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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
 2 years ago
 2 years 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
 2 years ago
 2 years ago

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

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
the magnification must be \[ \tan ({ \theta _i  \theta_r}) \times \sqrt{1  h^2 } + h \over h\]
 2 years 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 \[ \tan \left( \arcsin \left( {4 \over 3} \sin \left( \arctan \left( h \over \sqrt{1  h^2}\right)\right)\right) \arctan \left( h \over \sqrt{1  h^2}\right)\right ) \times \sqrt{1  h^2 } + h \over h\]
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
pls dont delete
 2 years 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?
 2 years 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
 2 years 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
 2 years ago

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

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

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

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

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

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