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.

 

[thomate1]

To calculate the critical angle, you can use the formula sin(theta) = 1/n, where n is the refractive index of the medium (in this case, water with a refractive index of 1.33). This will give you a critical angle of approximately 48.75 degrees, as you have correctly calculated.

To determine the distance from the edge of the pool that the penny can be and still be visible, you can use the trigonometric relationship tan(theta) = opposite/adjacent. In this case, the adjacent side is the distance from the edge of the pool to the observer (30cm), and the opposite side is the distance from the edge of the pool to the penny (x). So, you can rearrange the equation to solve for x: x = tan(theta)*30cm. Plugging in the critical angle of 48.75 degrees, you get x = tan(48.75)*30cm = 34.28cm. This means that the penny can be up to 34.28cm away from the edge of the pool and still be visible to the observer.

As for the second part of the question, yes, the light ray from the second penny will emerge from the water. This is because the light ray will be traveling at an angle less than the critical angle, so it will not undergo total internal reflection and will emerge from the water. To calculate the angle made by the light ray, you can use the same formula as before, but with the new distance from the edge of the pool (10 times farther away than the first penny). So, the angle made by the light ray will be tan^-1(10*tan(48.75)) = 84.96 degrees.

I hope this helps! Good luck on your test.