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woleraymond Group Title

A small tropical fish is at the center of a spherical fish bowl 1m in diameter. Determine the position and the lateral magnification of the image of the fish seen by an observer outside the bowl. The refractive index of water is 4/3

  • one year ago
  • one year ago

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  1. woleraymond Group Title
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    @benpen @RaphaelFilgueiras @experimentX pls any ideas

    • one year ago
  2. experimentX Group Title
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    Not sure about the answer ... this is just an attempt. |dw:1348160490381:dw|

    • one year ago
  3. experimentX Group Title
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    |dw:1348160596123:dw| i hope this picture is correct.

    • one year ago
  4. experimentX Group Title
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    |dw:1348160758615:dw|

    • one year ago
  5. experimentX Group Title
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    oops ... this is a unit circle. the the coordinate is \( (h, \sqrt{1 - h^2})\)|dw:1348161039200:dw|

    • one year ago
  6. experimentX Group Title
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    |dw:1348161529794:dw|

    • one year ago
  7. experimentX Group Title
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    the magnification must be \[ \tan ({ \theta _i - \theta_r}) \times \sqrt{1 - h^2 } + h \over h\]

    • one year ago
  8. experimentX Group Title
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    \[ {\sin \theta_i \over \sin \theta_r} = {4 \over 3} \\ \theta_r = \arctan \left( h \over \sqrt{1 - h^2}\right)\] Therefore \[ \theta_i = \arcsin \left( {4 \over 3} \sin \left( \arctan \left( h \over \sqrt{1 - h^2}\right)\right)\right)\] Hence the magnification is \[ \tan \left( \arcsin \left( {4 \over 3} \sin \left( \arctan \left( h \over \sqrt{1 - h^2}\right)\right)\right)- \arctan \left( h \over \sqrt{1 - h^2}\right)\right ) \times \sqrt{1 - h^2 } + h \over h\]

    • one year ago
  9. woleraymond Group Title
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    @experimentX in the diagram above pls make clear the angle \[\theta 1 - \theta 2\]

    • one year ago
  10. experimentX Group Title
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    |dw:1348162335702:dw|

    • one year ago
  11. woleraymond Group Title
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    Also make clear the angle of incidence and that of refraction

    • one year ago
  12. experimentX Group Title
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    theta_1 is angle of refraction and theta 2 is angle of incidence.

    • one year ago
  13. experimentX Group Title
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    it seems that magnification is dependent on size of the fish. I'm not sure on this though. do you have an answer?

    • one year ago
  14. woleraymond Group Title
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    im on it..trying to figgy it out

    • one year ago
  15. woleraymond Group Title
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    how did you derive the magnification formula

    • one year ago
  16. woleraymond Group Title
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    which should i follow the first or the second

    • one year ago
  17. experimentX Group Title
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    Image size/ object size ... isn't it? i haven't done this for quite a long time though.

    • one year ago
  18. woleraymond Group Title
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    yep u right

    • one year ago
  19. woleraymond Group Title
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    @experimentX how did u derive that magnificaton formular

    • one year ago
  20. experimentX Group Title
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    |dw:1348163866108:dw|

    • one year ago
  21. experimentX Group Title
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    all right ... I'm sleeping. carry on!!

    • one year ago
  22. woleraymond Group Title
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    @demitris yep i ve printed it.....but how would you define lateral magnification

    • one year ago
  23. woleraymond Group Title
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    so how did magnification becomes refractive index

    • one year ago
  24. woleraymond Group Title
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    pls dont delete

    • one year ago
  25. woleraymond Group Title
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    so you mean angle OA'A is the incidence angle while OAC is refracted angle?

    • one year ago
  26. woleraymond Group Title
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    im just trying to figgy why you are using sine L as ur incident or which formula are u using for n as refractive index

    • one year ago
  27. woleraymond Group Title
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    ur letters might be sending conflicting signals..it will be nice if they are legible...im just trying to make sure i get ur import

    • one year ago
  28. woleraymond Group Title
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    pls dont remove any thing...

    • one year ago
  29. woleraymond Group Title
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    igot it now @demitris thanks alot

    • one year ago
  30. Algebraic! Group Title
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    @woleraymond what did you get?

    • one year ago
  31. woleraymond Group Title
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    @demitris actually uploaded a document where he explained how the refractive index equated to magnification

    • one year ago
  32. Algebraic! Group Title
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    @woleraymond mind sharing it?

    • one year ago
  33. woleraymond Group Title
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    • one year ago
  34. woleraymond Group Title
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    pls can i delete now?

    • one year ago
  35. Algebraic! Group Title
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    just close it?

    • one year ago
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