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# Answer to Question #126476 in Mechanics | Relativity for Stefanie Hernandez-Mendez

Question #126476
A flashlight is held at the edge of a swimming pool at a height h = 2.5 m such that its beam makes an angle of θ = 38 degrees with respect to the water's surface. The pool is d = 3.75 m deep and the index of refraction for air and water are n1 = 1 and n2 = 1.33, respectively.

What is the horizontal distance, D, from the edge of the pool to the point on the bottom of the pool where the light strikes? Write your answer in m.
1
Expert's answer
2020-07-23T08:46:25-0400

Firs, let's find the distance "CE = BO". From the triangle "\\triangle AOB" expressing the side "BO", obtain:

"BO = \\dfrac{AB}{\\tan\\theta} =h\\cot\\theta = CE"

Now let's find "EF". According to the Snell's law:

"n_1\\sin\\theta_1 = n_2\\sin\\theta_2\\\\\n\\sin\\theta_2 = \\dfrac{n_1\\sin\\theta_1}{n_2}\\\\\n\\theta_2 = \\arcsin\\left(\\dfrac{n_1\\sin\\theta_1}{n_2}\\right)"

From the triangle "\\triangle ODF":

"EF = OE\\tan\\theta_2 = d\\tan\\theta_2 = d\\tan\\left(\\arcsin\\left(\\dfrac{n_1\\sin\\theta_1}{n_2}\\right) \\right)"

Finally:

"D = CE + EF = h\\cot\\theta + d\\tan\\left(\\arcsin\\left(\\dfrac{n_1\\sin\\theta_1}{n_2}\\right) \\right)"

Obtain the following number:

"D = 2.5\\cot38\\degree + 3.75\\tan\\left(\\arcsin\\left(\\dfrac{1\\cdot\\sin52\\degree}{1.33}\\right) \\right) \\approx 5.83m"

Answer. 5.83 m.

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