- #1
RoseCrye
- 16
- 0
there's a light source, S, sitting 2.60 metres below the surface of a pool 1.00 metres away from one side. I'm supposed to find the angle at which the light left the water and the difference between the apparent and actual depths of the light source.
i attempted using Snell's Law with n of water = 1.33 and the incedent angle = 90 degrees. 1.33(sin90)=1.00(sin_theta) I got theta=1.33. that can't be right.
I also tried trig. Theta is the angle of a rt triangle with the light source. 2.60 is the opposite, 1.00 is adjacent. Using the cot_theta, my answer was 69 degrees. neither of these answers was accepted by WebAssign.
HELP!
i attempted using Snell's Law with n of water = 1.33 and the incedent angle = 90 degrees. 1.33(sin90)=1.00(sin_theta) I got theta=1.33. that can't be right.
I also tried trig. Theta is the angle of a rt triangle with the light source. 2.60 is the opposite, 1.00 is adjacent. Using the cot_theta, my answer was 69 degrees. neither of these answers was accepted by WebAssign.
HELP!