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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Jan 2007 23:31:52 IST
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a fish rising vertically to the surface of a lake at 3m/s observes a bird diving vertically downat the rate of 9m/s. what is the actual velocity of the bird?
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14 Jan 2007 10:46:29 IST
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since this situation deals with normal incidence of light on the water air- interrface, no phenomena of refraction will occur.
hence actual velocity of bird=9-3=6m/s downwards
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Imagination is more important than knowledge
-------Albert Einsetein |
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14 Jan 2007 11:40:59 IST
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refractive index (n) = apparent depth/real depth
if displacement with respect to unit time is taken = (9/n)
actual velocity= (9/n) - 3
n is the refractive index of water
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The yardstick of human intelligence is the ability to overcome the last fallacy |
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14 Jan 2007 12:17:10 IST
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The answer will be 6/n : In fish frame; vb,f = vf + nvb
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you have to run faster and faster so as to remain where you are.. even if you are on the right track you will get run over if you just keep sitting there...
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14 Jan 2007 15:26:26 IST
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what is the esay way to solve numarical of refractive index through prism.
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15 Jan 2007 12:07:00 IST
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18 Jan 2007 04:46:56 IST
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If velocity of bird is v it wil appear 1.33v to an stationary fish (refractive index of water is 1.33). 1.33v+3=9 v=6/1.33 So Aakriti's answer is right
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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