Code (Text):
    f = 1/d = 1/20 = 0.05 m

    Using lens maker's equation:

    1/f = (n1/n2 - 1) (1/r1 - 1/r2)

    Let (1/r1 - 1/r2) = K.

    Solve for K in air (which will be the same in water).

    1/0.05 m = (1.50 / 1.00 - 1)K
    20 = (0.5)K
    K = 40


    Now plug in values for water:

    1/f = (1.50/1.33 - 1) (40)
    1/f = 5.1 diopters
    f = 0.20 m
    A priori, it seems to make sense that the focal length would increase and the power would decrease. Since the change of light going from water to the lens is less than the change of light going from air to the lens (in terms of indices of refraction), it seems that the light should be less bent in water and therefore travel further before converging to an image.
     
    Last edited: Mar 8, 2005
  2. jcsd
  3. Mar 8, 2005 #2
    I seem to have a third opinion. :smile:

    I don't recognize the equation you're using. The "lensmaker's equation" that I learned is:
    [tex]\frac{1}{f} = (n-1)\left ( \frac{1}{R_1} - \frac{1}{ R_2} \right ) [/tex]

    In your problem [itex] \left ( \frac{1}{R_1} - \frac{1}{ R_2} \right ) [/itex] is constant so call it k and we have just
    [tex]\frac{1}{f} = k(n-1)[/tex]
    [tex]20 = k(1.50-1)[/tex]
    [tex]k = 40[/tex]
    And then, in water:
    [tex]\frac{1}{f} = 40(1.33-1)[/tex]
    [tex]\frac{1}{f} = 13.2 \; \text{diopters} \: = .076 \; \text{m}[/tex]
     
  4. Mar 9, 2005 #3
    Actually, I wasn't too familiar with that form of the lens maker's equation either, but the place I got it from is located here:

    http://online.cctt.org/physicslab/content/Phy1/lessonnotes/thinlensequation/lensmakerequation.asp

    (And I also noticed it here: http://gecko.gc.maricopa.edu/~medgar/RayDiagram/

    and here

    http://astro1.panet.utoledo.edu/~rgb/lens.htm)

    I don't doubt that your answer is correct (considering a change of focal length of only a few centimeters is much more likely than about 15), but I'm still curious as to why exactly it works out. For example, when you solved for the value of K in air, you used the index of refraction of the lens, but when you solved for the focal length in the second part, you used the index of refraction of water. It just seems somewhat odd to at one point use the index of the material and in the second use the index of the medium. It seems like the indices of both the material and medium should be included together all the time, especially when the medium is changing (this is coming from a definitely NON-physics student, so forgive the egregious mistakes).


    Just out of curiousity, what would be mathematically wrong with the following (which is yet ANOTHER mutation of this problem I've come up with):

    Code (Text):
    The lens maker's equation more formally is:

    n1/f = (n2 - n1) (1/r1 - 1/r2), where (1/r1 - 1/r2) = K.

    In this case, n2 is the index of the material and n1 is the index of the medium

    Let n = index of lens.
    Let na = index of air.
    Let nw = index of water.

    In the case of air, we can solve for K by doing:

    na/f = (n - na)(K)
    1/0.05 = (1.5 - 1.0)(K)
    20 = (1.5 - 1.0)(K)
    K = 40

    But now in water...

    nw/f = (n - nw)(K)
    1.33/f = (1.5 - 1.33)(40)
    1.33/f = 6.8 diopters
    1/f = 5.11 diopters
    f = 0.20 m


    Which brings me back to the answer I originally got.
     
    This problem will be the death of me.
     
    Last edited: Mar 9, 2005
  5. Mar 9, 2005 #4
    Sorry, I guess my mind was somewhere else. You are right. I should have written:

    in water:
    [tex]\frac{1}{f} = 40 \left( \frac{1.50}{1.33} -1 \right )[/tex]
    [tex]\frac{1}{f} = 5.11 \; \text{diopters} \: = 0.20 \; \text{m}[/tex]

    What's important is not the index of refraction itself, but the ratio of the index of the lens to the index of the surrounding medium. In air the ratio is 1.50, in water the ratio is only 1.13. This accounts for the relatively large change in focal length.
     
  6. Mar 9, 2005 #5
    Excellent. Thanks!
     
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