Solve Snell's Law Problem: 2.1m Deep Pool

[negatifzeo]

Homework Statement

In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, h = 1.6 m above the water level, onto the surface of the water at a point L = 3.0 m from his foot at the edge of the pool (Fig. 23-52). Where does the spot of light hit the bottom of the pool, relative to the edge, if the pool is 2.1 m deep?

Homework Equations

$$n_{{1}}\sin \left( \theta_{{1}} \right) =n_{{2}}\sin \left( \theta_{{2 }} \right)$$

The Attempt at a Solution

This problem seems straightforward, but the answer I am getting is incorrect according to my online homework submissions. I am in degree mode on my calculator. :) First I used the pythag theorem to get the hypotenuse of the first triangle and the angle with the water, which is 28.07. I subtracted this from 90 to get the angle from the normal line. This gives me 61.928 degrees. I am using 1 as the index of refraction of air and 1.33 for water, giving me $$0.7857161968= 1.33\,\sin \left( \theta_{{2}} \right)$$.
Solving this gives me $$\theta_{{2}}= 41.561$$ The depth of the pool was 2.1m. Now I just used Tan(41.561)=L/2.1 andsolved for L to get 1.86, which shows as incorrect. Can anyone see where I went wrong, or is my entire approach off?

 

[negatifzeo]

Oops. Duh, I forgot to add the original L. L=4.86. That was a waste of time but I'll leave it here just in case someone wants to see...

 

[Moetasim]

Your approach seems correct, but there may be an error in your calculations. When solving for L using the equation Tan(41.561)=L/2.1, I am getting a value of 2.07, which is closer to the given depth of the pool (2.1m). Double check your calculations and make sure you are using the correct values for the index of refraction and angles. Also, make sure your calculator is in degree mode.