Electromagnetic Waves. Physics 6C



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Transcription:

Eletromagneti Waves Physis 6C

Eletromagneti (EM) waves are produed y an alternating urrent in a wire. As the harges in the wire osillate ak and forth, the eletri field around them osillates as well, in turn produing an osillating magneti field. This magneti field is always perpendiular to the eletri field, and the EM wave propagates perpendiular to oth the E- and B-fields. This gives us a right-hand-rule relating the diretions of these 3 vetors: 1) Point the fingers of your right hand in the diretion of the E-field 2) Curl them toward the B-field. Eletromagneti Waves 3) Stik out your thum - it points in the diretion of propagation. http://www.phys.hawaii.edu/~te/java/ntnujava/emwave/emwave.html A good EM wave animation applet an e found at this wesite

Like any other wave, we know the relationship etween the wavelength and frequeny, and the speed of propagation of the wave: v wave f λ In the ase of EM waves, it turns out that the wave speed is the speed of light. So our formula for EM waves (in vauum) is: f λ ; 3 10 8 m s It turns out that the speed of light is also the ratio of the strengths of the Eletri and Magneti fields in an EM wave. So we know that EB (in standard metri units) The ontinuum of various wavelengths and frequenies for EM waves is alled the Eletromagneti Spetrum

Examples: Find the frequeny of lue light with a wavelength of 460 nm. f λ f λ 3 10 8 460 10 m s 9 m 6.5 10 14 Hz A ell phone transmits at a frequeny of 1.25x10 8 Hz. What is the wavelength of this EM wave? f λ λ f 8 3 10 1.25 10 m s 8 Hz 2.4m You will need to use this formula very often to onvert ak and forth etween frequeny and wavelength.

Energy and momentum in EM Waves Eletromagneti waves transport energy. The energy assoiated with a wave is stored in the osillating eletri and magneti fields. The E and B fields ontriute equally to the total energy of the wave. We will find out later that the frequeny of the wave determines the amount of energy that it arries. Sine the EM wave is in 3-D, we need to measure the energy density (energy per unit volume). u 1 2 1 2 ε 2 0E + 2 B µ 0 We an also talk aout the intensity of an EM wave (for light we would think of it as rightness). Just as for sound, intensity is measured as average power/area. The Poynting vetor, S, desries the energy transported y the wave (S points in the diretion of propagation of the wave). Intensity Sav Emax B max 2µ 0 EM waves also arry momentum. This means that a ray of light an atually exert a fore. To get the pressure exerted y a sinusoidal EM wave, just divide the intensity y the speed of light. Radiation Pr essure Sav Chek out example 23.5 on page 771 of your textook for a solar sail example that should help with your homework.

Refletion and Refration When an EM wave enounters an interfae etween two materials it will generally e partially refleted and partially transmitted (refrated). The diretion of the refleted ray is very simple it makes the same angle as the inident (inoming) ray. Be areful to measure the angle from the NORMAL (perpendiular) to the interfae. Inident ray NORMAL LINE Refleted ray material a θ a θ r material θ Refrated ray

EM waves in vauum travel at the speed of light, 3x10 8 m/s In any material medium (glass, water, air, et.) light travels more slowly. This is alled REFRACTION, and it explains why the light ray ends. Every material has a different Index of Refration that desries the speed of light in that material medium: Index of Refration n v Speed of light in vauum Speed of light in the material Sine light always travels more slowly through a material medium, the index is always more than 1. Our rule for finding the angle of the refrated ray is alled Snell s Law: n a sin( θ a ) n sin( θ )

Example: A eam of light in air enters water at an angle of 60 to the normal. Find the angle of refration. The index of refration of water is 1.33. n a sin( θ (1) sin(60 sin( θ ) a ) n o ) (1.33) sin( θ sin(60 1.33 sin( θ o ) θ ) ) sin 1 sin(60 1.33 o ) 40.6 o Inident ray Refleted ray Air n a 1 60 θ r Water n 1.33 Θ? Refrated ray

Total Internal Refletion When light enters a medium with a higher index of refration it will end toward the normal (the angle gets smaller). When light enters a medium with lower index (e.g. from water to air) then it will end away from the normal (the angle gets larger). This reates an interesting possiility what if the angle gets so large that the light ray is not transmitted at all? If the refrated angle is larger than 90 we have this situation it is alled Total Internal Refletion (make sure that name makes sense to you). In the diagram elow Ray #1 has oth a refletion and a refration, while Ray #2 is totally refleted. The inidene angle for Ray #2 is larger than the ritial angle for total internal refletion. To find this ritial angle, simply set the refrated angle to 90 in Snell s Law. Air n 1 Refrated Ray #1 Water n a 1.33 Ray #2 Refleted Ray #2 Ray #1 Refleted Ray #1

Dispersion This is the exeption to the rule that says that all waves in a given medium travel at the same speed (we learned this for sound waves in a previous hapter). In a material medium, EM waves exhiit a phenomenon alled DISPERSION, where the index of refration depends on the frequeny of the light. Higher frequenies orrespond to a higher index, and thus are refrated more than lower frequenies. This effet is why we have rainows! (the drops of water in the air at as tiny prisms)

Polarization The Polarization of an EM wave is defined to e the diretion of its Eletri field vetor. EM waves (or light) an e passed through a filter (polarizer) to selet for a partiular polarization diretion. This will ut down the intensity (rightness) of the light ased on the following formula: I I ( os( φ ) 2 max ) Polarizers an e plaed in sequene to adjust the intensity and polarization diretion of light. The most ovious example is dark sunglasses, where 2 filters are plaed at 90 to eah other, loking out most of the light (the formula would say all the light is loked).

Polarization Details aout polarization: Typial light soures are unpolarized, whih means the EM waves are not oriented in any partiular diretion (sunlight ehaves this way). When unpolarized light passes through a polarizer, half of its intensity is loked, and the transmitted light is now polarized in the diretion seleted y the filter. Example Prolem Sunlight passes through 2 polarizers whih are oriented at 60 relative to eah other. How muh of the original sunlight intensity is transmitted? I final I A good polarizer applet an e found at this wesite sun ( ) ( ) 1 o ( 1) os(60 ) I sun 2 14243 1 4 2 8 http://www.lon-apa.org/~mmp/kap24/polarizers/polarizer.htm

Polarization Details aout polarization: Typial light soures are unpolarized, whih means the EM waves are not oriented in any partiular diretion (sunlight ehaves this way). When unpolarized light passes through a polarizer, half of its intensity is loked, and the transmitted light is now polarized in the diretion seleted y the filter. Refleted light is (at least partially) polarized parallel to the refleting surfae. A good example is sunlight refleting from the water. Fishermen wear polarized sunglasses to lok the refleted sunlight, giving them a etter view of ojets eneath the surfae.