# Physics 203 at Portland State 2014

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===Quick links=== ==This wiki== * [[start|Main classwiki page]] * [[CLASS INFORMATION|Class Info]] * {{:Kuzma_Syllab_Spr2014_v4.pdf|Syllabus}} * [[FREQUENTLY ASKED QUESTIONS|FAQ]] * [[CLASS MATERIALS|Class materials]] * [[PHYSICS LABORATORY|Labs]] * [[http://web.pdx.edu/~ralfw/physics/lab/index_files/LabScheduleSpring.pdf|Schedule]] * [[PHYSICS WORKSHOP|Workshop]] * {{:workshops:ph299_syllabus_14sp.pdf|W/S syllabus}} * [[Computational Projects|Projects]] * [[White noise project|White noise]] * [[Rainbow project|Rainbow]] * [[Digital sound project|Digital sound]] * [[Announcements]] ==Earlier material== * [[Chapter 13]] * [[Chapter 14]] * [[Exam 1 review]] * [[Chapter 25]] * [[Chapter 26]] * [[Chapter 27]] * [[Exam 2 review]] * [[Final exam review]] ==Previous wikis== * [[http://web.pdx.edu/~nkuzma/Ph202_2014_wiki|Ph202 - 2014]] ==Other learning tools== * [[http://d2l.pdx.edu|University D2L site]] * [[http://masteringPhysics.com|Text & homework]] \\ <sub><color magenta>PH203KUZMASPRING2014</color></sub> ==Knowledge & computation== * [[http://wolframalpha.com|Wolfram]] $\alpha$ * [[wp>Physics_portal|Wikipedia]] * [[http://physics.nist.gov/cuu/Constants/index.html|Physical constants]] * [[http://physics.info/| The Physics Hypertextbook]] * [[http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html| HyperPhysics]] ==Add more by editing:== * [[sidebar|This sidebar]] * [[Tasks to do]] ==Help for editors== * [[doku>wiki:syntax|Help on wiki codes]] * [[http://en.wikibooks.org/wiki/LaTeX/Mathematics|Help on wiki math]] * [[Tips on editing]] =="Sandboxes" for practice== * [[Draft page|Practice here]] * [[Draft page 2|Or here if locked-out]]

chapter_25
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======Ch.25: Electromagnetic waves (Apr 10-24)====== =====Concepts Ch.25===== * [[wp>Electromagnetic radiation]] * [[wp>Electric field]] * [[wp>Magnetic field]] * The [[wp>Doppler effect]] * The [[wp>Electromagnetic spectrum]] * [[wp>Color]]s ====After the 1st exam==== * [[wp>Polarization (waves)|Polarization]] and [[wp>Polarizer]]s * [[wp>Polarizer#Malus.27_law_and_other_properties|Law of Malus]] * [[wp>Polarization_(waves)#Unpolarized_light|Unpolarized light]] =====Units Ch.25===== ====Non-SI units==== * [[wp>Angstrom]] (Å) =====Constants Ch.25===== * [[wp>Speed of light]] $c=299792458\,\frac{\text m}{\text s}\approx 3\times 10^8\,\frac{\text m}{\text s}$ * [[wp>Vacuum permittivity|Electric constant]] $\epsilon_0$ $\approx 8.854\times 10^{-12}\,\frac{\text F}{\text m}$ * [[wp>Vacuum permeability|Magnetic constant]] $\mu_0$ $=\frac{4\pi}{10^7}\,\frac{\text N}{ {\text m}^2}\approx 1.257\times 10^{-6}\,\frac{\text N}{ {\text m}^2}$ =====Lecture notes Apr 15,24===== * Apr 15: {{:notes:notes05cor3.pdf|}} - please download <color red>the trice-corrected version</color>! * Apr 24: {{:notes:notes07s.pdf|}} - please download before class =====Equation Sheet Ch.25===== ====Hand-written notes and equations for Ch.25.1 - 25.4 ==== Available as PDF download {{:notes_equations_25.1_-_25.4_.pdf|}} (equations are boxed). ====Wiki version==== - Speed of light in vacuum: * $c=3\times 10^8\frac{\text m}{\text s}$ $= \frac{1}{\sqrt{\epsilon_0\mu_0}}$ $= f\lambda$ * $\epsilon_0 = 8.85 \times 10^{-12}\frac{\text F}{\text m}\;$ is the [[wp>Vacuum permittivity|electric constant]] * $\mu_0 = 4\pi\cdot 10^{-7}\frac{\text N}{\,{\text A}^2}\;$ is the [[wp>Vacuum permeability|magnetic constant]] - The Doppler effect for electromagnetic waves: * $f'$ $= f \left( 1 \pm \frac{u}{c}\right)$ * $u$ is the relative velocity between the source and the observer (difference in speed projected on the axis drawn through both of them) * Use the plus sign if the distance between the source and the observer is **decreasing** (they are moving **towards** each other) * Use the minus sign if the distance between the source and the observer is **increasing** (they are moving **away** from each other) - Total energy density of an electromagnetic wave = the energy density of the electric field + the energy density of the magnetic field * $u=u_E+u_B$ * $u_E=u_B\;\;$ (the two densities are equal at every point in space and in any moment in time, far away from the sources) * $u=2u_E=2u_B$ - For an electromagnetic wave described by an electric field of magnitude //E// and a magnetic field of magnitude //B//: * Total energy density at any point in space, some moment in time: * $u\,=\,\frac{1}{2}\epsilon_0E^2\!+\!\frac{1}{2\mu_0}B^2$ $\,=\,\epsilon_0E^2$ $\,=\,\frac{1}{\mu_0}B^2$ * Time-averaged energy density at any point in space: * $u_\text{av}$ $\,=\,\frac{1}{2}\epsilon_0E_\text{rms}^2\!+\!\frac{1}{2\mu_0}B_\text{rms}^2$ $\,=\,\epsilon_0E_\text{rms}^2$ $\,=\,\frac{1}{\mu_0}B_\text{rms}^2$ $\,=\,\frac{1}{2}u_\text{max}$ $\,=\,\frac{1}{4}\epsilon_0E_\text{max}^2\!+\!\frac{1}{4\mu_0}B_\text{max}^2$ $\,=\,\frac{1}{2}\epsilon_0E_\text{max}^2$ $\,=\,\frac{1}{2\mu_0}B_\text{max}^2$ * To find the average energy density simply plug in the rms value of the fields instead of //E// and/or //B//. * rms value = square root of 1/2 times the max value: * $E_\text{rms}=\frac{1}{\sqrt{2}}E_\text{max}$ $\approx 0.7071\!\cdot\!E_\text{max}$ * $B_\text{rms}=\frac{1}{\sqrt{2}}B_\text{max}$ $\approx 0.7071\!\cdot\!B_\text{max}$ - The energy density of the magnetic field and the energy density of the electric field are equal to each other * $u_E=u_B\;\;$ (at any point in space and time, far away from the sources) * $\frac{1}{2}\epsilon_0E^2=\frac{1}{2\mu_0}B^2$ * $E=\frac{1}{\sqrt{\epsilon_0\mu_0}}B=cB\;\;$ (at any point in space and time, in vacuum) * $E_\text{max}=cB_\text{max}\;\;$ (at any point in space, in vacuum) * $E_\text{rms}=cB_\text{rms}\;\;$ (at any point in space, in vacuum) - Intensity is the energy of the waves, per unit area of the receiver, per unit time: * $I = \frac{U}{A\cdot\Delta t}$ $=uc\;\;$ (at any moment in time, $\Delta t$ is much **shorter** than the period $\frac{1}{f}$) * $I_\text{av} = \frac{U}{A\cdot\Delta t}$ $=u_\text{av}c\;\;$ (average over time, $\Delta t$ is much **longer** than the period $\frac{1}{f}$) - Momentum = total energy absorbed divided by the speed of light: * $p=\frac{U}{c}$ - Pressure of the completely absorbed radiation is equal to the intensity divided by the speed of light * $P=\frac{I}{c}\;\;$ (at any given moment of time) * $P_\text{av}=\frac{I_\text{av}}{c}\;\;$ (average over time) - Pressure of the completely reflected radiation is equal to **twice** the intensity divided by the speed of light * $P=2\frac{I}{c}\;\;$ (at any given moment of time) * $P_\text{av}=2\frac{I_\text{av}}{c}\;\;$ (average over time) ===Equations Ch.25.5=== - Law of Malus (for a linearly polarized light incident normally (perpendicular to) an ideal polarizer): * $I = I_0\cos^2\theta$ $=I_0(\cos\theta)^2$ * $I_0$ is the average incident intensity * $I$ is the average transmitted intensity * $\theta$ is the angle between the $\vec{\mathbf E}$ of the incident light and the polarization axis of the polarizer * SI units: $\frac{\text W}{\,{\text m}^2}$ * The more polarizers that are used, hence the more smoothly the direction of polarization changes-the greater the final intensity - Transmitted intensity for an unpolarized beam incident normally at an ideal polarizer * $I = \frac{1}{2}I_0$ * $I_0$ is the average incident intensity * $I$ is the average transmitted intensity * SI units: $\frac{\text W}{\,{\text m}^2}$ =====Examples Ch.25===== ---- ====Problem 3.1==== Find frequencies of $\lambda=800\,$nm (<color darkred>deep red</color>) and $\lambda=400\,$nm (<color darkviolet>violet</color>) light. {{ :figs:lec03fg1.jpg?nolink|}} These roughly correspond to the limits of what our eyes can see. Steps: - Use the relationship between the wavelength, speed of light, and frequency: * $f=\frac{c}{\lambda}$ * For deep red, $\;\;f=\frac{3\times 10^8\,\frac{\text m}{\text s}}{800\times 10^{-9}\,{\text m}}$ $=3.75\times 10^{14}\,$Hz * For violet, $\;\;f=\frac{3\times 10^8\,\frac{\text m}{\text s}}{400\times 10^{-9}\,{\text m}}$ $=7.5\times 10^{14}\,$Hz ---- ====Problem 3.2==== Find the maximum //E// and //B// fields of a green light source with the average intensity corresponding to the "threshold of pain" ($I=1000\,\frac{\text W}{\,{\text m}^2}$). * $c=299792458\,\frac{\text m}{\text s}$ $\approx 3\times 10^8\,\frac{\text m}{\text s}$ * $\epsilon_0$ $\approx 8.854\times 10^{-12}\,\frac{\text F}{\text m}$ * $\mu_0$ $=\frac{4\pi}{10^7}\,\frac{\text N}{ {\text m}^2}$ $\approx 1.257\times 10^{-6}\,\frac{\text N}{ {\text m}^2}$ Steps: - Find $E_\text{rms}$ and $B_\text{rms}$ from the average intensity: * $I_\text{av}=u_\text{av}c$ $=\epsilon_0cE_\text{rms}^2$ $=\frac{1}{\mu_0}cB_\text{rms}^2$ * $E_\text{rms}=\sqrt{\frac{I_\text{av}}{c\epsilon_0}}$ * $B_\text{rms}=\sqrt{\frac{\mu_0 I_\text{av}}{c}}$ - Find $E_\text{max}$ and $B_\text{max}$ by multiplying the above values by $\sqrt{2}\approx 1.41$ * $E_\text{max}=\sqrt{2}\sqrt{\frac{I_\text{av}}{c\epsilon_0}}$ $= 868\,\frac{\text V}{\text m}$ * $B_\text{max}=\sqrt{2}\sqrt{\frac{\mu_0I_\text{av}}{c}}$ $=2.9\times 10^{-6}\,$T ===Questions about Pr. 3.2=== - threshold of pain, isn't it $I=1\,\frac{\text W}{\,{\text m}^2}$? * **Answer**: for the sound (audio), it is $1\,\frac{\text W}{\,{\text m}^2}$. But for the visible light, I (unofficially) take it to be the intensity of bright sunlight - it is quite painful to stare directly into the sun. After some absorption in the atmosphere, the intensity of bright sun on Earth is about $1000\,\frac{\text W}{\,{\text m}^2}$. --- //[[nkuzma@pdx.edu|Nicholas Kuzma]] 2014/04/19 21:45// * **Correction**: <color red>In the previous version of the notes and this wiki example, the threshold of pain for the **sound** was used by mistake instead of the threshold of pain for the **light** ($1000\frac{\text W}{\,{\text m}^2}$).</color> Please take a note of the corrected answers! * Also, please download the corrected class notes {{:notes:notes05cor3.pdf|}}. Sorry about this! --- //[[nkuzma@pdx.edu|Nicholas Kuzma]] 2014/04/21 10:37// ---- ====Problem 3.3==== [[wp>IKAROS|{{ :images:ikaros.jpg?nolink|}}]] Assume the [[wp>solar sail]] density is comparable to that of diamond $\big(\rho=3500\,\frac{\text{kg}}{\,{\text m}^3}\!\big)$, and that it completely reflects all incident sunlight. Find the maximum thickness of the sail that can be totally supported by the sunlight against the Sun's gravity (//Hint: both the sunlight intensity and gravity drop off with the square of the distance. Therefore, you can pick any position to do your calculation, such as the Earth orbit//.\\ //Hint #2: pick any area of the sail, such as// 1 m<sup>2</sup>. //In the end, it should cancel out//.) Equations and values: * Momentum * $p=\frac{U}{c}$ * Pressure exerted by the completely absorbed light * $P_\text{av}({\text{absorption}}) = \frac{I_\text{av}}{c}$ * Pressure exerted by the completely reflected light * $P_\text{av}({\text{reflection}}) = 2\!\frac{I_\text{av}}{c}$ * Average intensity of sunlight at Earth orbit: * $I_\text{av}$ $=1.36\times 10^3\frac{\text W}{\,{\text m}^2}$ Steps: - Acceleration due to the gravity of the Sun at the location of Earth * $g_\text{sun}$ $= \frac{v^2}{R}$ $= \omega^2R$ $= \left(\frac{2\pi}{T}\right)^2R$ $= \left(\frac{2\pi}{1\,{\text{year}}}\right)^2R$ $= \left(\frac{2\pi}{365.25\cdot 24\cdot 3600\,{\text s}}\right)^2\!\cdot\!1.5\times 10^{11}\,$m $=0.006\,\frac{\text m}{\,{\text s}^2}$ * Here $\omega= \frac{2\pi}{T}= 2\times 10^{-7}\frac{\text{rad}}{\text s}\;$ is the angular velocity of the Earth around the Sun * $T$ is the period of the Earth (1 year) in seconds * $R$ is the radius of Earth's orbit (= distance to the Sun, 93.3 million miles, $1.5\times 10^{11}\,$m) - Force due to the pressure of sunlight on the area of the sail of $A\!=\!1\,{\text m}^2$: * $F=P_\text{av}({\text{reflection}})\!\cdot\!A$ $=2\!\frac{I_\text{av}}{c}\!\cdot\!A$ $=2\!\frac{1.36\times 10^3\frac{\text W}{\,{\text m}^2}}{3\times 10^8\frac{\text m}{\text s}}\!\cdot\!1\,{\text m}^2$ $=9\times 10^{-6}\,$N - Balance of forces (gravity vs sunlight): * $mg_\text{sun}=F$ - Mass of 1 m<sup>2</sup> of the sail: * $m=\frac{F}{\;g_\text{sun}}=\frac{9\times 10^{-6}\,{\text N}}{0.006\,\frac{\text m}{\,{\text s}^2}}$ = 0.0015 kg - Density of diamond * $\rho= 3.5\,\frac{\text g}{\,{\text{cm}}^3}$ $=3500\,\frac{\text{kg}}{\,{\text m}^3}$ - Volume of 1 m<sup>2</sup> of the sail * $V=\frac{m}{\rho}=4.3\times 10^{-7}\,{\text m}^3$ - The thickness: * $d=\frac{V}{A}=4.3\times 10^{-7}\,{\text m}$ $=0.43\,\mu$m ---- ====Problem 3.4==== Unpolarized light of average intensity $I_0$ is passed (normally = perpendicular to) through two polarizers (one after another) with polarization axes at an angle $\alpha$ relative to each other. Find the average intensity $I_2$ of the transmitted light. Steps: - Since the first beam is unpolarized the average intensity after the 1<sup>st</sup> polarizer is: * $I_1 = \frac{1}{2}I_0$ - The general trigonometry formula of $\cos^2\alpha$: * $\cos^2 \alpha$ $=\frac{1}{2}\big(1+\cos(2\alpha)\big)$ - After passing through the first polarizer, the light becomes polarized with the electric field along the axis of the 1<sup>st</sup> polarizer: * $I_2 = I_1\cos^2 \alpha$ $=\frac{1}{2}I_0\cos^2\alpha$ $=\big(\frac{1}{4}+\frac{1}{4}\cos 2\alpha\big)I_0$ * SI units: $\frac{\text W}{\,{\text m}^2}$ ---- ====Problem 3.5==== Unpolarized light of average intensity $I_0$ is passed (normally) through polarizers 1 and 3 that have their polarization axes perpendicular $\left(\frac{\pi}{2}\right)$ relative to each other. Find the average intensity $I_3$ of the transmitted light. Then, polarizer 2 is inserted between polarizers 1 and 3. The polarization axis of the inserted polarizer 2 is at an angle $\beta$ relative to the axis of the first polarizer. Find the average intensity $I_3'$ in that case as a function of $\beta$. Steps: - After passing the 1<sup>st</sup> polarizer: * $I_1 = \frac{1}{2}I_0$ - With just polarizers 1 and 3: * $I_3=I_1\big(\!\cos\frac{\pi}{2}\!\big)^2$ $=0$ - With the 2<sup>nd</sup> polarizer inserted between polarizers 1 and 3: * $I_2 = I_1\cos^2\beta$ * $I_3' = I_2\cos^2\big(\!\frac{\pi}{2}-\beta\big)$ $=I_2\sin^2\beta$ * Put it all together * $I_3' = I_0\frac{1}{2}\cos^2\beta\sin^2\beta$ $=\frac{1}{8}I_0\sin^22\beta$ $=\frac{1}{16}\!\big(1-\cos4\beta\big)I_0$ * So, for example, $I_3'$ is maximized at $\beta=\frac{\pi}{4}$ * $I_3'=\frac{1}{8}I_0$ ---- =====Homework Questions Ch.25===== ---- ====Problem 50==== <color #5555BB>Electromagnetic wave 1 has a maximum electric field of $E_0=52\,\frac{\text V}{\text m}$, and electromagnetic wave 2 has a maximum magnetic field of $B_0=1.5\,\mu$T... Calculate the intensity of each wave.</color> * **Question**: For parts B and C, I am having trouble calculating the average intensity. * Part B: $\;\;I=c\cdot\epsilon_0\cdot E^2$ $\Rightarrow 3\times 10^8\cdot 8.85\times 10^{-12}\cdot 52^2=7.179\,\frac{\text W}{\,{\text m}^2}$ * Part C: same calculations, but using 600 V/m for $E$, my answer is $956\,\frac{\text W}{\,{\text m}^2}$ * Mastering Physics says this is incorrect, where is my problem? * **Answer**: I can spot two potential pitfalls: * If they are asking to find the average intensity, you need to use rms values for the fields, but the given numbers are maximum values. You need to divide them by $\sqrt{2}\approx 1.41$ first, or divide the intensity by 2. * I am not sure how you converted $\;B_0\!=\!1.5\,\mu{\text T}\;$ to $\;E_0\!=\!600\,\frac{\text V}{\text m}$. Multiplying it by the speed of light should give $450\,\frac{\text V}{\text m}\;$ (unless you had a different number for the magnetic field on the website). --- //[[nkuzma@pdx.edu|Nicholas Kuzma]] 2014/04/18 23:29// ----