- #1
ReMa
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Homework Statement
You are looking up from under the water in a swimming pool. If you are 2m below the surface, what is the radius of the "hole" at the water surface through which you can see out of the pool?
Homework Equations
n.sin(theta)1 = n.sin(theta)2
The Attempt at a Solution
Could it be assumed that looking straight up through the water? So, sin(theta)1 = sin theta(90) = 1 ?
I'm assuming a hole would have an area of pi.r2, but don't know how to include that in the former equation (Snell's Law).
I also know n = c/v
c = 3x10^8 m/s (speed of light in a vacuum)
v = velocity
These values weren't given, but if applicable:
n air = 1.00
n water = 1.33