A coin is lying at the bottom of a pool of water that is 6.5 feetdeep. View from

Question

A coin is lying at the bottom of a pool of water that is 6.5 feetdeep. View from directly above the coin, how far below thesurface of the water does the coin appear to be? ((The coinis assumed to be small in diameter; therefore, we can use the smallangle approximation sin theta = tan theta = theta.)
Question: The solution to this problem does not make sense. Physics (3rd) by James Walker in Step 3 and 4 it states thatyou find the angular value finding theta 2 / theta 1 but why don'tthey do theta 1 / theta 2. And in Step 4 the equation doesnot make sense. Please explain.
Question: The solution to this problem does not make sense. Physics (3rd) by James Walker in Step 3 and 4 it states thatyou find the angular value finding theta 2 / theta 1 but why don'tthey do theta 1 / theta 2. And in Step 4 the equation doesnot make sense. Please explain.

Explanation / Answer

refractive index of water=1.33
refractive index=real depth of the coin / apparent depth
so apparent depth=real depth/refractiveindex=6.5/1.33=4.8872feet.
so the coin appears to be at a depth of 1.44feet from thetop.

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