anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@benpen @RaphaelFilgueiras @experimentX pls any ideas
experimentX
  • experimentX
Not sure about the answer ... this is just an attempt. |dw:1348160490381:dw|
experimentX
  • experimentX
|dw:1348160596123:dw| i hope this picture is correct.

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

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