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anonymous
 3 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
anonymous
 3 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@benpen @RaphaelFilgueiras @experimentX pls any ideas

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Not sure about the answer ... this is just an attempt. dw:1348160490381:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1348160596123:dw i hope this picture is correct.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1348160758615:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1oops ... this is a unit circle. the the coordinate is \( (h, \sqrt{1  h^2})\)dw:1348161039200:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1348161529794:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1the magnification must be \[ \tan ({ \theta _i  \theta_r}) \times \sqrt{1  h^2 } + h \over h\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1\[ {\sin \theta_i \over \sin \theta_r} = {4 \over 3} \\ \theta_r = \arctan \left( h \over \sqrt{1  h^2}\right)\] 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 \[ \tan \left( \arcsin \left( {4 \over 3} \sin \left( \arctan \left( h \over \sqrt{1  h^2}\right)\right)\right) \arctan \left( h \over \sqrt{1  h^2}\right)\right ) \times \sqrt{1  h^2 } + h \over h\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX in the diagram above pls make clear the angle \[\theta 1  \theta 2\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1348162335702:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Also make clear the angle of incidence and that of refraction

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1theta_1 is angle of refraction and theta 2 is angle of incidence.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1it seems that magnification is dependent on size of the fish. I'm not sure on this though. do you have an answer?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im on it..trying to figgy it out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how did you derive the magnification formula

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which should i follow the first or the second

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Image size/ object size ... isn't it? i haven't done this for quite a long time though.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX how did u derive that magnificaton formular

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1348163866108:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1all right ... I'm sleeping. carry on!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@demitris yep i ve printed it.....but how would you define lateral magnification

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so how did magnification becomes refractive index

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you mean angle OA'A is the incidence angle while OAC is refracted angle?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im just trying to figgy why you are using sine L as ur incident or which formula are u using for n as refractive index

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ur letters might be sending conflicting signals..it will be nice if they are legible...im just trying to make sure i get ur import

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0pls dont remove any thing...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0igot it now @demitris thanks alot

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@woleraymond what did you get?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@demitris actually uploaded a document where he explained how the refractive index equated to magnification

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@woleraymond mind sharing it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0pls can i delete now?
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