Physics 150 Electromagne5c waves Chapter 22 Electromagne5c waves around us Physics 150, Prof. M. Nikolic 2 Energy from the Sun – EM waves
Physics 150, Prof. M. Nikolic 3 History – Maxwell’s theory In 1865, James Clerk Maxwell developed a theory about electricity and magne5sm. His star5ng points were: 1.  Electric field lines originate on + charges and terminate on – charges. 2.  Magne5c field lines form closed loops. 3.  A 5me varying magne5c field induces an electric field 4.  A magne5c field is created by a current. Physics 150, Prof. M. Nikolic 4 Charges and fields summary è Sta5onary charges produce only electric fields. è Charges moving at constant veloci5es produce electric and magne5c fields. Charges that are accelerated will produce variable electric and magne5c fields. These are electromagne5c (EM) waves. If the charges oscillate with a frequency f, then the resul5ng EM wave will have a frequency f. Physics 150, Prof. M. Nikolic 5 Maxwell’s theory and equa5ons Maxwell’s theory is a mathematical formulation that relates electric and
magnetic phenomena. He combined 4 basic laws of electromagnetism:
1.  Gauss’s Law: -­‐ Electric fields (not induced) must begin on + charges and end on -­‐ charges. qenc
Φ E = EA cosθ =
ε0
2. Gauss’s Law for magne5sm: -­‐ There are no magne5c monopoles (a magnet must have at least one north and one south pole). Φ B = BA cosθ = 0
Physics 150, Prof. M. Nikolic 6 Maxwell’s theory and equa5ons 3. Faraday’s law: -­‐ A changing magne5c field creates an electric field. ΔΦ B
ε = −N
Δt
ΔB
Ed = −NA cosθ
Δt
3. Ampere-­‐Maxwell’s law: -­‐ A current or a changing electric field creates a magne5c field. $
ΔΦ E '
∑ B||Δl = µ0 &% Ienclosed + ε0 Δt )(
There are no electric or magne5c waves è there are only electromagne5c waves. Physics 150, Prof. M. Nikolic 7 Antennas If an ac voltage is applied to an antenna, the charges will be accelerated up and down and radiate EM waves. A short 5me later the electric field at P is s5ll downward, but now with a reduced magnitude. At t = 0 the electric field at point P is downward. Physics 150, Prof. M. Nikolic A]er one quarter of a cycle, at t = 1/4 T, the electric field at P vanishes. The decreasing electric field at point P creates a magne5c field at point Q poin5ng into the graph 8 Antennas Changing electric field produces magne5c field At a 5me when the electric field produced by the antenna points downward, the magne5c field points into the page. The B-­‐field of an EM wave is perpendicular to its E-­‐field and also to the direc5on of travel. Physics 150, Prof. M. Nikolic 9 Propaga5on of the EM waves Physics 150, Prof. M. Nikolic 10 The EM spectrum EM waves can be generated in different frequency bands: radio, microwave, infrared, visible, ultraviolet, x-­‐rays, gamma rays Physics 150, Prof. M. Nikolic 11 Visible light The human eye does not see all parts of the visible spectrum equally. It is most sensi5ve to colors around green part of the spectrum. Physics 150, Prof. M. Nikolic 12 Notes on EM spectrum Infrared radia-on (IR) • 
• 
• 
• 
Wavelengths longer than visible light Incorrectly called “heat waves” Produced by hot objects and molecules Readily absorbed by most materials Microwaves •  Part of EM spectrum lying between radio waves and IR •  Wavelengths from about 1 mm to 30 cm •  Well suited for radar systems •  Microwave ovens are an applica5on Physics 150, Prof. M. Nikolic 13 Notes on EM spectrum Radio waves •  Used in radio and television communica5ons Physics 150, Prof. M. Nikolic Ultraviolet light •  Covers about 400 nm to 0.6 nm •  Sun is an important source of uv light •  Most uv light from the sun is absorbed in the stratosphere by ozone 14 Notes on EM spectrum X rays •  Most common source is accelera5on of high-­‐energy electrons striking a metal target •  Used as a diagnos5c tool in medicine Gamma rays •  Emiked by radioac5ve nuclei •  Highly penetra5ng and cause serious damage when absorbed by living 5ssue •  Provide indirect evidence on gigan5c black holes Physics 150, Prof. M. Nikolic 15 Speed of EM waves -­‐ vacuum Maxwell managed to derive the speed of electromagne-c waves in vacuum 1
c=
= 3.00 ×108 m/s
ε0µ0
Indeed, c=
1
=
ε0µ0
1
(8.85 ×10
−12
C 2 /Nm 2 ) ( 4π ×10 −7 Tm/A )
= 3.00 ×108 m/s
Maxwell noted that this value was the same as the measured speed of light, so he proposed that light was an electromagne-c wave. Verified a]er his death by Heinrich Hertz. Physics 150, Prof. M. Nikolic 16 Speed of EM waves -­‐ vacuum Light is an electromagne-c wave! •  The speed of light in vacuum is c = 3x108 m/s According to Chapter 11, last semester: c= fλ
Wavelength – λ is the distance wave travels in order to make one period Units of wavelength: m Physics 150, Prof. M. Nikolic 17 Conceptual ques5on – EM waves Q1
Light having a certain frequency, wavelength and speed is traveling through empty space (vacuum). If the frequency of the light were doubled, then its new wavelength and speed would be (choose the number corresponding to the combina5on of wavelength A. 
B. 
C. 
D. 
c/2 and λ/2 2c and λ c and λ/2 c and 2λ You cannot change the speed of light in vacuum. It’s always c=3x108 m/s. Thus, wavelength has to be halved. è Longer frequency – shorter wavelength Physics 150, Prof. M. Nikolic 60 18 Exercise: Radio waves As you drive by an AM radio sta5on, you no5ce a sign saying that its antenna is 142 m high. If this height represents one quarter-­‐wavelength of its signal, what is the frequency of the sta5on? What’s given: h = 142 m h = ¼ λ c
f=
λ
c= fλ
3⋅108 m / s
f=
= 528⋅10 3 Hz
568m
= 528kHz
λ = 4⋅h
λ = 4 ⋅142 = 569m
How long would it take the radio wave to travel from the sta5on to your house 6 km away? d = ct
Physics 150, Prof. M. Nikolic d
t=
c
t=
6000m
= 20µ s
8
3⋅10 m / s
19 Speed of EM waves -­‐ maker When light travels though a material medium, its speed is less than c: c
v=
n
v is the speed of light in the medium n is the refrac-ve index of the medium When a wave passes from one medium to another the frequency stays the same, but the wavelength is changed. f=
Physics 150, Prof. M. Nikolic c
λvacuum
=
v
λmedium
λvacuum
λmedium =
n
20 Conceptual ques5on – speed of light Q2
If the speed of light through glass is 2x108 m/s, what is the index of refrac5on of glass? A. 
B. 
C. 
D. 
0.333 1 1.5 2 Physics 150, Prof. M. Nikolic c
v=
n
c 3⋅108 m / s
n= =
= 1.5
v 2 ⋅108 m / s
60 21 Proper5es of EM waves •  All EM waves in vacuum travel at the “speed of light” c. •  Both the electric and magne5c fields have the same oscilla5on frequency f. •  The electric and magne5c fields oscillate in phase (both reach a maximum and minimum at the same 5me). Emax = cBmax
Emax and Bmax are amplitudes (peaks) of the electric and magne5c field respec5vely •  The E and B fields are perpendicular to the direc5on of travel (transverse waves) Physics 150, Prof. M. Nikolic 22 Energy transport by EM waves The EM waves carry energy. The average energy density uave (energy per unit volume) in a region of empty space where electric and magne5c fields are present is U
1
1 2
1 2
2
2
uave =
= ε 0 Eave +
Bave = ε 0 Eave = Bave
Volume 2
2µ 0
µ0
In the case of sinusoidal electric and magne5c fields, we can use RMS values (pretend that you remember this from Chapter 21): Erms =
Emax
2
Brms =
Bmax
2
Physics 150, Prof. M. Nikolic 2
uave = ε 0 Erms
=
1 2
Brms
µ0
23 Intensity of EM waves It’s difficult to measure energy density, but we CAN measure intensity. Intensity is a measure of how bright the light is. Back in Chapter 11 we said that intensity is: Power P
I=
=
Area A
I is not the current here, it’s intensity! It turns out that, intensity is also: I = cuave = cε 0 E
Physics 150, Prof. M. Nikolic 2
rms
1
2
= cBrms
µ0
24 Exercise: Energy from the sun The intensity of the sunlight that reaches Earth’s upper atmosphere is 1400 W/m2. (a) What is the total average power output of the Sun on the Earth (at what rate does the sun radiate EM energy), assuming it to be an isotropic source? I=
Power P
=
Area A
P = IA
Area of an isotropic source is an area of a sphere whose radius is the distance between the source of EM waves and the final point. A = 4π R 2
R is the distance between Sun and Earth: R = 1.5x1011 m 2
A = 4π (1.50 ×10 m ) = 28.26 ×10 22 m 2
11
P = 1400 W/m 2 ⋅ 28.26 ×10 22 m 2 = 3.96 ×10 26 W
Physics 150, Prof. M. Nikolic 25 Exercise: Energy from the sun The intensity of the sunlight that reaches Earth’s upper atmosphere is 1400 W/m2. (b) What is the average energy density? I
uave =
c
I = cuave
1400 W/m 2
6
3
uave =
=
4.76
×10
J/m
3×108 m/s
(c) What are the rms and peak values of the electric field? I = cε 0 E
2
rms
I
Erms =
cε 0
1400 W/m 2
Erms =
= 812 V/m
3×108 m/s ⋅ 8.85 ×10 −12 C 2 (N ⋅ m 2 )
Emax = Erms 2
Physics 150, Prof. M. Nikolic Emax = 730 V/m 2 = 1144.8 V/m
26 Polariza5on A source of EM waves is unpolarized if the E-­‐fields are in random direc5ons. If all of the electric fields are oscilla5ng in planes parallel to each other, then the light is said to be polarized. For an EM wave, the direc5on of polariza5on is given by the direc5on of the E-­‐field. Physics 150, Prof. M. Nikolic 27 Polarizers They are materials that cause light that pass through them to become virtually perfectly polarized. unpolarized before polarized a]er Long, organic molecules aligned to be parallel to each other Physics 150, Prof. M. Nikolic 28 Polarizers When completely unpolarized light passes through a perfect polarizer, exactly half of the intensity is lost. 1
I p = I0
2
When already-­‐polarized light with intensity Ip passes through a polarizer, the intensity of the light that passes through is I = I p cos2 θ
θ is the angle between the ini5ally polarized light and the direc5on of the transmission angle of the polarizer (Malus’s law) Physics 150, Prof. M. Nikolic 29 Exercise: Polariza5on Unpolarized light of intensity 250 W/m2 passes through two polarizers in turn with axes at 45° to each other. What is the frac5on of the incident light intensity that is transmiked? A]er passing through the first polarizer, the intensity of the light drops by a factor of 2 to I1 = ½ I0 = 250/2 = 125 W/m2. (It does not maker what is the angle of the polariza5on axis for the unpolarized light). A]er passing through the second polarizer, the intensity of the light becomes I 2 = I1 cos2 θ
1
I 2 = I 0 cos2 θ
2
I 2 = 125W / m 2 cos2 450 = 62.5 W/m 2
Physics 150, Prof. M. Nikolic 30 Conceptual ques5on – Polariza5on I Q3
If you have one ideal polarizing filter, it will reduce the intensity of light by a factor of two from I to I/2. If a second ideal polarizing filter is placed perpendicular to the first, what intensity of light will pass the combina5on? A. 
B. 
C. 
D. 
I/4 I/2 0 Between I/4 and I/2 A]er passing through the first polarizer, the intensity of the light drops by a factor of 2 to I1 = ½ I0. A]er passing through the second polarizer I2 = I1cos2900 = 0 60
Conceptual ques5on – Polariza5on II Q4
If you have two ideal polarizing filters at right angles, no light will get through. If a third polarizing filter is placed between the two at 45o to both of them, what intensity of light will pass the combina5on? A. 
B. 
C. 
D. 
No light gets through. Some light gets through but less than I0 All the light gets through More than I0 light gets through A]er passing through the first polarizer, the intensity of the light drops by a factor of 2 to I1 = ½ I0. A]er passing through the second polarizer I2 = I1cos2450 = ¼ I0 A]er passing through the third polarizer I3 = I2cos2450 = 1/8 I0 60
The Doppler effect A Doppler Effect occurs for EM waves, but differs from that of sound waves fo = f s
vrel
1+
c
vrel
1−
c
•  fs is the frequency emiked by the source •  fo is the frequency received by the observer •  vrel is the rela5ve velocity of the source and the observer •  c is the speed of light If the source and observer are approaching each other, then vrel is posi5ve, and vrel is nega5ve if they are receding. When v/c << 1, the previous expression can be approximated as: " vrel %
fo ≈ fs $1+
'
#
c &
Physics 150, Prof. M. Nikolic 33 Exercise: Doppler shi] " vrel %
fo ≈ fs $1+
'
#
c &
Light of wavelength 659.6 nm is emiked by a star. The wavelength of this light as measured on Earth is 661.1 nm. How fast is the star moving with respect to the Earth? Is it moving toward Earth or away from it? What’s given: λs = 659.6 nm λo = 661.1 nm " vrel %
fo ≈ fs $1+
'
#
c &
vrel fo
= −1
c
fs
" fo %
vrel = c $ −1'
# fs &
First, we need to convert wavelengths into frequencies: c
fs =
λs
c
fo =
λo
3⋅108 m / s
14
fs =
=
4.55⋅10
Hz
−9
659.6 ⋅10
3⋅108 m / s
14
fo =
=
4.538⋅10
Hz
−9
662.2 ⋅10
# 4.538⋅1014 &
5
vrel = 3⋅10 m / s %
−1
=
−6.8⋅10
m/s
(
14
$ 4.55⋅10
'
8
Physics 150, Prof. M. Nikolic 34 Review Speed of EM waves (light) in vacuum 1
c=
= 3.00 ×108 m/s
ε0µ0
c= fλ
Speed of EM waves (light) in any medium c
v=
n
Emax = cBmax
Average energy density U
1
1 2
1 2
2
2
uave =
= ε 0 Eave +
Bave = ε 0 Eave = Bave
Volume 2
2µ 0
µ0
uave = ε 0 E
2
rms
1 2
= Brms
µ0
Intensity I = cuave = cε 0 E
Physics 150, Prof. M. Nikolic 2
rms
1
2
= cBrms
µ0
35 Review Polariza5on of the unpolarized light 1
I p = I0
2
Polariza5on of the polarized light 2
I = I p cos θ
Doppler effect fo = f s
vrel
1+
c
vrel
1−
c
•  fs is the frequency emiked by the source •  fo is the frequency received by the observer •  vrel is the rela5ve velocity of the source and the observer •  c is the speed of light " vrel %
fo ≈ fs $1+
'
#
c &
Physics 150, Prof. M. Nikolic 36