Sight and Waves Part 2
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Transcript of Sight and Waves Part 2
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Sight and Waves Part 2
Problem Solving
Mr. KlapholzShaker Heights
High School
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Problem 1What is the angular position of the first minimum (in degrees and in radians) when light of wavelength 500 nm is diffracted through a single slit of width 0.05 mm?
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Solution 1 (slide 1 of 2) • Let’s start with the full equation [b sinq = l] and
then see if the small-angle approximation [bq= l] would have been good enough.
b sinq = l (0.05 x 10-3 m) sinq = 500 x 10-9 m
sinq = 500x10-9 ÷ 0.05x10-3
Please practice typing this calculation.sinq = 0.01q = sin-1(0.01)
q = 0.6˚q = 0.6˚ { 2p rad / 360˚ } = 0.01 radians
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Solution 1 (slide 2 of 2) • And now the small-angle approximation…
b q = l (0.05 x 10-3 m) q = 500 x 10-9 m
q = 500x10-9 ÷ 0.05x10-3
q = 0.01 radiansThe small-angle approximation gave the same
answer that the full equation gave.
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Problem 2You are managing a spy satellite, and you want to know if a person is having a conversation in a park with another person. Your satellite is orbiting 180 km above the surface of the earth. The camera lens has a diameter of 45.0 cm. If the light has a wavelength of 500 nm, can you tell if it is one person or two people?
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Solution 2 q = 1.22 l / b
q = 1.22 (500 x 10-9) ÷ (0.450 m) q = 1.36 x 10-6 radians
How big a distance is this?For q in radians, q = arclength / radius.
q = s / rs = r q
s = (180 x 103 m) × (1.36 x 10-6 rad) s = 0.24 m
You can see that it is two people because they will be more than 0.24 meters apart.
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Problem 3Warm, life-giving, sunlight with an intensity of 6.0 W m-2, is incident on two polarizing filters. The light goes through one filter, and then goes through the other. The transmission axes of the filters differ by 60.0˚. What is the intensity of the light that emerges from the second filter?
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Solution 3
The light that emerges from the first filter has half the intensity of the light that came in.
So we’re down to 3.0 W m-2.Now we use Malus’s law to finish:
I = IO cos2 q
I = (3.0 W m-2) cos2 60.0˚Please perform this calculation on your own.
I = 0.75 W m-2
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Problem 4 …
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Problem 4At what angle of reflection off of water does light get completely polarized? The index of refraction of water is 1.33.
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Solution 4
Light will reflect off of water at many angles, but only one angle produces pure polarized
light: the Brewster Angle.tan qB = n
tan qB = 1.33
qB = tan-1(1.33)
qB = 53.1˚
This angle is measured from a line perpendicular to the surface. So, the angle is
about 37˚ off of the surface...
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Where is the 53˚ angle?
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Tonight’s HW:
Go through the Sight and Waves section in your textbook and scrutinize the
“Example Questions” and solutions.Bring in your questions to tomorrow’s
class.