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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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malay

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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

rahul_c

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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

AAKRITI

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14 Jan 2007 12:17:10 IST

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The answer will be 6/n :

In fish frame; v_b,f = v_f + nv_b

you_saket

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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.

mohit1

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15 Jan 2007 12:07:00 IST

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Krishna Gopal Singh

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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

vijay kharya

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5 Oct 2010 22:44:49 IST

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yes, to remove the effect of rare/ denser medium in this case we can just multiply the velocity of the bird by ( u { water} / u { air } ) * v and then apply usual relation of relative velocity

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