1. Phy2005 Applied Physics II Spring 2016 Announcements: Test 2 Wednesday, March 23 2 practice tests posted on course Tests page Review session in class March 21 in class + March 21 6pm NPB 2205 PH travel next week APS Spring meeting; Prof. Lee subs

2. Science news page Link to NYT article “Dino pet” “My Dinosaur’s Jet Lag Helps Explain Why a Time Change Is Hard”

3. Last time E B EM wave

4. Last time All electromagnetic waves such as light transmit energy. Light intensity = power flowing through the area / area (power density) [w/m 2 ]  R A  = A/R 2 [steradian] Total solid angle:  = 4  R 2 /R 2 = 4  (sr) Solid angle Intensity

5. 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 a incandescent light bulb, only 10% is luminous flux.

6. 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)

7. 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

8. Shadows

9. Eclipses

10. The eye

15. Q1 The intensity of the sun's radiation at the position of the earth is 1340 W/m 2. What is the total power put out by the sun? The Earth-to-Sun distance is 1.5 x 10 11 m. I=1340 W/m 2 = P/(4  R 2 )  P = (1340)(4  (1.5 x 10 11 ) 2 ) = 3.8×10 26 W

16. ACADEMIC HONESTY Each student is expected to hold himself/herself to a high standard of academic honesty. Under the UF academic honesty policy. Violations of this policy will be dealt with severely. There will be no warnings or exceptions.UF academic honesty policy

17. Q2 The intensity of the sun's radiation at the position of the earth is 1340 W/m 2. The Earth-to-Sun distance is 1.5 x 10 11 m. What is the intensity of light falling on Saturn? The Saturn-to-Sun distance is 1.4 x 10 12 m. RSRS RERE 1)87 W/m 2 2)15 W/m 2 3) 143 W/m 2 4) 87  W/m 2 5) 15/  W/m 2

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

20. For unpolarized light, a polaroid sheet reduces its intensity to half (sunglasses). θ I = I 0 cos 2 θ Intensity after polaroid Intensity before polaroid

22. Long wavelength Short wavelength

23. x Ax B Starting from A, make one touch at the wall and finish at B. Which path takes the shortest time? θ

24. David Hasselhoff Jeremy Jackson Pamela Anderson Who has the better chance to save this drowning guy? Pam Anderson

25. Reflection and Mirrors ii rr  i =  r Law of reflection Specular ReflectionDiffuse Reflection

26. When we talk about an image, start from an ideal point light source. Every object can be constructed as a collection of point light sources. p q IMAGE Image forms at the point where the light rays converge. When real light rays converge  Real Image When imaginary extension of L.R. converge  Virtual Image VIRTUAL Only real image can be viewed on screen placed at the spot.

27. p q IMAGE VIRTUAL For plane mirror: p = q How about left-right? Let’s check?