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

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 [b sin  = ] and then see if the small-angle approximation [b  = ] would have been good enough. b sin  = (0.05 x 10 -3 m) sin  = 500 x 10 -9 m sin  = 500x10 -9 ÷ 0.05x10 -3 Please practice typing this calculation. sin  = 0.01  = sin -1 (0.01)  = 0.6˚  = 0.6˚ { 2  rad / 360˚ } = 0.01 radians

4. Solution 1 (slide 2 of 2) And now the small-angle approximation… b  = (0.05 x 10 -3 m)  = 500 x 10 -9 m  = 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  = 1.22 / b  = 1.22 (500 x 10 -9 ) ÷ (0.450 m)  = 1.36 x 10 -6 radians How big a distance is this? For  in radians,  = arclength / radius.  = s / r s = r  s = (180 x 10 3 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.

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 = I O cos 2  I = (3.0 W m -2 ) cos 2 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  B = n tan  B = 1.33  B = tan -1 (1.33)  B = 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.