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??
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.