critical angle and area

mogley76

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??

lightgrav

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.