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Sight and Waves Part 2

Mr. Klapholz Shaker Heights High School. Sight and Waves Part 2. Problem Solving. Problem 1. What 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?. Solution 1 (slide 1 of 2).

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Sight and Waves Part 2

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  1. Mr. Klapholz Shaker Heights High School Sight and Waves Part 2 Problem Solving

  2. Problem 1 What 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?

  3. Solution 1 (slide 1 of 2) • Let’s start with the full equation [bsinq = l] and then see if the small-angle approximation [bq= l] would have been good enough. bsinq = l (0.05 x 10-3m) sinq = 500 x 10-9m sinq = 500x10-9 ÷ 0.05x10-3 Please practice typing this calculation. sinq = 0.01 • = sin-1(0.01) • = 0.6˚ • = 0.6˚ { 2prad / 360˚ } = 0.01 radians

  4. Solution 1 (slide 2 of 2) • And now the small-angle approximation… bq = l (0.05 x 10-3m) q = 500 x 10-9m • = 500x10-9 ÷ 0.05x10-3 • = 0.01 radians The small-angle approximation gave the same answer that the full equation gave.

  5. Problem 2 You 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?

  6. Solution 2 q= 1.22 l / b • = 1.22 (500 x 10-9) ÷ (0.450 m) • = 1.36 x 10-6 radians How big a distance is this? For q in radians, q = arclength / radius. • = s / r s = rq s = (180 x 103m) × (1.36 x 10-6rad) s = 0.24 m You can see that it is two people because they will be more than 0.24 meters apart.

  7. Problem 3 Warm, 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?

  8. 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 cos2q I = (3.0 W m-2) cos2 60.0˚ Please perform this calculation on your own. I = 0.75 W m-2

  9. Problem 4 …

  10. Problem 4 At what angle of reflection off of water does light get completely polarized? The index of refraction of water is 1.33.

  11. 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...

  12. Where is the 53˚ angle?

  13. 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.

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