| Never enter units in your answers. Be
sure to use the units indicated on the HW submission page.
Enter numbers without commas. 23,546 should be 23546 Always enter a * for multiplication. g(H+h) should be g*(H+h). |
| You are now obtaining input data for CID = |
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Submitted Answers will be graded W 10/22, Th 10/23, F 10/24, and M 10/27 at 12:00 noon.
| Problem 6.1 | |||||||||||||
| What you should learn: For reflection, the angle of incidence equals the angle of reflection. | |||||||||||||
| Problem: Two flat mirrored surfaces are joined with an angle φ between them, as shown in the figure. Light is incident from the right at an an angle of 60° with respect to the normal Find the angle psi (ψ) that light leaves the mirror. Note that ψ is measured with respect to the x axis, as shown in the figure. |
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Variables that must be changed with each
submission:
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| Hints:
You have to be careful about angles, but this is just an exercise in geometry. |
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Results of previous submissions:
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| Answer Range: 70.0 - 135° Submit HW answers. |
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| Problem 6.2 | |||||||||||||
| What you should learn: Snell's law is used for refraction. Remember that the angles are measured with respect to the normal and that the n1 is the index of refraction where θ1 is the angle of the ray in medium 1, etc. Snell's Law is n1sinθ1 = n2sinθ2. | |||||||||||||
| Problem: A ray of light is incident at an angle (measured with respect to the normal) of theta1 on the surface of a pond. What is the angle of refraction in the water. You may call this angle theta2. | |||||||||||||
| Constants and fixed variables:
The index of refraction of water is n = 1.33 |
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Variables that must be changed with each
submission:
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Results of previous submissions:
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| Answer Range: 21.0 - 33.0° Submit HW answers. |
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| Problem 6.3 | |||||||||||||
| What you should learn: This is an application Snell's Law. | |||||||||||||
| Problem: A turtle's
head pokes above the surface of the water in an aquarium. Fred, looking
directly from in front of the aquarium, sees the head of the turtle
lined up with its body. He puts a purple dot
at the center of the turtle's head. Sally, who is off to the side, sees
the head of the turtle above the water as if it's not attached to the
turtle's neck. Sally has Fred place a red dot at the location of
the turtle's head and a blue dot at its neck. Sally is located a distance a in front of the aquarium. (Note that this is the "y distance" from the glass of the aquarium to Sally, as indicated in the figure.) The turtle is located a distance d behind the front of the aquarium. The distances between pairs of dots are b and c as labeled in the figure. Given a value for b, find the distance c. Ignore the thickness of the glass in this problem.
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| Constants and fixed variables:
Index of refraction of the water: n = 1.33 a = 70.0 cm b = 4.00 cm |
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Variables that must be changed with each
submission:
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| Hints:
I need help with the geometry. |
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Results of previous submissions:
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| Answer Range: 1.40 - 3.30 cm Submit HW answers. |
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| Problem 6.4 | |||||||||||||
| What you should learn: This is a problem involving total internal reflection. The critical angle is the smallest incident angle for which no refracted ray can be produced. This means that the refracted ray is at 90°. When the second medium is air with an index of refraction of 1, we have: n sin θc = 1 x sin 90° = 1. | |||||||||||||
| Problem: The critical angle for total internal reflection in a gemstone is measured to be thetac. What is the speed of light v in that material? | |||||||||||||
Variables that must be changed with each
submission:
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| Hints:
How is the index of refraction related to the speed of light? |
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Results of previous submissions:
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| Answer Range: 1.25 - 1.70
x 108
m/s Submit HW answers. |
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| Problem 6.5 | |||||||||||||
| What you should learn: This is just an application of total internal reflection. To find the critical angle, you can use Snell's Law with the angle of refraction equal to 90°, but with neither index of refraction equal to 1. | |||||||||||||
| Problem: Light travels 30 m in a vacuum along a straight path in about 0.10 μs. An optical fiber has a square cross section. The fiber material is surrounded with a cladding. Light is shined into the fiber so it travels at the critical angle for total internal reflection, reflecting from the top and bottom surfaces of the fiber. The indices of refraction for the fiber and the cladding (the material surrounding the fiber) are given below. How long does it take light to travel along 30.0 m of the fiber? | |||||||||||||
| Constants and fixed variables:
Index of refraction of the cladding: nc =1.24 |
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Variables that must be changed with each
submission:
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| Hints:
How do I account for the fact that the light in the fiber is bouncing around at an angle and not going straight down the fiber? |
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Results of previous submissions:
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| Answer Range: 0.180 - 0.240 μs Submit HW answers. |
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