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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
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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woleraymond Group TitleBest ResponseYou've already chosen the best response.0
@benpen @RaphaelFilgueiras @experimentX pls any ideas
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Not sure about the answer ... this is just an attempt. dw:1348160490381:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1348160596123:dw i hope this picture is correct.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1348160758615:dw
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1348161529794:dw
 one year ago

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

experimentX Group TitleBest 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 \[ \tan \left( \arcsin \left( {4 \over 3} \sin \left( \arctan \left( h \over \sqrt{1  h^2}\right)\right)\right) \arctan \left( h \over \sqrt{1  h^2}\right)\right ) \times \sqrt{1  h^2 } + h \over h\]
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
@experimentX in the diagram above pls make clear the angle \[\theta 1  \theta 2\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1348162335702:dw
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
Also make clear the angle of incidence and that of refraction
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
theta_1 is angle of refraction and theta 2 is angle of incidence.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
im on it..trying to figgy it out
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
how did you derive the magnification formula
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
which should i follow the first or the second
 one year ago

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

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
yep u right
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
@experimentX how did u derive that magnificaton formular
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1348163866108:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
all right ... I'm sleeping. carry on!!
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
@demitris yep i ve printed it.....but how would you define lateral magnification
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
so how did magnification becomes refractive index
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
pls dont delete
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
so you mean angle OA'A is the incidence angle while OAC is refracted angle?
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
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

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
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

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
pls dont remove any thing...
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
igot it now @demitris thanks alot
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
@woleraymond what did you get?
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
@demitris actually uploaded a document where he explained how the refractive index equated to magnification
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
@woleraymond mind sharing it?
 one year ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
pls can i delete now?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
just close it?
 one year ago
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