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One bright summer day a swimmer is floating with her head just above the surface of a large, clear lake where the water has index of refraction n = 1.43. She comes upon a tall wooden post emerging from the water. When she looks into the water at just the right angle she can see the bottom of the post, which is at a depth h = 5.50 m below the water's surface. As she floats away from the post, eventually she can no longer see the bottom of it. At what distance d from the post does this occur? (in m)

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boffeemadrid

Answer:

5.38035 m

Explanation:

[tex]n_w[/tex] = Refactive index of water = 1.43

h = Depth = 5.5 m

Critical angle is given by

[tex]\theta_c=sin^{-1}\dfrac{1}{n_w}\\\Rightarrow \theta_c=sin^{-1}\dfrac{1}{1.43}\\\Rightarrow \theta_c=44.37^{\circ}[/tex]

d = horizontal distance from the post where she no longer see the bottom of wooden post

So,

[tex]tan\theta_c=\dfrac{d}{h}\\\Rightarrow d=tan\theta_c\times h\\\Rightarrow d=tan44.37\times 5.5\\\Rightarrow d=5.38035\ m[/tex]

The distance d is 5.38035 m

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