1. Announcement Exam #2 grades are posted on E-learning Exam #2 grades are posted on E-learning Exam #2 average is 6.6 out of 10 (66%) Exam #2 average is 6.6 out of 10 (66%) Master copy of exam #2 (correct answer 1) posted on E-learning Master copy of exam #2 (correct answer 1) posted on E-learning

2. Chapter 25 Light Waves

3. Maxwell’s Rainbow Electromagnetic waves Electromagnetic waves Wave of EM fieldsWave of EM fields Frequency f Frequency f WavelengthWavelength Speed c in vacuumSpeed c in vacuum EM spectrum EM spectrum Radio wavesRadio waves InfraredInfrared VisibleVisible UltravioletUltraviolet X-raysX-rays Gamma raysGamma rays

4. EM Spectrum

5. C = f  C: speed of light C = 3 x 10 8 m/s in vacuum In other medium, the speed of light is smaller than in vacuum. EM wave can travel in the absence of medium.

6. http://laxmi.nuc.ucla.edu:8248/M248_99/iphysics/spectrum.gif nano = 10 -9 3 x 10 8 = f

7. Human eye is sensitive 380 < < 700 nm  < 300 nm (UV) is absorbed mostly by cornea and > 1200 nm by aqueous humor. Green cell (most sensitive for 520 nm), and 2 other types of cells sensitive to shorter and longer wavelength sense light and send signal to brain.

8. Q. What frequency should a radio station have if its waves are to have a wavelength of 200m?

9. Light Intensity All electromagnetic waves such as light transmit energy. Light intensity = power flowing through the area / area (power density) [w/m 2 ] Q. A laser sends out a beam 3 mW power. The diameter of the beam is 3 mm. What is the intensity of the laser beam in w/m 2 ? Intensity = (0.003 w)/( x 0.0015 2 m 2 ) = 424.4 w/m 2 My laser point produces about 5 mW power!!!

10. 1 lumen (lm) = light power of 1/683 W in yellow (lighting industry) When a light bulb is on, it radiates out broad spectrum of EM wave. The total radiant energy emitted by the lamp per unit time is called radiant flux. Only a small portion of this is in visible range (400 – 700 nm). The portion of radiant energy in visible range is called luminous flux (F). ( in textbook) In a incandescent light bulb, only 10% is luminous flux.

11. Solid Angle 2 Dimension  R s = s/R [radian]  R A  = A/R 2 [steradian] Total solid angle:  = 2R/R = 2(rad) Total solid angle:  = 4R 2 /R 2 = 4(sr)

12. Q. What solid angle is subtended at the center of 6 m diameter sphere by a 2 m 2 area on its surface?  = A/R 2 = (2 m 2 )/(3 m) 2 = 0.22 sr R1 R2  Luminous intensity: I = F/  (luminous flux per unit solid angle) [lm/sr] = candela (cd)

13. Q. A spotlight is equipped with a 30 cd bulb. This light beams onto a vertical wall in a circular shape with 3 m 2 area. The spotlight is 10 m from the wall. Calculate the luminous intensity for the spotlight. 3 m 2 10 m Total luminous flux coming out of the bulb, F = I 4 = (30 cd)(4 sr) = 377 lm This amount of luminous flux is concentrated on the 3 m 2 spot. To get the luminous intensity, we need to know the solid angle that bright spot subtend.  = A/R 2 = (3 m 2 )/(10 m) 2 = 0.03 sr I (spot) = (377 lm)/(.03 sr) = 1.26 x 10 4 cd

14. Illuminance E = F/A Luminous flux per unit area [lm/m 2 ] = Lux (lx) R1 R2  1 2 I 1 = I 2 However, E 1 ≠ E 2 E 1 = F/A 1 = F/(R 1 2 ), E 2 = F/A 2 = F/(R 2 2 )  = A/R 2

15. Relation between I (luminous intensity) and E (illuminance) I = F/  E = F/A  = A/R 2 E = I/R 2  A  = Acos  /R 2 I = F/  = F/(Acos  /R 2 ) E = F/A E = Icos  /R 2

16. Q. A 300 cd lamp is located 3 m from a surface with 0.25 m 2 area. The luminous flux makes an angle of 30 deg with the normal to the surface. (a) What is the illuminance? (b) what is the luminous flux shining on the surface?   Flux thru the surface: f F= I = A/R2 = (0.25)cos30/3 2  = 0.024 sr F = (300 cd)(0.024 sr) = 7.2 lm E = F/A = (7.2 lm)/(0.25 m 2 ) = 28.8 lx

17. http://atom.physics.calpoly.edu/Graphics/em_surface.gif

18. - + E B Polarization For this type of EM wave, E field oscillates in a fixed direction, and so does B field. Linearly polarized!! x y z Direction of E-field determines polarization direction: This wave is polarized in x-direction.

19. Natural light is “unpolarized”. E We can produce polarized light from unpolarized light!! Polaroid sheet allows light pass through inly if the E-field vector is in a specific direction.

21. For unpolarized light, a polaroid sheet reduces its intensity to half (sunglasses).