30. Electromagnetic waves and optics
www.ariel.ac.il Ԧ 1 푊 Power per unit area and points Poynting vector: 푆 = 퐸 × 퐵 has units [ 2] 휇0 푚 in the direction of propagation 1 푆Ԧ = 퐸 × 퐵 휇0 퐸 퐸 퐵 푆Ԧ
http://www.ariel.ac.il/ 퐵 푆Ԧ Intensity is time averaged norm of Poynting vector: 1 푇 퐼 = 푆 = 푆Ԧ 푑푡 푎푣 푇 0
퐸 푥, 푡 = 퐸 cos 푘푥 − 휔푡 푦 0 1 푆Ԧ = 퐸 × 퐵 퐸0 휇 퐵 푥, 푡 = cos 푘푥 − 휔푡 0 푧 푐 Intensity is time averaged norm of Poynting vector: 2 1 푇 Ԧ 퐸0 푇 2 퐼 = 푆푎푣 = 푆 푑푡 = cos 푘푥 − 휔푡 푑푡 푇 0 푐휇0 0
퐸 푥, 푡 = 퐸 cos 푘푥 − 휔푡 푦 0 1 푆Ԧ = 퐸 × 퐵 퐸0 휇 퐵 푥, 푡 = cos 푘푥 − 휔푡 0 푧 푐 Intensity is time averaged norm of Poynting vector: 푇 푇 1 퐸2 퐸2 푐퐵2 Ԧ 0 2 0 0 푊 퐼 = 푆푎푣 = න 푆 푑푡 = න cos 푘푥 − 휔푡 푑푡 = = units [ ൗ 2] 푇 푐휇0 2푐휇0 2휇0 푚 0 0 1 2 1 1 2 Energy density 푢 = 휀표퐸 + 퐵 2 2 휇0 퐸푦 푥, 푡 = 퐸0cos 푘푥 − 휔푡
퐸 Electric field Magnetic field 0 퐵푧 푥, 푡 = cos 푘푥 − 휔푡 energy density energy density 푐 1 2 1 1 2 2 Energy density 푢 = 휀표퐸 + 퐵 = 휀표퐸 2 2 휇0 퐸푦 푥, 푡 = 퐸0cos 푘푥 − 휔푡
퐸 Electric field Magnetic field 0 퐵푧 푥, 푡 = cos 푘푥 − 휔푡 energy density energy density 푐 퐸 (퐵 = = 휀 휇 퐸) 푐 표 0 1 2 1 1 2 2 Energy density 푢 = 휀표퐸 + 퐵 = 휀표퐸 2 2 휇0 퐸푦 푥, 푡 = 퐸0cos 푘푥 − 휔푡
퐸 Electric field Magnetic field 0 퐵푧 푥, 푡 = cos 푘푥 − 휔푡 energy density energy density 푐
1 푇 1 Average energy density 푢 = 휀 퐸2푑푡 = 휀 퐸2 푎푣 푇 0 표 2 표 0 1 2 1 1 2 2 Energy density 푢 = 휀표퐸 + 퐵 = 휀표퐸 2 2 휇0 퐸푦 푥, 푡 = 퐸0cos 푘푥 − 휔푡
퐸 Electric field Magnetic field 0 퐵푧 푥, 푡 = cos 푘푥 − 휔푡 energy density energy density 푐
2 1 푇 2 1 2 퐸0 ( = Average energy density 푢푎푣 = 휀표퐸 푑푡 = 휀표퐸0 = 휀표휇0푐퐼 (Intensity = 퐼 푇 0 2 2푐휇0 퐼 퐽 푢 = units: [ ] 푎푣 푐 푚3 1 2 1 1 2 2 Energy density 푢 = 휀표퐸 + 퐵 = 휀표퐸 2 2 휇0 퐸푦 푥, 푡 = 퐸0cos 푘푥 − 휔푡
퐸 Electric field Magnetic field 0 퐵푧 푥, 푡 = cos 푘푥 − 휔푡 energy density energy density 푐
2 1 푇 2 1 2 퐸0 ( = Average energy density 푢푎푣 = 휀표퐸 푑푡 = 휀표퐸0 = 휀표휇0푐퐼 (Intensity = 퐼 푇 0 2 2푐휇0 퐼 퐽 = units: [ ] 푐 푚3
퐹표푟푐푒 Radiation pressure: 푃 = 푟푎푑 퐴푟푒푎 1 2 1 1 2 2 Energy density 푢 = 휀표퐸 + 퐵 = 휀표퐸 2 2 휇0 퐸푦 푥, 푡 = 퐸0cos 푘푥 − 휔푡
퐸 Electric field Magnetic field 0 퐵푧 푥, 푡 = cos 푘푥 − 휔푡 energy density energy density 푐
2 1 푇 2 1 2 퐸0 ( = Average energy density 푢푎푣 = 휀표퐸 푑푡 = 휀표퐸0 = 휀표휇0푐퐼 (Intensity = 퐼 푇 0 2 2푐휇0 퐼 퐽 = units: [ ] 푐 푚3 momentum
푑푝 퐹표푟푐푒 Radiation pressure: 푃 = = 푑푡 푟푎푑 퐴푟푒푎 퐴푟푒푎 1 2 1 1 2 2 Energy density 푢 = 휀표퐸 + 퐵 = 휀표퐸 2 2 휇0 퐸푦 푥, 푡 = 퐸0cos 푘푥 − 휔푡
퐸 Electric field Magnetic field 0 퐵푧 푥, 푡 = cos 푘푥 − 휔푡 energy density energy density 푐
2 1 푇 2 1 2 퐸0 ( = Average energy density 푢푎푣 = 휀표퐸 푑푡 = 휀표퐸0 = 휀표휇0푐퐼 (Intensity = 퐼 푇 0 2 2푐휇0 퐼 퐽 = units: [ ] 푐 푚3 momentum Energy=momentum*c=푝푐 푑푝 푑퐸 퐹표푟푐푒 1 Radiation pressure: 푃 = = 푑푡 = 푑푡 푟푎푑 퐴푟푒푎 퐴푟푒푎 푐 퐴푟푒푎 1 2 1 1 2 2 Energy density 푢 = 휀표퐸 + 퐵 = 휀표퐸 2 2 휇0 퐸푦 푥, 푡 = 퐸0cos 푘푥 − 휔푡
퐸 Electric field Magnetic field 0 퐵푧 푥, 푡 = cos 푘푥 − 휔푡 energy density energy density 푐
2 1 푇 2 1 2 퐸0 ( = Average energy density 푢푎푣 = 휀표퐸 푑푡 = 휀표퐸0 = 휀표휇0푐퐼 (Intensity = 퐼 푇 0 2 2푐휇0 퐼 퐽 = units: [ ] 푐 푚3 momentum Energy=momentum*c=푝푐 푑푝 푑퐸 퐹표푟푐푒 1 1 Radiation pressure: 푃 = = 푑푡 = 푑푡 = 푆 푟푎푑 퐴푟푒푎 퐴푟푒푎 푐 퐴푟푒푎 푐 푑퐸 Power/Area and =power 푑푡 1 2 1 1 2 2 Energy density 푢 = 휀표퐸 + 퐵 = 휀표퐸 2 2 휇0 퐸푦 푥, 푡 = 퐸0cos 푘푥 − 휔푡
퐸 Electric field Magnetic field 0 퐵푧 푥, 푡 = cos 푘푥 − 휔푡 energy density energy density 푐
2 1 푇 2 1 2 퐸0 ( = Average energy density 푢푎푣 = 휀표퐸 푑푡 = 휀표퐸0 = 휀표휇0푐퐼 (Intensity = 퐼 푇 0 2 2푐휇0 퐼 퐽 = units: [ ] 푐 푚3 momentum Energy=momentum*c=푝푐 푑푝 푑퐸 퐹표푟푐푒 1 1 Radiation pressure: 푃 = = 푑푡 = 푑푡 = 푆 푟푎푑 퐴푟푒푎 퐴푟푒푎 푐 퐴푟푒푎 푐 푑퐸 푎푣 푆푎푣 퐼 푁 Power/Area and =power 푃 = = units: [ ] 푑푡 푟푎푑 푐 푐 푚2
The pressure of light on a mirror is larger than on a black piece of paper
A. True B. False C. They are the same units: [푊ൗ ] Intensity is time averaged norm of Poynting vector: 퐼 = 푆푎푣 푚2
Radiation pressure: 푃푟푎푑 = 퐼/푐 (absorbing surface) Intensity is time averaged norm of Poynting vector: 퐼 = 푆푎푣
Radiation pressure: 푃푟푎푑 = 퐼/푐 (absorbing surface)
photon Absorbing wall momentum momentum and energy
Ei, pi ET pi
absorbing material Intensity is time averaged norm of Poynting vector: 퐼 = 푆푎푣
Radiation pressure: 푃푟푎푑 = 퐼/푐 (absorbing surface)
Radiation pressure: 푃푟푎푑 = 2퐼/c (reflecting surface)
photon photon mirror Absorbing wall momentum momentum momentum momentum and energy
pi Ei, pi ET 2pi pi -pi
mirror absorbing material Which will make the radiometer turn more? Shining the laser on the A. Black side B. Metal side C. Both equal
Radiation pressure: 푃푟푎푑 = 퐼/푐 (absorbing surface)
Radiation pressure: 푃푟푎푑 = 2퐼/c (reflecting surface) photon photon mirror Absorbing wall momentum momentum momentum momentum and energy
pi Ei, pi E 2pi pi T -pi
Crookes radiometer mirror absorbing material This would make the radiometer turn more if the laser were pointed on the metallic side
BUT this is a very small effect because 푃푟푎푑 = 퐼/풄 Large Radiation pressure: 푃푟푎푑 = 퐼/푐 (absorbing surface)
Radiation pressure: 푃푟푎푑 = 2퐼/c (reflecting surface) photon photon mirror Absorbing wall momentum momentum momentum momentum and energy
pi Ei, pi E 2pi pi T -pi
absorbing Crookes radiometer mirror material No energy absorbed Large energy absorbed will radiate E&M waves and also heat the gas. Both will push on the black side. Hot objects emit so-called Black body radiation and the color emitted depends on the object temperature
commons.wikimedia.org Hot objects emit so-called Black body radiation and the color emitted depends on the object temperature
commons.wikimedia.org
http://www.physast.uga.edu/~rls/1010/ch5/ovhd.html
Hot bodies emit radiation
Color
Temperature [K] Hot objects emit so-called Black body radiation and the color emitted depends on the object temperature
commons.wikimedia.org Blackbody radiation: Based on wave/particle in a box
Planck (early 1900s)
wiki
Energy=hfrequency=hc/λ (frequency=speed of light/wavelength) Light behaves like individual particles (Planck didn’t believe it yet, but Einstein did)!
http://physics.bgsu.edu/~stoner/P202/quantum1/sld008.htm
www.everythingmaths.co.za