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## Homework Statement

In the figure attached, a 2.00m long vertical pole extends from the bottom of a swimming pool to a point 50.0cm above the water. Sunlight is incident at 55.0° above the horizon. What is the length of the shadow of the pole on the level bottom of the pool?

## Homework Equations

See figure attached for my drawing.

n1 = 1 for air

n2 = 1.33 for water

sin(θ2) = (n1/n2)*sin(θ1)

## The Attempt at a Solution

Since N2 > N1, we know that the beam bends towards the normal.

θ1 = 90° - 55° = 35°

θ2 = arcsin[(1/1.33)*sin(35°)) = 25.55°

From here, it's geometry:

Tan(55°) = X1 / 50cm

X1 = 0.71m

Tan(25.55°) = X2/1.5m

X2 = 0.71m

X = X1 + X2 = 1.43m

According to the back of the book, this is not correct. Where am I going wrong? This seems like a really straight forward problem. Thanks in advance, MrMoose