*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **Physics 212
Lecture 23 Physics 212 Lecture 23, Slide 1 Plane Waves from Last Time E and B are perpendicular and in phase Oscillate in time and space Direction of propagation given by E X B E0 = cB0 Argument of sin/cos gives direction of propagation
Physics 212 Lecture 23, Slide 2 Understanding the speed and direction of the wave
Ex = Eosin(kzt)
Ex z t=0 Ex sin( kz - ) = - cos( kz)
2 z t = /2 What has happened to the wave form in this time interval? It has MOVED TO THE RIGHT by /4 /4 speed = c = = = f / 2 2
Physics 212 Lecture 23, Slide 3 Checkpoint 1a No moving in the minus z direction No has Ey rather than Ex Physics 212 Lecture 23, Slide 4 Checkpoint 2a c=3.0 x 108 m/s Wavelength is equal to the speed of light divided by the frequency. c 300, 000, 000 1 = = = f 900, 000, 000 3
Check: Look at size of antenna on base unit
Physics 212 Lecture 23, Slide 5 Doppler Shift The Big Idea As source approaches: Wavelength decreases Frequency Increases Physics 212 Lecture 23, Slide 6 Doppler Shift for em Waves
What's Different from Sound or Water Waves ?
Sound /Water Waves : You can calculate (no relativity needed) BUT Result is somewhat complicated: is source or observer moving wrt medium? Electromagnetic Waves : You need relativity (time dilation) to calculate BUT Result is simple: only depends on relative motion of source & observer 1+ f= f 1- 1 2 = v/c > 0 if source & observer are approaching < 0 if source & observer are separating
Physics 212 Lecture 23, Slide 7 Doppler Shift for em Waves f' f
Source approaching observer v
or f f' Observer approaching source v The Doppler Shift is the SAME for both cases ! f'/f ONLY DEPENDS ON THE RELATIVE VELOCITY 1+ f= f 1- 1 2 Physics 212 Lecture 23, Slide 8 Red Shift Wavelengths shifted higher wavelength Frequencies shifted lower Star separating from us (Expanding Universe) Our Sun Star in a distant galaxy Physics 212 Lecture 23, Slide 9 Example Police radars get twice the effect since the EM waves make a round trip: f f ( 1 + 2 )
If f = 24,000,000,000 Hz (k-band radar gun) c = 300,000,000 m/s v 30 m/s (67 mph) 31 m/s (69 mph) 1.000 x 10-7 1.033 x 10-7 f' 24,000,004,800 24,000,004,959 f'-f 4800 Hz 4959 Hz
Physics 212 Lecture 23, Slide 10 Checkpoint 2b
A) B) C) ficlicker = 900 MHz Need to approach i>clicker Need to shift frequency UP How fast would you need to run to see the i>clicker radiation?
f 1014 1+ = 9 = 105 = f 10 1- 1/ 2 ( > 0) 1 + 10 = 1- 10 1010 - 1 1 - 10-10 = 10 = 10 + 1 1 + 10-10 Approximation Exercise: 1 - (2 10-10 )
Physics 212 Lecture 23, Slide 11 Doppler Shift for em Waves
A Note on Approximations
1 2 1+ f= f 1- << 1 f f ( 1+ )
Remember > 0 for approach < 0 for separation around = 0 WHY ?? Taylor Series: Expand
1+ F ( ) = 1- 1/ 2 F ( ) = F ( 0) + F (0) F (0) 2 + + ... 1! 2! F (0) = 1 Evaluate: F ( ) 1 + NOTE: F ( ) = (1 )n
F ( ) 1 n
Physics 212 Lecture 23, Slide 12 F (0) = 1 Waves Carry Energy Physics 212 Lecture 23, Slide 13 Intensity
Intensity = Average energy delivered per unit time, per unit area 1 dU I A dt Length = c dt Area = A dU = u volume = u Acdt I =c u
Sunlight on Earth: I ~ 1000J/s/m2 ~ 1 kW/m2
Physics 212 Lecture 23, Slide 14 Waves Carry Energy Physics 212 Lecture 23, Slide 15 Comment on Poynting Vector
Just another way to keep track of all this Its magnitude is equal to I Its direction is the direction of propagation of the wave Physics 212 Lecture 23, Slide 16 Light has Momentum!
Momentum of light is E/c light can exert pressure!!
Analogy from mechanics: p2 E= 2m d E 2 p dp mv dp = = dt 2m dt m dt = vF For EM waves: dE dU = IA dt dt vc
IA = cF
I F = c A I P= c
Radiation pressure pressure
Physics 212 Lecture 23, Slide 17 Checkpoint 1b
Which of the following actions will increase the energy carried by an electromagnetic wave? A. Increase E keeping constant C. Both of the above will increase the energy B. Increase keeping E constant D. Neither of the above will increase the energy Is that all there is to it? WHAT ABOUT PHOTONS? Physics 212 Lecture 23, Slide 18 PHOTONS
We believe the energy in an em wave is carried by photons Question: What are Photons? Answer: Photons are the "quanta" of the electromagnetic field. Photons possess both wave and particle properties Particle: Energy and Momentum localized Wave: They have definite frequency & wavelength (f = c) Connections seen in equations: E = hf p = h/ Planck's constant h = 6.63e34 Js Question: How can something be both a particle and a wave? Answer: It can't (when we observe it) What we see depends on how we choose to measure it ! The mystery of quantum mechanics: More on this in PHYS 214
Physics 212 Lecture 23, Slide 19 Calculation 1
An electromagnetic wave is described by: ^ where is the unit vector in the +y direction. j y E = ^ 0 cos(kz - t ) jE x z Which of the following graphs represents the z-dependence of B x at t = 0? X (A) (B) (C) (D) E = ^ 0 cos(kz - t ) jE
E and B are "in phase" (or 180o out of phase) X Wave moves in +z direction y E E B points in direction of propagation x B ^ B = -iB0 cos(kz - t )
z
Physics 212 Lecture 23, Slide 20 Calculation 2
An electromagnetic wave is described by: What is the form of B for this wave?
^ j i+^ E= E0 cos(kz + t ) 2 y x z (A) ^ ^ (B) B = i - j ( E0 / c)cos(kz + t ) 2 i^ + ^ j E = E0 cos( kz + t ) 2 y E x B E B points in negative z-direction ^ j i+^ B= ( E0 / c )cos(kz + t ) 2 -i^ - ^ j (D) B = ( E0 / c )cos(kz + t ) 2 Wave moves in z direction +z points out of screen z points into screen ^ j -i + ^ ( E0 / c) cos(kz + t ) (C ) B = 2 Physics 212 Lecture 23, Slide 21 Calculation 3
An electromagnetic wave is described by: E = ^ 0 sin(kz + t ) jE Which of the following plots represents Bx(z) at time t = /2 ? (A) (B) (C ) (D)
Wave moves in negative zdirection y E B x B = i^( E0 / c)sin(kz + t ) Bx = ( E0 / c)sin(kz + / 2) Bx = ( E0 / c){sin kz cos( / 2) + cos kz sin( / 2)} Bx = ( E0 / c) cos(kz )
Physics 212 Lecture 23, Slide 22 +z points out of screen z points into screen a t t = /2 : E B points in negative z-direction A certain unnamed physics professor was arrested for running a stoplight. He said the light was green. A pedestian said it was red. The professor then said: "We are both being truthful; you just need to account for the Doppler effect !" Calculation Is it possible that the professor's argument is correct? ( green = 500 nm, red = 600 nm) (A) Y ES (B) NO As professor approaches stoplight, the frequency of its emitted light will be shifted UP The speed of light does not change Therefore, the wavelength (c/f) would be shifted DOWN If he goes fast enough, he could observe a green light ! Physics 212 Lecture 23, Slide 23 A certain unnamed physics professor was arrested for running a stoplight. He said the light was green. A pedestian said it was red. The professor then said: "We are both being truthful; you just need to account for the Doppler effect !" How fast would the professor have to go to see the light as green? ( green = 500 nm, red = 600 nm) (A) 5 4 0 m /s (B) 5 .4 X 1 0 4 m /s (C ) 5 .4 X 1 0 7 m /s (D) 5 .4 X 1 0 8 m /s Relativistic Doppler effect: f 600 1+ = = f 500 1- f= f 1+ 1- 36(1 - ) = 25(1 + )
f = f (1 + ) = Note approximation for small is not bad: c = 3 X 108 m/s v = 5.4 X 107 m/s 11 = 0.18 61 1 = = 0.2 5 Change the charge to SPEEDING!
Physics 212 Lecture 23, Slide 24 ...

View
Full
Document