A person is standing at the edge of a pool 2.1 m deep. The pool is filled with w

Question

A person is standing at the edge of a pool 2.1 m deep. The pool is filled with water of index of refraction n = 1.33. Light traveling from an object at the bottom of the pool emerges from the water and travels to the person's eye. The light traveling in the air to the person's eye makes a ? = 45o angle with the normal to the surface of the water. (See the figure.) The person's eyes are 2.1 m above the ground. (so y1 = y2 = 2.1 m)

1) Find the distance x of the object from the edge of the pool.

x =

A person is standing at the edge of a pool 2.1 m deep. The pool is filled with water of index of refraction n = 1.33. Light traveling from an object at the bottom of the pool emerges from the water and travels to the person's eye. The light traveling in the air to the person's eye makes a ? = 45o angle with the normal to the surface of the water. (See the figure.) The person's eyes are 2.1 m above the ground. (so y1 = y2 = 2.1 m) 1) Find the distance x of the object from the edge of the pool. x =

Explanation / Answer

since angle is 45 the distance of the person from the point of ray striking the water is 2.1 m

let the distance be x1 = 2.1m

theta1 = 45

using snells law

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

1*sin45 = 1.33*sin(theta2)

0.531 = sin(theta2)

theta2 = 32.1176 degrees

the distance from the normal to the ray striking the bottom is x2

tan(theta2) = x2/y1

tan(32.1176) = x2/2.1

x2 = 1.3182 m

total distance from the perso x = x1+x2

x = 2.1 + 1.31822 = 3.4182 m

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