Electromagnetic Waves - PDF Free Download

Eletromagneti Waves Physis 6C

Eletromagneti (EM) waves are produed by an alternating urrent in a wire. As the harges in the wire osillate bak 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 both 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 thumb - it points in the diretion of propagation. Clik here for an EM wave animation

Like any other wave, we know the relationship between the wavelength and frequeny, and the speed of propagation of the wave: v wave f λ

Like any other wave, we know the relationship between 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)

Examples: Find the frequeny of blue light with a wavelength of 460 nm.

Examples: Find the frequeny of blue 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

Examples: Find the frequeny of blue 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 3 10 8 1.25 10 m s 8 Hz 2.4m You will need to use this formula very often to onvert bak and forth between 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. 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 about the intensity of an EM wave (for light we would think of it as brightness). Just as for sound, intensity is measured as average power/area. The Poynting vetor, S, desribes the energy transported by the wave (S points in the diretion of propagation of the wave). Intensity Sav Emax Bmax 2 µ 0 EM waves also arry momentum. This means that a ray of light an atually exert a fore. To get the pressure exerted by a sinusoidal EM wave, just divide the intensity by the speed of light. Radiation S av Chek out example 23.5 on page 771 of your Pressure textbook for a solar sail example that should help with your homework.

Refletion and Refration When an EM wave enounters an interfae between two materials it will generally be partially refleted and partially transmitted (refrated). The refleted ray makes the same angle as the inident (inoming) ray. Measure the angles from the NORMAL (perpendiular) to the interfae. Inident ray NORMAL LINE Refleted ray material a θ a θ r material b θ b 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 bends. Every material has a different Index of Refration that desribes 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 greater than 1. Our rule for finding the angle of the refrated ray is alled Snell s Law: n a sin( θ a ) n b sin( θ b )

Example: A beam 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.

Example: A beam 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. Inident ray Refleted ray Air n a 1 60 θ r Water n b 1.33 Θ b? Refrated ray

Example: A beam 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 a ) n b sin( θ b ) Inident ray Refleted ray Air n a 1 60 θ r Water n b 1.33 Θ b? Refrated ray

Example: A beam 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 (1) sin(60 sin( θ sin( θ b ) a ) o ) n b sin(60 1.33 sin( θ o ) b ) (1.33) sin( θ θ b b ) sin 1 sin(60 1.33 o ) 40.6 o Inident ray Refleted ray Air n a 1 60 θ r Water n b 1.33 Θ b? Refrated ray

Total Internal Refletion When light enters a medium with a higher index of refration it will bend toward the normal (the angle gets smaller). When light enters a medium with lower index (e.g. from water to air) then it will bend away from the normal (the angle gets larger). This reates an interesting possibility 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 below Ray #1 has both 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 b 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 exhibit 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 rainbows! (the drops of water in the air at as tiny prisms) Dispersion of EM waves - aural interpretation by Pink Floyd

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

Polarization Details about polarization: Typial light soures are unpolarized, whih means the EM waves are not oriented in any partiular diretion (sunlight behaves this way). When unpolarized light passes through a polarizer, half of its intensity is bloked, and the transmitted light is now polarized in the diretion seleted by the filter. Example Problem Sunlight passes through 2 polarizers whih are oriented at 60 relative to eah other. How muh of the original sunlight intensity is transmitted?

Polarization Details about polarization: Typial light soures are unpolarized, whih means the EM waves are not oriented in any partiular diretion (sunlight behaves this way). When unpolarized light passes through a polarizer, half of its intensity is bloked, and the transmitted light is now polarized in the diretion seleted by 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 blok the refleted sunlight, giving them a better view of objets beneath the surfae.