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woleraymond

  • 2 years ago

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

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  1. woleraymond
    • 2 years ago
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    @benpen @RaphaelFilgueiras @experimentX pls any ideas

  2. experimentX
    • 2 years ago
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    Not sure about the answer ... this is just an attempt. |dw:1348160490381:dw|

  3. experimentX
    • 2 years ago
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    |dw:1348160596123:dw| i hope this picture is correct.

  4. experimentX
    • 2 years ago
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    |dw:1348160758615:dw|

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

  6. experimentX
    • 2 years ago
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    |dw:1348161529794:dw|

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

  8. experimentX
    • 2 years ago
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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 \[ \tan \left( \arcsin \left( {4 \over 3} \sin \left( \arctan \left( h \over \sqrt{1 - h^2}\right)\right)\right)- \arctan \left( h \over \sqrt{1 - h^2}\right)\right ) \times \sqrt{1 - h^2 } + h \over h\]

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

  10. experimentX
    • 2 years ago
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    |dw:1348162335702:dw|

  11. woleraymond
    • 2 years ago
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    Also make clear the angle of incidence and that of refraction

  12. experimentX
    • 2 years ago
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    theta_1 is angle of refraction and theta 2 is angle of incidence.

  13. experimentX
    • 2 years ago
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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?

  14. woleraymond
    • 2 years ago
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    im on it..trying to figgy it out

  15. woleraymond
    • 2 years ago
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    how did you derive the magnification formula

  16. woleraymond
    • 2 years ago
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    which should i follow the first or the second

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

  18. woleraymond
    • 2 years ago
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    yep u right

  19. woleraymond
    • 2 years ago
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    @experimentX how did u derive that magnificaton formular

  20. experimentX
    • 2 years ago
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    |dw:1348163866108:dw|

  21. experimentX
    • 2 years ago
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    all right ... I'm sleeping. carry on!!

  22. woleraymond
    • 2 years ago
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    @demitris yep i ve printed it.....but how would you define lateral magnification

  23. woleraymond
    • 2 years ago
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    so how did magnification becomes refractive index

  24. woleraymond
    • 2 years ago
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    pls dont delete

  25. woleraymond
    • 2 years ago
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    so you mean angle OA'A is the incidence angle while OAC is refracted angle?

  26. woleraymond
    • 2 years ago
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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

  27. woleraymond
    • 2 years ago
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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

  28. woleraymond
    • 2 years ago
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    pls dont remove any thing...

  29. woleraymond
    • 2 years ago
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    igot it now @demitris thanks alot

  30. Algebraic!
    • 2 years ago
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    @woleraymond what did you get?

  31. woleraymond
    • 2 years ago
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    @demitris actually uploaded a document where he explained how the refractive index equated to magnification

  32. Algebraic!
    • 2 years ago
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    @woleraymond mind sharing it?

  33. woleraymond
    • 2 years ago
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  34. woleraymond
    • 2 years ago
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    pls can i delete now?

  35. Algebraic!
    • 2 years ago
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    just close it?

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