You are looking up from underwater in a swimming pool radius of circle?

AI Thread Summary
To determine the radius of the "hole" visible from underwater in a swimming pool when 2 meters below the surface, one must consider Snell's Law and the concept of critical angle for total internal reflection. The refractive indices for air and water are essential, with air at 1.00 and water at 1.33. The critical angle can be calculated using the formula sin(theta)c = n1/n2, where n1 is the refractive index of water and n2 is that of air. By finding the angle at which light can no longer escape the water and calculating the corresponding horizontal distance from the initial position, the radius of the visible circle can be established. This approach integrates geometry and optics to solve the problem effectively.
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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
 
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I suspect they are looking for what angle from straight up will the ray from your eye hit the surface and be reflected back into the water (total internal reflection)
 
So that deals with critical angle then?
sin(theta)c = n1/n2

where n1 = lesser indice
n2 = greater indice

Hows does that relate to radius of a circle??
 
ReMa said:
So that deals with critical angle then?
sin(theta)c = n1/n2

where n1 = lesser indice
n2 = greater indice

Hows does that relate to radius of a circle??

You can only move your head so far until the only light reaching your eye is light TIRing from the swimming pool's bottom.

Find what angle you can tilt your head until this happens and find the corresponding horizontal distance moved from a initial position of looking straight up.
 
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