What is the critical angle and area for light at a water and air interface?

AI Thread Summary
The critical angle for light at the water (n=1.33) and air (n=1) interface is calculated to be approximately 48.75 degrees. A fish located 2 meters below the water surface can see an area of about 2.28 square meters at the surface. This area is described as a circular disk of bright blue sky directly above the fish, extending out to the critical angle of 48 degrees. The calculations involve basic trigonometry to determine the dimensions of the triangle formed by the light path. The solution confirms the fish's viewable area at the water's surface.
mogley76
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Homework Statement


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.

Homework Equations



none given

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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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.
 
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