A submarine is 2.81 10 2 m horizontally from shore and 1.00 10 2 m beneath the s

Question

A submarine is 2.81 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the submarine so that the beam strikes the surface of the water 2.00 102 m from the shore. A building stands on the shore, and the laser beam hits a target at the top of the building. The goal is to find the height of the target above sea level. (The index of refraction of water is n = 1.333.)

(b) Find the angle of incidence of the beam striking the water–air interface. (Give your answer to at least one decimal place.)

(c) Find the angle of refraction. (Give your answer to at least one decimal place.)

(d) What angle does the refracted beam make with respect to the horizontal? (Give your answer to at least one decimal place.)

(e) Find the height of the target above sea level.

Explanation / Answer

angle of incidence is the angle formed by ray with normal

angle = tan^-1(horizontal distance / vertical distance)

horizontal distance = 2.81 * 10^2 - 2 * 10^2

horizontal distance = 81 m

vertical distance = 1 * 10^2

angle = tan^-1(81 / (1 * 10^2)

incidence angle = 39.01 degree

by snell's law

n1 * sin(theta1) = n2 * sin(theta2)

1.333 * sin(39.01) = 1 * sin(theta2)

angle of refraction theta2 = 57.04 degree

angle refracted beam will make with horizontal = 90 - 57.04

angle refractef beam will make with horizontal = 32.96 degree

tan(32.96) = height / horizontal distance

tan(32.96) = height / (2 * 10^2)

height = 129.683 m

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