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Critical angle and area 
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#1
Aug2411, 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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