(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Critical angle and area

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