
#1
Apr2108, 05:32 PM

P: 66

1. The problem statement, all variables and given/known data
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. 2352). Where does the spot of light hit the bottom of the pool, relative to the edge, if the pool is 2.1 m deep? 2. Relevant equations [tex] n_{{1}}\sin \left( \theta_{{1}} \right) =n_{{2}}\sin \left( \theta_{{2 }} \right) [/tex] 3. 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. Im using 1 as the index of refraction of air and 1.33 for water, giving me [tex]0.7857161968= 1.33\,\sin \left( \theta_{{2}} \right) [/tex]. Solving this gives me [tex]\theta_{{2}}= 41.561[/tex] 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? 



#2
Apr2108, 05:41 PM

P: 66

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....



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