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Light refraction problem in water & air

  • Thread starter MrMoose
  • Start date
  • #1
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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
 

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Answers and Replies

  • #2
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10,277
From here, it's geometry:

Tan(55°) = X1 / 50cm
This is not the correct angle.
 
  • #3
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Hi mfb, please elaborate, I still don't see the error. Tangent of the angle is equal to the length of the opposite side over the length of the adjacent side.
 
  • #4
34,234
10,277
The opposite side of your 55°-angle is the .5m of the pole, and not the length of the shadow. If you want to calculate this part in the same way as you did the part under water, you need the corresponding angle (relative to the vertical, 35°).
 
  • #5
23
0
Oh wow, that was a silly mistake. So...

Tan(55°) = 50cm/X1

X1 = 0.35m

And X = X1 + X2 = 1.07m, which is the correct answer.

Thanks so much for your help.
 

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