B Fundamental Physical Constants A Vnits for Physical Quantities I Base Units Symbol Vnlt Meter - m /, x Length m,M Mass Kilogram - kg T Temperature Kelvin - K C Conversion factors and alternative units Symbol Base Units Unit Mass of electron me llx I0- kg Mass of proton I11p 1.67x I0- 27 kg , Avogadro Constant NA 6.022x I0 23 mol I Elementary charge e 1.602xl0 Faraday Constant F 96,4 85 C/mol Speed of light c x 0~ Molar Gas Constant R 8.3141 mol- I K I 19 Unit C Derived Units Ampere - A (Ci s) / Electric Current Unit Symbol a m/s2 Boltzmann Constant k 1.3 8x I0- 23 K- I Ang Acce! a radian/s Gravitation Constant G 67x I0- 11 m kg- I s Ang Momentum L kg m2/s i Permeability of Space 110 4n:x1 Ang Velocity w radian/sec Permittivity of Space Eo 8.85x10 12 F/m e, Angle , ~~ Farad F (CN ) ~ isplacement Q, q, e Coulomb C (A s) p kg/m d, h meter - m Electric Field E Vi m Electric Flux +" Vm Electromotive Force (EMF) S, ;r F Force Frequency Iv c Length of A IA I = I 'AC-:,'-+-AC-:,':-+- A:-C:,' = , Sample Addition and Length Calculations: Volt - V Tesla (Wb/m2) Magnetic Flux +m Weber Wb (kg m 21A s2) P Resistance j Voltage V (l/C) Torque Pascal - Pa (N /m2) R Ohm r Nm n (VIA) Volume v m/s V m3 - -,- Wavelength A Work W (A,B, ) + (A,.By ) + (A=B=) Note: is the angle between A and B; Sample: Scalar product: = 5i + 2j B = 3i + 5j A- B = x + x = 15 + 10 = 25 A I cV = 1.602x 1O- 19 Force Dyne CGS unit (g em/s = erg/em) I dyne = 10 N Volume Liter I L = I dm Pressure Bar I Bar = 10 Pa Angstrom I A = Ix lO ILength 10m ' meter - m louIe - (N m) b Cross or Vector Product: C = A x B = V4 II Blsin6 e - Angle between A and B, vector e is perpendicular to A and B i j k A x B=A,AA IB I=-/9 +25=134= 5.831 A-B cos6 25 IA II B I 5.385 x5.83 0.796 fig -a m ~ Sample: Vector Product: A = 2i + j B = i + 3j i j k ~ Q Z B, B, B I A x B = I = (6 - I) k = 5k I 130 -If A and B are in x-y plane, A x B is along I the + z direction -6 is the angle formed by A B: sin6 = I AC B Given: IAI = IS IBI = /iO ICi = sin6= 5/(1S x /iO )~51!50 = 1/ /1 < AB: = 45" = n:/2 radians c The R ight-Hand Rule gives the orientation of vector e flt!.t B Trigonometry I Basic relations for a triangle fi" sin = 2:' Values Of sin cos and tan e rad ["] f sin!i cos tJ tan tJ [O"] l -OOO 1.00 0.00 r cos = f tan = ~ IAI=/25 +4 = /29 = 5.385 = cos·I(0.796) = 37° = 0.2n: fad - ~- Velocity Note: V41 + IBI ~ IA + BI Multiply A & B: a Dot or scalar product: A - B = IAI IB lcos6 = A-B = 0,if6 = n: /2 tig2 kg m/s P,,/' ' Watt - W (1/s) Pressure 165 = 8.06 V4+BI = II +100 +4 = /lOS = 10.25 = A+B=i+ IOj+2k B Power IBI = /4+36+25 Hertz - Hz (cycle/s) Magnetic Field V = 134 =5.83 Newton - N (kg m/s2 = 11m) louIe - Potential Electron Volt IAI= -/9 + 16 +9 B= -2i+6j+5k JIK Q p CGS unit (g cm 2/s2) I erg = 10 J N/A2 Addition of vectors A & B, add components: A + B = (A,+B,)i + (A,,+B,,)j + (A=+BJk Heat Momentum Erg A Vector Algebra I Vector: Denotes directional character using (x,y, z) components fig I a V nit vectors: i along x, j along y, k along z b Vector A = Axi + A"j + A=k A= 3i+4j - 3k S Entropy Energy MATHEMATICAL CONCEPTS E, U,K Joule J (kg m s -2) Energy 180" = n: rad radian C Capacitance Charge II' " (degree) tM~Y m s- I Acceleration Angle I Second - s Time Description ~~~ = + COS = Sinand Cos waves fig (, sin l f-;;H,[J(J"t 0.50 0.866 0.577 100 rr/4l45"J 0.707 '0.707 f-;;;'J 1,6l fj ~O'J 0.866 50 , 1732 1.00 000 oc 1.0010·00 I ~O"] O.t)~ I " Q -a ~A: L 1e:~ ~'.: t~ ~~Gf: 90° = n/2 Right-Hand Rule x sin6 - cos6 MATHEMA11CAL CONCEPI'S (cont.) C Geometry Circle: Area = rrr2 ; Circumference = 2rrr Sphere: Volume = 4/) rrr\ Area = 4rrr2 • Electrostatic potential energy: I Vector vs scalar a Vector: Has magnitude and direction b Scalar: Magnitude only, no direction Number and unit a Physical data, constants and equations have numerical values and units b A correct a nswer must include the correct numerical value PLUS the correct unit Significant figures (sigfig) a The # of sigfigs reflects the accuracy of experimental data; calculations must accommodate this uncertainty b For multiplication: The # of sigfigs in the final answer is limited by the entry with the fewest sigfigs c For addition: The # ofdecimal places in the final answer is given by the entry with the fewest decimal places d Rules for "rounding sigfigs" ·I f the last digit is >5, round up ·If the last digit is cosS, y = r sinq> sine, y = rsine, z = r cosq>, ? = x + y2 + z2 fill II r2 = x2 + y2 'Calculate (r, e, q» from (x, y, z) 'Calculate (x, y, z) from (r, e , q» jJ Hint: Follow the strategy tor 2-d Spherical polar coordinates E Use of Calculus in Physics I Methods from calculus are used In physics definitions , and the derivations of equations and laws Physical meanings of calculus expressions: x x = r sinq> cose a Derivative - slope of the curve: d~~X) y = r sinq> sine b Integral - area under the curve: IF(x)d~ z = r cosq>, My energy: Joule I = kg Velocity: vex) = dF( x ) dt Acceleration: a = dr(x ) • Power and work: dt p = dW dt • Energy and force: E=IFd~ Other useful expressions: iR.,- -=Fex ) I x2 Xl 2x a/ax + aldy + dldz e Integration by parts: ludv = uv - Ivdu f Symbol for integration of closed surface or volume: f· 1,3 - F dG b ' ~=G dx - G' dx x" _ I _ x" II +1 , -I x" In x In x x In x - x eX eX eX sin(x) cos(x) -{;os(x) x = IJ X Il - I, x Sample: 54 cm x IOI(;; m = 0.54 m • Use the 2nd to convert "m" to "cm" - "'+v ~ Constant: g = 9.8 m /s fig 1'3 Pitfall - not simply search for the "right" equation in your notes or text Describe the physics using a diagram, with mathematical appropriate symbols and a coordinate system Obtain the relevant physical constants Do you have all the essenti al data'! jJ Hint: You may hayc extra information The hard part: Deri ve or obtain a mathematical express io n for the problem; use dimcnsional analysis to check the equation, constants and data The easy part: Plug numbers into the: equatio n and usc the calculator to obtain the numerical an swer C heck the tin a l answer usin g the: ori g in al stateme nt o r the problem, your sketch an d com m on sense; are the· units & si gn corrcce' II x I • Use the l ;t factor to convert "cm" to "m" , dF(x) 'IF(x) dx dx constant x I I dF c Partial derivative: aF(x, y,z) =dF ax dx' hold y & z constant d Gradient Operator V (Del) Ug = m g h m in kg II in //I Units of Ug = kg m 2/52 = J I ~mv2 Units of K = kg m in kg, v in m/s m 2/ 52 = J fig 12 • Gravitation a l potential energy: Iconstant d(F·G) =FdG +C.dF a dx dx dx d(F+G) - Pitfall: If the units are wrong, the answer is wrong! Kinetic: K = _ IIIL & 100('1/1 1m 100('m = kg m/s1 Dimensional a nalysis Verify that constants and variables in an equation result in the correct overall unit Samples: The energy unit is Joul es for ki netic, gravitational and Coulombic energy = x2 + y2 + z1 Common derivatives and integrals forc e: Newton N * Check all constants and variable units * Take special care if you derive the equation Ny r2 I acceleration: 111/,2 111 2/s2 Units of Uc = (J mlC )C 2/m = J Using Conversion Factors a Purpose: Modify experimental data to match the units of co nstants and equations b SI units: MKS (m-kg-s) and CGS (cm-g-s) c Common English units : Foot, pounds, BTU calories d Conversion factors are obtained from an equality of two units Sample: 100 em I m • This equality gives two conversion tactors: B Solve the Problem Strategically a Two key issues: I Understand the physics principles Have a correct mathematical strategy b Useful steps in problem solYing: I Prepare a rough sketch of the problem ldentify rel evant physical \ari abl.:s physical co ncepts and constants HI jJ Hint: Before doing the calculation: CD Samples: • Position: x or F(x) meter Temperature: K velocity : m/s constant : 1/4rrEo; units are J m ' C2 r in m, q] and q2 in Coulo mb Sample: 2.3 m x 101°('1/1 = 230 em m volume: mJ CD! q,q, Uc = 1/(4rr£o) - , A Motion along a Straight Line B Motion in Two and Three Dimensions I Goal: Determine position, velocity, acceleration I Goal: Similar to " A." with or dimensions Key concept: Select Ca rtes ian , polar Key terms: Acceleration: a = dvldl ; ve loc ity: v = dddt spherical coordinatcs, depending on the type Key Equations: x = Vi' + ~aI 1) = Vi + al motion x(/) v(/) for varia ble a fig 15 Sa mple: A proj ectile is launched at angle with 'i'i; how we set up the problem? Step I Detine x as horiLontal and)' vertical Step Detennine initial ",i and "r, 1'" 16 e " xi = V r; cosS cos(x ) - sin (x) sin(x) Step IdentifY a, - Gravitational force => a l · = -g Step Identify a" - No horizontal force => a, = Step Develop x- and y -equations of motion ? X = Vi,,1 + 2axt- = viI c Power = Work/time: W = Power~t or fp(l) dt d Woet = K tinal - K initial ; K is converted to work Sample: Determine the work expended in lifting a 50 kg box 10 m; given: a = g = 9.8 m/s I? I? )' = Vi)'t + a,.l - = viJ - 2gl Equations: F = III g => W = III g d Calculation: W = 50 kg x 9.8 m/s2 X 10m = 4,900 J C Newton 's LawS of Motion F Potential Energy & Energy Conservation Goal: Examine fo rce and acceleration I Goal: Use energy conservation to study the interplay Key concepts : Newton's Laws: of potential and kinetic energy Law #1 A body remains at rest or in motion unless Key Equations influenced by a force a Potential energy: Energy of position: VCr); Law #2 Forces acting on a body equal the mass gravitation (V = mgh), mu ltiplied by the acceleration; force and electrostatic (V ex qq/r) acceleration detemline motion b E = K + U Conservative system: No external force Law #3 Every action is countered by an opposing Sample: Examine K & V for a launched rocket action Initial: h = 0, therefore, V = m g h = 0 Key equations: a Law #2 : F = III a or LF = III a E=K= , 12 111 vI2 Hint: Forces are vectors! Next, resolve into x and y components: K yi & Kl'i b Types of forces: Body - gravity: Fg = III g ote: K, is constant d uring ~ • Surface - ti'iction: = Fj' = flF" y ", the flight Sample : F, exerted on At max height: K, = 0; V = object on a horizontal plane III gh = K'i @ F(= flF" = fl rg =fl III g Final state: Rocket hits the x Net force = F, - Fr lig 17 ground: V = 0, K = Ki 40 ~ Sample: Object on plane fig 20 inclined at angle 8; G Collisions and Linear Momentum fig 21 examine Fg & Ff Goal: Examine momentum of colliding bodies F" = Fg cos8 = 111 g cos8 Hint: For 2-D or 3-D, use Fj'= fl F Il = fl III g cos8 Cartesian components F, = III g si n8 fig I R Key Variables and Equations e Law #3: a Types of collisions: FI2 = - F21 or 1111 a l = - 1112 al • Elastic: Conserve energy Sample: Examine recoil of • Inelastic: Energy lost as heat or bullet fired from a rifle L -'_ _" ' - _ - - ' defomlation b Relative motion and frames of reference: A body Rifle recoil = ,,(bullet) x III (bullet) moves with velocity v in fram e S; in frame S', the D Circular Motion fig 19 velocity is v'; if V, is the ve locity of frame S' I Goal: Examine body moving in a circular path; use 2-d polar coordinates: (r, ) relative to S, then V = J{, + v' Key variables: c Linear Momentum: p = m v d Conserve K & p for conservative system (no distance from ! r m external forces): rotation center 2 l Ll Lm Vi = LmVj m v·I = L l1lv/ "f' a, angle with rad ~ reference (x) axis Sample I-d problem: Two bodies collide, stick a ' s together and move away ITom the collision site lig 22 00 rad/5 angular velocity Conservation of momentum: \8 \ P P 2i el (' I rad/5 s I m ml Vii angular acceleration P Hint: fUllctions as the effective mass for rotational energy and momentum Twirling thin rod of length, L 1=li mL2 " "' "'2 .~ vf 7-; Equation: V, = rOO Data: r = 6.378 x 106 m 00 = 2IT Tad/day; day = 24 x 60 x 60 sec = 86,400 Convert 00 to SI: 00 = 21t rad/day x I day/86,400 s = 7.3 X 10 rad/s Calculate V,: X after collision 106 m x 7.3 x 10 rad/s v, = 465 m/s E Energy and Work I Goal: Examine the energy and work associated with forces acting on an object Key equations: a Kinetic energy: 7l1lvl; energy of motion b Work: Force acting over a distance • For F(x): Work = fF(x) dx ·For a constant force: W = F d cos8 = F x r ·8 is the angle between the F and r • W maximum for = (note: sin(8 = 0) = 1) f Impulse: g Momentum change: PH" = Pini' + H Rotation of a Rigid Object I Goal: Examine the rotation of a rigid body of a defined shape and mass Key variables and equations: a Center of mass: XO!l' Ycm Zcm I LIIl,X, X Lm = Em, LIIl,y, Y,."nI Lm, = Zem I =l.mR2 Rotating sphere of radius, R 1=1.mR2 Sample: J for bodies of mass Ill: fIg 2~ Sample: Determine the r for a spherical Earth assume uniform M ; Data: M = x I 24 kg, r = 6.4 x 106 m r = 1M r = = 9.8 P LIIl, = 3.00kg III 66 III Hint: The center of mass is nearer the heavier O.66m ,······ ·····················1 t I kg X 1037 kg m 11 00 f Torque: 1: = 100 = r x F (ang acceleration force) I Angular Momentum I Goal: Quantify the force, cnergy '( ~ e Rotational Energv = and momentum of rotating objects Key variables and equations a Angular momentum: L = /oo = rxp = frxvdlll ~ F b Torque: 1: = r x F = dLldl; note: vector cross product lig 25 J Static Equilibrium and Elasticity Case I: Examine several forces acting on a body • Guiding principles: Equilibrium is detined as: L force = & Ltorq ue = The point of balance is the center of mass P Hint: Evaluak each component; any net torce moves the object, any net torque rotates the object Sample: Beam balance For equilibrium: fig 26 Case Examine deformation of a solid body Key Equation: Stress = clastic modulus x strain: modulus: stress/strain = force/change (Hooke's Law) y = F,/A L11/ I" fig 27a ~ ,./ (' ~ e 1_/o-IMI- Note: Force F I is longiludina l • Shape Stress: Shear Modulus S co = •• F,/A Axlh fig 27b Note: Force F, is tangential to face A • Volume Stress: Bulk Modulus B B= F"IA L1V/V lig '27c v , _L ,'1 ,,,-!:.:' ' f F" O.33m center of mass x 6xl0 24 kg X (6.4xI0 m)2 - Em; + 1112 x2 = 0.00 + 2.00 = 2.00 kg = LIIl,Xi = 2.00kg III Xcm ~/ LIIl,0 , Sample: Calculate the center of mass for a I kg & a kg ball connected by a 1.00 m bar ball I: XI = 0.00, 1111 = I kg; 1111 XI = 0.00 kg m ball 2: Xl = 1.00 III = kg; 1111 x = 2.00 kg III Lm i = I kg + kg = kg Lilli xi = 1111 XI ~ Rotating cylinder of radius, R " " [= FM or fF(t)dl Hint: For a full rotation, s = 2ITr = circumference of a circle of radius I' Tangential acceleration & velocity: v, = r oo ; a , = r ex; along path of motion arc , Centripetal acceleration: ac = directed towar~s the center fig 19 Sample: Determine V, at the Earth's equator G _ ~ L-~ ~ • Linear (Tensile) Stress: Young 's Modulus Y before collision motion arc; s = r8 (8 in rad) roo = 6.378 + m2 V2i = (1111 + m2)vr ~ P V, = = Lilli I}, with ri about the ce nter of mass along a specific axis D a l ex ball ftg 13 b Moment of inertia: kg Note: Force F" is nomlal to face MECHANICS (continued) K , Universal Gravitation I Goal: Examine gravitational energy and force fig 28 Case 1: Bodies of mass MI & M2 separated by, Key equations: a GravitatIOnal Energy: U, r ·_· ~ MI··· ·Mz ~ GM, M , ' ,orce: ~ F.,=~ GMIM b Uravltatlona i> i ln ll = ; I I I A Descriptive Variables I Types: Transverse, longitudinal, traveling, standing harmonic a General form for transverse travding wavc: y = f(x - I'/) (to the right) or y =I(x + 1'1) (to the left) b General fornl of harmonic wave: y = Asin(kx - WI) orv =Acos(kx - w() :/t c Standing wave: Integral multiples of.1 fit the length of the '11 I OSCI atmg materia d' Y I d' Y d General wave equatIOn: dx' =pI dl ' I I I I c Acceleration due to gravity: g = G M(earth)/r = _ 6.67X 10 "m'kg ' X6 XI 0"kg _ Calculation (6.4 X I 0" In)' - 9.8 ms II b Weight 'Peri for a " I! oExact for monatomiC gas, modify for natural process molecular gast!s = Note: q in Coulombs and r in meters Superposition Principle: Forces and tields are composites of contributions from each charge r b Carnot's Law: For ideal gas: Cp - C v = R oLl.E=C Ll.TLl.H=C-Ll.T '~" = % r(!n ) b Kinetic energy for Ideal Gas: K = lRT c For real gas: Add terms for vibrations and rotations Ll.Suniv d Electric Field: E = q;.?,e F = x 109N q, (C )q, !C ) T(K) = T(°C) + 273.15 = 35 + 273.15 = 308.15 K Thermal Expansion of Solid, Liquid or Gas F Entropy & 2nd Law of Thermodynamics a Goal: Determine the change in the length I Goal: Examine the driving force for a process (L) or volume (V) as a function of Key Variables: a Entropy: S, thermal disorder; dS = dF temperature b S(univ) = S(system) + S(thermal bath) b Solid: LlL = uLl.T L Guiding Principle: nd Law of c Liquid: LlV = ~Ll.T Thermodynamics: V For any process, Ll.Suniv > 0; one exception : d Gas' Ll.V = (7; - T.)nR b e - charge on an electron; 1.6022 x 10 19 C c Coulomb' Law - electrostatic force: F = 4it: oVector direction defined bye " j) Hint: Calculation shortcut: B Temperature & Thermal Energy A Electric Fields and Electric Charge I Goal: Examine the nature of the field gt!nerated by an electric charge, and forces between charges Key Variables and Equations a Coulomb C: "ampere sl!c" of charge Conducting Sphere V(K) = tV(VaCllllm l F(lC) = IF(vacuum) K D Capacitance and Dielectrics I Goal: Study capacitors, plates with charge Q separated by a vacuum or dielectric material fi~ · ) Key Equations: a Capacitance, C = ~ , V is the measured voltage b Parallel plate capacitor, vacuum, with area A , spacing d: C = EO A ; E = Q A d t:u c Parallel plate capacitor, die lectric lC, with area A, A spacing d: C = Kto1 Capacitors in series: t Lt \AI ;: J C V l = 'r Z 1( Capacitors in parallel: C tot = I.Ci fI'l43 Ll.T=O C ~ V a Four steps in the cycle: two isothermal, two adiabatic; for overall cycle: Ll.E = and Ll.S = b Efficiency = I - ~." 1'", G) Two Capacitors in Series Tcold CJ I I I I I ctot = c I + c Cz CICz or C tot = C +C Two Capacitors in Parallel lCJc;l ill Ctot - cl+CZ I Z ELECTRICITY Be MAGNETISM (continued) ~ E Current and Resistance tIIIII 'liliiii I Goal: Examine the current, I, quantity of charge, Q, resistance, R; determine the voltage and power dissipated Key Equations: a Total charge, Q = It b V= IR , or R = T V c Resistors in Series: Rtot = ~Ri rt)! 44 d Resistors in Parallel: Two Resistors in Parallel lR.J!Q I I R to ; R;+Tz e Power = IV = 12R F Direct Current Circuit I Goal: Examine a circuit containing battery, resistors and capacitors; determine voltage and current properties Key Equations and Concepts: tlJ a EMF: Circuit voltage; if = Vh + IR; battery voltage Vh = 1r, r internal battery resistance b Junction: Connection of or more conductors c Loop: A closed conductor path d Resistors in series or parallel => replace with Rtot e Capacitors in series or parallel => replace with Ctot Kirchoff's Circui~ Rules a ForanyLoop:~V = ~/R; P Hint: Conserve energy b For any Junction: ~ I = 0; " b Magnetic Flux: +/11 = b.lndex of refraction: 1/; Reflection and Refraction Incident Ra) f, = speed of light in medium Renectcd Ru)' c Light as electromagnetic wave: A{= C Light characterized by "color" or wavelcngth d Light as particle: e = hI; energy of photon Reflection and Refraction of Light fig 47 a Law of Reflection: III = Ilr fig 48 b Refraction: Bending of light ray as it passes fi'om 1/ I to 112 ·Snell's Law: IIlsinlll = 1l2sinIl2.1lJ, 172 are the indices of refraction of two materials fig 49 Normal ; , 1,/ ~ Normal c Internal Reflectance: sinll, = !!2 II, Light passing from material of higher n to a lower n may be trapped in the material Polarized light: E field is not spherically symmetric a Examples: Planellinear polarized, circularly polarized b Polarization by reflection from a dielectric surface at angle Il, ; Brewster's Law: tanll, = _'!.! III B Lenses and Optical Instruments I Goal: Lenses and mirrors generale images of objects Key concepts and variables Lens and Mirror Properties a Radius of curvature: R = 2I + sign - sign b Optic axis: Line from base of object Parameters through center of lens or mirror f - - - - f - - - - - j - - - - - converging lens diverging lens I focal length concave mirror convex mirror c Magnification: M = 1+.1=.1 2.=_l! S I' f' s' h' real object virtual object real image virtual image h object size erect invcrt.:d h' image size erect inverted e Combination of thin lenses: I I I =7,+]; ~ c Force on charge, q and v, moving in B: F= qv x B = qvBsinll ; v parallel to B => F = 0; v perpendicular to B => F = qvB ~ d Magnetic Moment of a Loop: M = 1A e Torque on a loop: 't = M x B f Magnetic Potential Energy: U = - M • B g Lorentz Force: Charge interacts with E and B; F=qE+qvxB H Faraday's Law and Electromagnetic Induction Key Equations: = if = s object dis! s' image dist d Laws of Geometric Optics: f BdA I Faraday's Law: Induced EMF: A Basic Properties of Light I Goal: Examine Iight and its interaction with matter Key variables: a c: speed of light in a vacuum f PHint: Conserve charge; define "+" flow Ii~ 4" G Magnetic Field: B I Key concepts: • a Moving ch Magnetic Field B Z ~ BEHAVIOR OF LIGHT f E ds = {J,J; or = J, +}; ~ Sample Guidelines for ray tracing: a Rays that parallel optic axis pass through "j''' b Rays pass through center of the lens unchanged c Image forms at convergence of ray tracings Sample ray tracings: fig 50, a,b,c Plane mirror: Law of Reflection Converging Lens s'_1 ',: :, ~ " -d J m ldt Biot-Savart Law: Conductor induces B; current 1, ).ill r length dL : dB = 47f1dL x ? f; f Sample: Long conducting wire: B(r) = I Electromagnetic Waves- Key Equations and Concepts: I Transverse Band E fi elds; £ = c 2.c = _IB ,j J L oeu Electromagnetic Wave: c = fA )' EIB Wave x J Maxwell's Equations: E • dA = ~ ; I Gauss's Law for Electrostatics: key: Charge gives rise to E ' " Gauss's Law for Magnetism: B • dA = 0; key: Absence of magnetic charge Ampere-Maxwell Law: d4J ~ B • ds = flo! + flo£o df; ¢ rf key: Current + change in electric flux => B Faraday's Law: 1E • dS = _ d: Spherical Concave Mirror C Interference of Light Waves I Goal: Examine constructive and destructive interference of light waves Key Variables and Concepts: a Constructive interference: fig 51 o b Destructive interference: fig 52 c Huygens' Principle: Each portion of wave front acts as a _ _ _ _-*' _ _ ">j source of new waves DifIraction of light, from a grating with spacing d, produces an interference pattern; dsinll = inA; (m = 0, 1,2,3, ) Single-slit experiment, slit width a; destructive interference for sinll = I7~A; (111 = 0, ±J, ±2 ) X-ray diffraction from a crystal with atomic spacing d; 2dsinll = 111"-; (m ~ Y J' J' YI J'2 e ~ / / ~r_ -t-i~ -r ; -x key: Change in magnetic flux => E 0, 1,2,3, ) : ~ y -+ ~ ~ ~~~ x " m ; ~ / / = ~~ o