## anonymous 4 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

1. anonymous

@benpen @RaphaelFilgueiras @experimentX pls any ideas

2. experimentX

Not sure about the answer ... this is just an attempt. |dw:1348160490381:dw|

3. experimentX

|dw:1348160596123:dw| i hope this picture is correct.

4. experimentX

|dw:1348160758615:dw|

5. experimentX

oops ... this is a unit circle. the the coordinate is $$(h, \sqrt{1 - h^2})$$|dw:1348161039200:dw|

6. experimentX

|dw:1348161529794:dw|

7. experimentX

the magnification must be $\tan ({ \theta _i - \theta_r}) \times \sqrt{1 - h^2 } + h \over h$

8. 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 $\tan \left( \arcsin \left( {4 \over 3} \sin \left( \arctan \left( h \over \sqrt{1 - h^2}\right)\right)\right)- \arctan \left( h \over \sqrt{1 - h^2}\right)\right ) \times \sqrt{1 - h^2 } + h \over h$

9. anonymous

@experimentX in the diagram above pls make clear the angle $\theta 1 - \theta 2$

10. experimentX

|dw:1348162335702:dw|

11. anonymous

Also make clear the angle of incidence and that of refraction

12. experimentX

theta_1 is angle of refraction and theta 2 is angle of incidence.

13. experimentX

it seems that magnification is dependent on size of the fish. I'm not sure on this though. do you have an answer?

14. experimentX
15. anonymous

im on it..trying to figgy it out

16. anonymous

how did you derive the magnification formula

17. anonymous

which should i follow the first or the second

18. experimentX

Image size/ object size ... isn't it? i haven't done this for quite a long time though.

19. anonymous

yep u right

20. anonymous

@experimentX how did u derive that magnificaton formular

21. experimentX

|dw:1348163866108:dw|

22. experimentX

all right ... I'm sleeping. carry on!!

23. anonymous

@demitris yep i ve printed it.....but how would you define lateral magnification

24. anonymous

so how did magnification becomes refractive index

25. anonymous

pls dont delete

26. anonymous

so you mean angle OA'A is the incidence angle while OAC is refracted angle?

27. 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

28. 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

29. anonymous

pls dont remove any thing...

30. anonymous

igot it now @demitris thanks alot

31. anonymous

@woleraymond what did you get?

32. anonymous

@demitris actually uploaded a document where he explained how the refractive index equated to magnification

33. anonymous

@woleraymond mind sharing it?

34. anonymous

35. anonymous

pls can i delete now?

36. anonymous

just close it?