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## 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

[tex] n_{{1}}\sin \left( \theta_{{1}} \right) =n_{{2}}\sin \left( \theta_{{2

}} \right) [/tex]

## 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?

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