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Modern Optics problems

  1. Sep 12, 2011 #1
    These problems are from my JUNIOR-level class. Don't move this one. I'm very rusty in Optics. It's supposed to be taken after University Phys II, but I took that three years ago.

    For A1, I'm not sure how to setup this problem. Do I need to take into account all three index of refractions, water, lens, and air?

    For A2, it appears to be a straightforward thin-lens system. However, I'm not sure how to get the focal length of the other lens with the given information.

    http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-2.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 12, 2011 #2

    rude man

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    A1: I would start with the basics. Draw your "fish" so that its middle is coaxial with the tank lens. That ray would be a straight line from the fish's midsection to you r eye's center. Then use Snell's law to determine the path of two additional rays, one emanating from each end of the fish, refracting at each lens surface, and defining a new angle upon entering the air. You then (mentally) move the fish towards or away from you such that the angle the fish would have subtended in the absence of a lens equals your refracred angle. The resulting distance is your answer.

    A2: there are standard formulae available for a microscope, e.g. at Wikipedia.
     
  4. Sep 12, 2011 #3
    For A2, MT = fo/fe.

    fo = focal length of the objective

    fe = focal length of the eyepiece

    So, fo = 80 cm.
     
  5. Sep 12, 2011 #4
    How do I get any angles if I'm not given any angles? I have the radius of curvature to worry about, too. I think I have to use more than just Snell's Law.
     
  6. Sep 13, 2011 #5

    rude man

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    Forget what i said. That way you'd have to pretty much rederive all of refractive optics. I also forgot to mention tat you have to simplify Snell's law to
    n1*theta1 = n2*theta2 (paraxial rays).

    OK, the following is my first stab at this & it's late, so I may be sending a later edit. Hopefully it'll jump-start you in any case.

    Consider the first interface, betw. the water and the lens. The equation for that surface is
    n1/p - n2/q' = (n2 - n1)/R ... from resnick & halliday chapt. 42
    where p = distance of fish to lens = 20 cm. , q' = image distance. q' > 0 in this form even though the image is virtual, R = 30cm, n1 = 1.33, n2 = 1.5.
    For second interface, betw. lens & air,
    n2/q' + 1/q = (1-n2)/R where q = final image distance to lens.
    Combined gives
    n1/p - (n2 - n1)/R = (1 - n2)/R - 1/q
    Solve for q. Then |q| is your answer since that is the distance the image is behind the lens (virtual image, erect).
     
  7. Sep 13, 2011 #6
    Hecht has a similar equation:

    nm/so1 + nl/si1 = (nl - nm)/R1

    nl = 1.5
    nm = 1.33
    R1 = 30 cm
    so1 = 20

    That yields si1 = -24.6 cm. Of course, that is a virtual image inside the tank. Why do you say it's positive?
     
  8. Sep 13, 2011 #7

    rude man

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    Because the answer calls for a magnitude. It asks for the apparent distance between you and the fish. A distance cannot be negative (unless maybe you have eyes in the back of your head? :-)
     
  9. Sep 13, 2011 #8
    My point is don't I have to maintain the sign from the first interface to the second when calculating how the light is refracted through the lens? I certainly see why I would find the magnitude of the second image distance.
     
  10. Sep 13, 2011 #9

    rude man

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    Also, your equation seems to have only two indices of refraction in it. It looks like my first or second equation (they're essentially the same form), but you need to combine them to get the overall lensing effect.

    Did you solve q using my last equation?
     
  11. Sep 13, 2011 #10
    I'm breaking it up into two situations. I'm using si1 as so2 in the second equation.

    Let me see what I end up with as si2.
     
  12. Sep 13, 2011 #11

    rude man

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    My first equation would be the same as my second and your Hecht's if I hadn't changed q' to be > 0. It happened that resnick & halliday did it that way. I had enough trouble extending their derivation to the three-n case without deviating from their choice of signs.
     
  13. Sep 13, 2011 #12

    rude man

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    In any case I think we're on the right track. Sorry I don't have the Hecht text, my optics knowledge is limited to introductory engineering-level physics (viz. resnick & halliday).
     
  14. Sep 13, 2011 #13
    I calculated si2 as 22.5 cm.
     
  15. Sep 13, 2011 #14
    Well, this material is fresher on your mind than mine. Hopefully, one the pros can chime in here.
     
  16. Sep 13, 2011 #15

    rude man

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    For what it's worth, I got 12.9cm.
     
  17. Sep 13, 2011 #16

    rude man

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    I just noticed that my q' = 24.7cm. But that is not the end of the story. That just determines the image distance of the first interface. You also have continue on thru the second interface.
     
  18. Sep 13, 2011 #17
    I did. I have -24.6 and 22.5.
     
  19. Sep 14, 2011 #18

    rude man

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    OK, we agree on si1 = so2. Could you show me the equation you used to compute si2, which is the answer we're looking for?
     
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