|Aug24-11, 08:29 PM||#1|
critical angle and area
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
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??
|Aug24-11, 08:47 PM||#2|
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.
|area, critical angle, optics water|
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