Problem 1 A pole stands vertically in a lake, projecting above the water surface

Question

Problem 1

A pole stands vertically in a lake, projecting above the water surface by 2.18 m. The Sun is at an angle of 32.4° above the horizon. If the water is 3.23 m deep, what is the length of the shadow cast by the pole on the flat lake bottom?

Got 8.5 meters and its wrong

Problem 2

A glass prism with an index of refraction of 1.51 sits on a horizontal surface, as shown. A ray of light, traveling horizontally, enters the prism on its vertical face, as shown. What is the maximum angle so that there is total internal reflection at the slanted face?

Explanation / Answer

1)

This is a fairly simple problem to solve, but hard to explain without a diagram, but I'll do my best without having one.

The top of the pole, the point on the pole at the water surface, and the point on the water surface where the sun casts a shadow of the top of the pole comprise a right triangle. The side opposite the triangle angle of 32.4 deg is the top 2.18m of the pole. The adjacent side to this angle is the distance along the surface of the water from the pole to the top of the pole's shadow on the water surface. This is the first unknown we need to find. Let's call it A:

A = 2.18m / tan32.4 = 3.44m

So the shadow of the top of the pole hits the water surface 3.44m away from the pole at an angle from the vertical = 90deg – 32.4deg = 57.6deg

Now we can use Snell's law to find how the shadow of the top of the pole is refracted under the water.

Snell's law says (refractive index of medium 1) * (sine of angle in medium 1)

= (refractive index of medium 2) * (sine of angle in medium 2)

Medium 1 is air, with refractive index = 1.00

Medium 2 is water, with refractive index = 1.33

So, the next unknown we need to find is the angle in medium 2. Let's call it B:

1 * sin(57.6deg) = 1.33 * sin(B)

Solving for B: B = arcsin(sin(57.6deg)/1.33) = 39.4deg

Now we have another triangle, with one angle = 39.4deg, the adjacent side = depth of water = 3.23m, and an unknown opposite side we'll call C:

C = 3.23m * tan(39.4deg) = 2.65m

The length of the pole's shadow at the bottom of the water = A + C

= 3.44m + 2.65m = 6.09m

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