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Critical angle and area

by mogley76
Tags: area, critical angle, optics water
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Aug24-11, 08:29 PM
P: 19
1. The problem statement, all variables and given/known data
calculate the critical angle for light at a water (n=1.33) and air (n=1) interface.
if a fish is 2m below the surface of the water , calculate the area at the surface through which the fish can viw the world above the water.

2. Relevant equations

none given

3. The attempt at a solution

critical angle = sin theta c= n2/n1 = 48.75 deg

area of surface=

using basic trig find out all the sides and angles of triangle ..
then area of tribgle is .5*2.28*2 = 2.28 m^2

am i right in all this??
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Aug24-11, 08:47 PM
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lightgrav's Avatar
P: 1,124
the Area on the SURFACE that the fish sees thru ...
it looks like a circular disk of bright blue sky directly above the fish ... out to 48deg.

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