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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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

Physics
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Not sure about the answer ... this is just an attempt. |dw:1348160490381:dw|
|dw:1348160596123:dw| i hope this picture is correct.

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|dw:1348160758615:dw|
oops ... this is a unit circle. the the coordinate is \( (h, \sqrt{1 - h^2})\)|dw:1348161039200:dw|
|dw:1348161529794:dw|
the magnification must be \[ \tan ({ \theta _i - \theta_r}) \times \sqrt{1 - h^2 } + h \over h\]
\[ {\sin \theta_i \over \sin \theta_r} = {4 \over 3} \\ \theta_r = \arctan \left( h \over \sqrt{1 - h^2}\right)\] 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 \[ \tan \left( \arcsin \left( {4 \over 3} \sin \left( \arctan \left( h \over \sqrt{1 - h^2}\right)\right)\right)- \arctan \left( h \over \sqrt{1 - h^2}\right)\right ) \times \sqrt{1 - h^2 } + h \over h\]
@experimentX in the diagram above pls make clear the angle \[\theta 1 - \theta 2\]
|dw:1348162335702:dw|
Also make clear the angle of incidence and that of refraction
theta_1 is angle of refraction and theta 2 is angle of incidence.
it seems that magnification is dependent on size of the fish. I'm not sure on this though. do you have an answer?
http://www.wolframalpha.com/input/?i=plot+%28tan%28sin%5E%28-1%29%284%2F3+sin%28tan%5E%28-1%29%28h%2Fsqrt%281-h%5E2%29%29%29%29-tan%5E%28-1%29%28h%2Fsqrt%281-h%5E2%29%29%29+sqrt%281-h%5E2%29%2Bh%29%2Fh+from+0+to+0.2
im on it..trying to figgy it out
how did you derive the magnification formula
which should i follow the first or the second
Image size/ object size ... isn't it? i haven't done this for quite a long time though.
yep u right
@experimentX how did u derive that magnificaton formular
|dw:1348163866108:dw|
all right ... I'm sleeping. carry on!!
@demitris yep i ve printed it.....but how would you define lateral magnification
so how did magnification becomes refractive index
pls dont delete
so you mean angle OA'A is the incidence angle while OAC is refracted angle?
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
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
pls dont remove any thing...
igot it now @demitris thanks alot
@woleraymond what did you get?
@demitris actually uploaded a document where he explained how the refractive index equated to magnification
@woleraymond mind sharing it?
pls can i delete now?
just close it?

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