Help with critical angle calculation

[shar_p]

Homework Statement

A penny sits at the bottom of a pool of water (n=1.33) at a depth of 3.0m. If the observer 1.8m tall stands 30cm away from the ledge, how close to the side can the penny be and still be visible to the observer. Suppose there is another penny 10 times farther away than the 1st one, will a light ray going from this new penny to the top edge of the pool emerge from the water ? And if yes, what is the angle made by the light ray?

Homework Equations

To still be visible, the angle should be less than critical angle, is that correct? but I can't figure this out .. please help

The Attempt at a Solution

real depth = 3.0 so apparent depth can be calculated to be 3/1.33 = 2.25 m.
I can also calculate the critical angle with Snell's sin(theta) = 1.33 (since sin 90 = 1). So theta = 48.75 . Which means that the angle has to be less that 48.75 for the penny to be visible.
Now how do I get the distance from this?

Please help I have a test on Wednesday. Thanks

 

[Carid]

Why not turn the problem around? Imagine a light ray coming from the observer's eye, grazing the edge of the pool, and going down into the water. Where would it reach the bottom?

Now you've got that distance sorted out you'll be able to attack the next part of the problem.