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Magnetostatics: Stationary Charges ➾ Constant electric fields: electrostatics. Steady Currents ➾ Constant magnetic fields: magnetostatics. Static electric fields characterized by E or D and related according to D = εE. Static magnetic fields characterized by H or B and related according to B = μH. B=Magnetic flux density, H = Magnetic field intensity two major laws governing magnetostatics: 1. Bio-Savart law 2. Ampere’s circuital law Just as Gauss’s law is special case of Coulombs law, Ampere’s law is a special case of Bio-Savart’s law and is easily applied in problems involving symmetrical current distribution. Biot-Savart Law • Currents, i.e. moving electric charges, produce magnetic fields. There are no magnetic charges Ampere’s Circuital Law** in Integral Form Ampere’s Circuital Law -“the circulation of the magnetic flux density in free space is proportional to the total current through the surface bounding the path over which the circulation is computed.” B dl 0 I encl C The line integral is around any closed contour bounding an open surface S. I encl is current through S: I encl J dS S Where J is defined as current density B 0 J Note: ** Ampere’s circuital Law is Maxwell’s IV equation. H J 0 This shows that magnetostatic field is nonconservative in nature. Electricity Magnetism ELECTRIC FLUX Electricity Magnetism MAGNETIC FLUX Magnetic lines of force produced in the medium surrounding electric currents or magnets and is expressed as surface integral of the magnetic flux density. B dS s weber (Tesla m2) B dS 0 B .B 0 Gauss’ law for magnetic fields** says there can not be a net magnetic flux through the surface since there can be no net magnetic charge enclosed by the surface. magnetic monopoles do not exist. ** Maxwell’s II equation 1 If the magnetic flux density in a medium is given by B cos aˆr , r what is the flux crossing the surface defined by Solution: 4 4 ,0 z 2m. 1 B cos âr , r B dS Where dS r d dz âr s 1 cos âr rd dz âr sr 2 4 cos d dz 0 4 = 2.83 wb Laws of Electromagnetism 1. Gauss Law of electrostatics Electric Flux E E dS S Qenclosed o Over any closed surface S. 2. Gauss Law of magnetostatics Magnetic Flux B B dS 0 Over any closed surface S. S Consider a surface S around one of the poles of a magnet. Magnetic lines of force travel inside the magnet from one pole and emerges out from the other pole, so the flux enters through the surface and the same flux emerges out of the surface. Therefore the net flux over the surface S will be zero. 3. Ampere’s LAw B dl 0 I encl C Faraday’s law: Changing magnetic field induces electrical current (a) When a magnet is moved toward a loop of wire connected to a galvanometer, the galvanometer deflects as shown, indicating that a current is induced in the loop. (b) When the magnet is held stationary, there is no induced current in the loop, even when the magnet is inside the loop. (c) When the magnet is moved away from the loop, the galvanometer deflects in the opposite direction, indicating that the induced current is opposite that shown in part (a). 3. Faraday’s Law ** Changing magnetic field gives rise to electric current. Induced emf in the loop, due to changing magnetic flux. B t i.e. rate of change of magnetic flux is the e.m.f. induced in the circuit. E ΦB If q0 is the charge taken around the loop.Then Force F q0 E Now work done in taking the charge around the loop will be E B (Magnetic field inward) dB dt dW F dl W q 0 E 2 r q 0 q 0 E 2 r E dl P B t B t E dl P E dl P B t t B.ds S Integral form Note: ** Faraday’s Law is Maxwell’s III equation. or E B t Differential form 1. 2. E dS S Qenclosed o B dS 0 S 3. 4. B E dl t P B dl 0 i P Asymmetry in the above laws 1. In equation s1 and 2, the R.H.S. of equation 2 contains no value, because of the non existence of the magnetic monopole. 2. In equation 4 R.H.S. contains μ0 i → μ0(dq/dt), but there is no such term in equation 3 corresponding to μ0 i. This asymmetry is also because of nonexistence of magnetic monopole that is no magnetic currents. 3. In equation 3 changing magnetic flux gives rise to electric field. There is no such corresponding term relating to changing electric flux producing magnetic field in equation 4. This led Maxwell to introduce a new term corresponding to changing electric flux in equation 4. The corresponding term E B dl 00 t P ΦE B E 0 ( i id ) B dl 0i 00 t P id displacement curent 0 E t Maxwell’s Equations in Integral Form 1. Qenclosed E d S o 3. B dS 0 S E dl P S 2. E B 4. B t E B d l i 0 0 t P Modification to Ampere’s Law: -Ampere’s law must be wrong! B dl I 0 enc - it depends on what “enclosed” means Surface S1 encloses a current Surface S2 does not! What if we moved S1 into the gap? How can we modify the rule to handle all situations? Ampere’s Law (constant currents): B.dl I Ampere’s Law for constant currents. 0 enc What about currents which are not continuous? Displacement current in a capacitor E-field increasing as Q increases! I I +Q charge deposits on the plate The capacitor holds a charge Q over the two plates. How can there be a current emerging from the capacitor plates? -Q charge induced by E-field Modified Ampere’s Law d B . dl I I E . dA B . dl I 0 enc d 0 0 0 enc dt To see how the displacement current comes about, one has to consider the electric flux through the capacitor’s plate (Gauss’s Law). E.dA E Qenc 0 Qenc 0 E.dA Q increases on the capacitor, the electric flux also increases at the same rate. dQenc d d 0 E 0 dt dt dt E.dA I d CONTINUITY EQUATION From the principle of charge conservation, the time rate of decrease of charge with in a given volume must be equal to the net outward current flow through the surface of the volume. Thus current Iout coming out of the closed surface is Qin is the total charge dQin enclosed by the closed I out J dS dt surface. Using the divergence theorem J dS Jdv S v dQin d v and v dv dv dt dt v t v v dv Jdv v v t v J t This equation is called continuity equation. It is derived from principle of conservation of charge and shows that there can be no accumulation of charge at any point. v For steady current 0 t J 0 Hence the total charge leaving a volume is the same as the total charge entering it. Maxwell’s modification of Ampere’s Law B dl I From Ampere's Law 0 encl where I encl P J dS S Applying Stroke's theorem to left-hand side of above equation 0 I encl B dl B dS L S B dS 0 J dS S Differential form of Ampere’s Law S B 0J Taking divergence of above equation B 0 0 J Since R.H.S. of above equation is not zero**, so let us apply Continuity equation and Gauss’s law (Electrostatics) we get v E J 0 E 0 t t t E J 0 0 t Hence B 0 J 0 0 E or . J 0 0 t E t E E The term 0 is called as displacement current density . J d 0 t t Hence modified Ampere’s Law (differential form) B 0 J J d **Note: Fundamental theorem of vector analysis; B 0 Exercise: Write Maxwell’s equations in integral form and obtain their differential form by using vector analysis. Write statement and physical significance of every equation. Maxwell’s Equations in Integral Form 1. E dS Qenclosed S 2. B dS 0 S o “Gauss Law in Electrostatics ” : Electric flux coming out from the surface of the body is equivalent to the charge enclosed by the body “Gauss Law in Magnetostatics ” : Magneticflux coming out from the closed surface of the body is zero as no magnetic monopole exist . “Faraday'sLaw”:Changing Magnetic flux produce B electric current or field in a closed loop.where 3. E dl t L Magnetic flux φB = B.ds S “Modified Ampere’s Law”or “Maxwell-Ampere’s Law”: Changing electric flux can produce magnetic E 4. B dl i 0 field in a discontinuous circuit to hold Ampere’s t L circuital Law.where Electric flux φE = E.ds S 0 Proof: Differential form Use equation 1 and apply Gauss divergence theorem E dS ( E) dv V S ( E) dv Qenclosed o V dv V 0 E 0 Use equation 2 and apply Gauss divergence theorem B dS ( B) dv 0 V S B 0 Use equation 3 and apply Stoke’s theorem E dl ( E ) dS P S B ( E) dS - t t dS S S E - B t Use equation 4 and apply Stoke’s theorem B dl ( B ) dS P S E ( B ) d S JA 0 0 t S ( EA ) ( B ) dS 0 JA 0 t S E ( B ) d S J dS 0 0 t S S E B 0 J 0 t Since D 0E Where A dS D B 0 J t Curl of B is due to current flow and a changing electric field. Maxwell's Equations with Physical Interpretation 1. Relates net electric flux to net enclosed electric charge. Qenclosed E dS o E S v 0 2. Relates net magnetic flux to net enclosed magnetic charge. B dS 0 B 0 S 3. Relates induced electric field to changing magnetic flux. B E dl t P E - B t 4. Relates induced magnetic field to changing electric flux to and to current. E B dl 0 i 0 t P D B 0 J t Case 1: Maxwell equations in free space* (no free charges and no currents) q 0 0, i 0 J 0, E 0 / 0 B 0 B E t E E B 00i0 0 0 t t * Helpful to understand Electromagnetic waves in free space. Maxwell’s Equations in linear media (Perfect dielectric) Inside matter, but in regions where there is no free charge or free current, 1. D 0 2. B 0 If the medium is linear 3. 4. 1 B E t H D t D E, H B That is homogeneous ( and do not vary from point to point), and isotropic (in which μ and ε are invariant with field orientation )Maxwell’s equation reduces to B 1. E 0 E 3. t 2. B 0 4. B E t •Derivation of Electromagnetic wave equation •Properties of E.M. Waves. •E is prependicular to k (direction of propagation); k.E=0 •B is prependicular to k (direction of propagation); k.B=0 •E is prependicular to B too; k E=? •Wave Impedance E0/B0=c •Poynting Theorem. Electromagnetic waves in free space or vacuum Take curl of equation 3 B t E E 2E E - B 2 E t 2E B t 2E o o 2 t 2E 2 E oo 2 t Take curl of equation 4 E B 00 t B B 2B E B o o t 2B oo E t 2B o o 2 t 2 B 2B oo 2 t 2 26 2 E 2E o o 2 t 2B 2 B o o 2 t 1 v 2 oo Standard Wave equation: 2 1 f 2 f 2 2 v t 1 v oo o 4 107 weber / amp m o 8.542 1012 Farad / m v 3 108 m / sec c Thus we conclude that light is electromagnetic in nature with electric vector E and magnetic vector B oscillating as a wave and propagating with a velocity of light in free space . Maxwell’s equations also imply that empty space supports the propagation of electromagnetic waves, traveling with speed of light. 27 •Derivation of Electromagnetic wave equation •Properties of E.M. Waves. •E is prependicular to k (direction of propagation); k.E=0 •B is prependicular to k (direction of propagation); k.B=0 •E is prependicular to B too; k E=? •Wave Impedance E0/B0=c •Poynting Theorem. Solution of Electromagnetic waves in free space 2E E o o 2 t 2B 2 B o o 2 t 2 The harmonic solutions to the wave equations E Eoeikr t i kr t B B e o Where, k=2π/λ is propagation vector, ω=2πν is angular frequency where Eo and Bo (amplitudes of wave) space and time independent vectors but may in general be complex. E(r, t) Eo e ikr t E o E ox aˆ x E oy aˆ y E oz aˆ z k k x aˆ x k y aˆ y k z aˆ z r xaˆ x yaˆ y yaˆ z Prove-Transverse Nature k is perpendicular to E and B. k E 0 or k B 0 and E , B and k are orthogonal 29 .E (aˆ x aˆ y aˆ z ) Eoeikr t x x x k .r (k x aˆ x k y aˆ y k z aˆ z ).( xaˆ x yaˆ y yaˆ z ) kx x k y y kz z .E i (k x Eox k y Eoy k z Eoz )eikr t i (k x aˆ x k y aˆ y k z aˆ z ).( Eox aˆ x Eoy aˆ y Eoz aˆ z )e ikr t ik .E Similarly, B 0 B ik B o eikr t ik B =0 Now Prove E , B and k are orthogonal Proof: B E i k E and , iB t ik E iB aˆ x E x Ex aˆ y y Ey aˆ z z Ez where, E x E ox e i ( k .r t ) E y E oy e E z E oz e Ez E y Ez Ex E aˆ x aˆ y z z x y i ( k .r t ) i ( k .r t ) Similarly from equation 4, we can show that E B 0 0 t ik B i 0 0E or k H D i.e. E is perpendicular to the Plane formed by k and B !! Thus, k, E and B vectors are mutually perpendicular to each other. k .E 0 (i.e. E is perpendicular to k) k .B 0 (i.e. B is perpendicular to k) (i.e. E, B and k are orthoganal) k B o o E k E B we conclude that electromagnetic field vectors E and B are both perpendicular to the direction of propagation vector k. This shows that Electromagnetic waves are transverse waves. 35 Plane Waves Ey Ey Bz kx x direction of propagation Bz Plane wave because the fields are uniform over every plane perpendicular to the direction of propagation (i.e. x = constant plane) as shown in the figure below. E y E o cos kx t E y Eo ei kx t Bz Bo ei kx t or B z B o cos kx t E y E o sinkx t or B z B o sinkx t 36 Show that E and B of plane wave are in same phase at any time in space E y Eoei kxt From equation 3 Since But E Ey B Bz k k x Bz Boei kxt E k x E y Bz E y Bz kx Since /k is a real number, the electric and magnetic vectors are in phase; thus if at any instant, E is zero then B is also zero, similarly Bo k E B 2, k E i kx t i kx t Bz Boe E y Eoe B t y cBz Eo c Eo Eo o o c H o Bo 2 2 c 2 k Eo 1 c Bo k o o o o o 376.72 o o o when E attains its maximum value, B also Here ηo is universal const.and called as attains its maximum value, etc. Both Ey and Bz are in same phase. characterstic or intrinsic waveimpedance of the freespace. 37 Numerical In free space the Electric field is given as E 10 sin (2 x 100t ) ˆj. Determine D, B and H by using Maxwell’s equations. Sol: Wave is propagating along x direction. (1) (2) (3) D 0 E 10 0 sin (2 x 100t ) ˆj B Using E , or k E B t B ˆ 20 Cos (2 x 100t ) k , t 1 B Sin (2 x 100t ) kˆ 5 B 1 H Sin (2 x 100t ) kˆ 0 5 0 Ey Bz kx Energy in EM Waves: Poynting Theorem and Poynting Vector Energy in EM Waves The energy densities (energy per unit volume) associated with electric field and magnetic fields are: Electric field Energy Density Magnetic field Energy Density UE UB 1 0 E 2 2 1 B2 2 0 J/m 3 J/m 3 In vacuum at any moment of time UB 1 B2 1 E2 1 2 = E =U E 0 2 2 0 2 0 c 2 Therefore, total energy density UTot = 0 E 2 0 EBc J/m3 How much energy passes through 1 m2/sec??? Like energy flux!!!!! Energy Carried by Electromagnetic Waves Total energy passes from volume c m3 through a closed surface of 1m2 per unit time Utot c =0EBc2 =EB/0 J/m2 sec C m Which is poynting vector 1m2 1 S=EXB/0 Thus, S represents power per unit area. The direction of S is along the direction of wave propagation. The units of S are J/(s·m2) =W/m2. The magnitude S represents the rate at which energy flows from the volume through a unit surface area perpendicular to the direction of wave propagation. Energy Carried by Electromagnetic Waves Electromagnetic waves carry energy, and as they propagate through space they can transfer energy to objects in their path. The rate of flow of energy in an electromagnetic wave is described by a vector S, called the Poynting vector.* 1 S = E B 0 EB or S = 0 E2 cB2 or S = = 0 c 0 y E B EM waves are sinusoidal. E y E o cos(kx ωt) B z B o cos(kx ωt) S c x z *J. H. Poynting, 1884. Energy Carried by Electromagnetic Waves The average of S over one or more cycles is called the I Saverage wave intensity I. E o Bo E o2 cBo2 cμ o H o2 c o E o2 S 2μ o 2μ o c 2μ o 2 2 The time average of cos2(kx - t) is ½ The magnitude of S is the rate at which energy is transported by a wave across a unit area at any instant: Thus, energy power time S= = area area instantaneous instantaneous power I= S = area average Relation between S and U When we average this instantaneous energy density over one or more cycles of an electromagnetic wave, we again get a factor of from the time average of cos2(kx - t). 1 2 uE = 0Emax , 4 2 max 1B uB = , and 4 0 2 2 1 Emax 1 cBmax Saverage = S = = 2 0 c 2 0 uTotal 2 1 1 Bmax 2 = 0Emax = 2 2 0 S =c u . The intensity of an electromagnetic wave equals to the ‘c’ times of the average energy density. Momentum Transport- Radiation Pressure Relativistic energy and Momentum 2 E= Energy 2 (pc) momentum + m c 2 2 o rest mass energy For light mo=0 E=U=pc When em wave strikes a surface , its momentum changes. Rate of change of momentum is equal to Force. This force acting per unit area of the surface exert a pressure, called RADIATION 1 dp 1 d U A dt A dt c PRESSURE dU 1 1 dt S c A c Total Absorption dp 1 dU 1 F= IA dt c dt c F 1 P I (radiation pressure) A c Total Reflection dp 2 dU 2 F= IA dt c dt c F 2 1 pascal (Pa) = 1 N/m2 = P I (radiation pressure) 1 kg/(m·s2) A c Total Absorption dp 1 dU 1 F= IA dt c dt c F 1 P I (radiation pressure) A c Total Reflection dp 2 dU 2 F= IA dt c dt c F 2 1 pascal (Pa) = 1 N/m2 = P I (radiation pressure) 1 kg/(m·s2) A c Ex: a radio station on the surface of the earth radiates a sinusoidal wave with an average total power of 50 kW. Assuming the wave is radiated equally in all directions above the ground, find the amplitude of the electric and magnetic fields detected by a satellite 100 km from the antenna. All the radiated power passes through the hemispherical surface* so the average power per unit area (the intensity) is P power I= = 2 area av 2R = 7.96 10-7 W m2 Satellite R Station 2 1 Emax I= S = Emax = 20 cI 2.45 10-2V / m Exercise: calculate the average 2 0 c energy densities associated with the electric and magnetic field. Emax = 8.17 10-11 T B max = -15 3 u = u 1.33 10 J / m B E c *In problems like this you need to ask whether the power is radiated into all space or into just part of space. If the earth receives 2cal min-1cm-2 solar energy, what are the amplitudes of electric and magnetic fields of radiation? E0=1026.8V/m H0=2.726A-turn/m The electric field in an em wave is given by E=E0sin(t-x/c), where E0=100 N/Coul. Find the energy contained in a cylinder of cross section 10-3m2 and length 100 cm along xaxis. U=4.42510-11 Joule Poynting’s Theorem “Conservation of Energy” Chapter 8, Page 346, Griffith Rate of work in the system System of q and I Applied Lorentz force E J dv Power flow V 1 1 E 2 H 2 dv t 2 2 V Rate of decrease in stored energy E H ds S Poynting’s Theorem “Conservation of Energy” Chapter 8, Page 346, Griffith “The work done on the charges by the electromagnetic force is equal to the decrease in energy stored in the field, less the energy that flowed out through the surface” 1 2 1 2 V E J i dv t V 2 E 2 H dv S E H d s the total electromagnetic represent the total energy generated (Rate of electromagnetic energy flow of instantaneous power out of the volume stored in the volumeV. V through the surfaceS. work done) by the sources in the volumeV. Differential form of Poynting’s Theorem 1 2 1 2 V E J dv t V 2 E 2 H dv U mec V t dv t V U EM dv E H ds S S ds S V t (U mec U EM )dv V .S dv .S (U mec U EM ) t Shows Conservation of Energy Refractive index of the medium is defined as In a dielectric medium, r o Since dielectric is non magnetic, n r n r r 0 0 r o r 1 v 1 1 c μo n μo o n That is the speed of electromagnetic waves in an isotropic dielectric is less than the speed of electromagnetic waves in free space. Exercise: The relative permittivity of distilled water is 81. Calculate refractive index and velocity of light in it . Boundary conditions, Page 333, Ch. 7 • If there is no free charge or free current at the interface of two medium, then 1. 1 E 2 E 1 1 2 3. E E // 2 2 Incident E E// 2. B B // 1 4. // 1 B 1 // 2 B 2 E┴ Reflected Medium-1(ε1 , µ1) Medium-2(ε2 , µ2) Refracted Reflection and Transmission at Normal incidence Reference Introduction to Electrodynamics By D. J. Griffith Reflection and Transmission at Normal incidence Incident wave Transmitted wave 1 1 2 2 Reflected wave Suppose yz plane forms the boundary between two linear media. A plane wave of frequency ω traveling in The x direction (from left) and polarized along y direction, approaches the interface from left (see figure) Reflection and Transmission at Normal incidence E I ( x, t ) E0 I exp i k1 x t ˆj B I ( x, t ) 1 E0 I exp i k1 x t kˆ v1 1 1 Reflected wave Transmitted wave 2 2 Reflection and Transmission at Normal incidence E I ( x, t ) E0 I exp i k1 x t ˆj B I ( x, t ) 1 E0 I exp i k1 x t kˆ v1 1 1 E R ( x, t ) E0 R exp i k1 x t ˆj B R ( x, t ) 1 E0 R exp i k1 x t kˆ v1 Transmitted wave 2 2 Reflection and Transmission at Normal incidence E I ( x, t ) E0 I exp i k1 x t ˆj B I ( x, t ) 1 E0 I exp i k1 x t kˆ v1 1 1 E R ( x, t ) E0 R exp i k1 x t ˆj B R ( x, t ) 1 E0 R exp i k1 x t kˆ v1 E T ( x, t ) E0T exp i k2 x t ˆj BT ( x, t ) 1 E0T exp i k2 x t kˆ v2 2 2 Reflection and Transmission at Normal incidence Incident wave E I ( x, t ) E0 I exp i k1 x t ˆj 1 B I ( x, t ) E0 I exp i k1 x t kˆ v1 E R ( x, t ) E0 R exp i k1 x t ˆj Reflected wave 1 B R ( x, t ) E0 R exp i k1 x t kˆ v1 E T ( x, t ) E0T exp i k2 x t ˆj Transmitted Wave 1 BT ( x, t ) E0T exp i k2 x t kˆ v2 Reflection and Transmission at Normal incidence At x=o the combined fields to the left EI+ER and BI+BR, must join the fields to the right ET and BT in accordance to the boundary condition. Since there are no components perpendicular to the surface so boundary conditions (i) and (ii) are trivial. However last two [(iii) & (iv)] yields: 1 D B1 E // B1 1 // x 0 D2 x 0 (i) x 0 B2 x 0 (ii) x 0 E x 0 (iii) // // B2 x 0 x 0 2 (iv) E0 R E0 I E0T 1 1 1 1 1 E0 I E0 R E0T 1 v1 v1 2 v2 1 1 1 1 1 E0 I E0 R E0T 1 v1 v1 2 v2 E0 R E0 I E0T (1) or , E0 I E0 R E0T (2) 1v1 where 2v2 Using (1) and (2) E0 I E0 R 1 1 1v1 E0T 2 v2 1 2 1n2 21 2 n1 2 E0 I and E0T 1 E0 R E0 I v1 If r 1(nonmagnetic media) then = v2 thus we have, E0 R v2 v1 2v2 E0 I and E0T E0 I v2 v1 v2 v1 The reflected wave is in phase if v2>v1 or n1>n2 and out of phase if v2<v1.or n1<n2 The real amplitudes are related by 2v2 v2 v1 E0 R E0 I and E0T E0 I v2 v1 v2 v1 c in terms of refractive index n= v 2n1 n1 n2 E0 R E0 I and E0T E0 I n2 n1 n2 n1 Reflected wave is 180o out of phase when reflected from a denser medium. This fact was encountered by you during Last semester optics course. Now you have a proof!!! Reflectance (R) and Transmittance (T) • Wave Intensity (average power per unit area is given by): 1 B0 E0 1 I vE02 2 2 • If μ1= μ2 = μ0, i.e μr=1 , then the ratio of the reflected intensity to the incident intensity is 2 2 I R E0 R n1 n2 R I I E0 I n1 n2 Where as the ratio of transmitted intensity to incident intensity is 2 2 IT 2 v2 E0T n2 2n1 4n1n2 T I I 1v1 E0 I n1 n1 n2 (n1 n2 ) 2 Use ε α (n)2 Traveling E and H waves in free space ( region 1) are normally incident on the interface with a perfect dielectric (region 2) for which r=3.0. Compare the magnitudes of the incident, reflected and transmitted E and H waves at the interface. Find out Reflectance and Transmittance . Check the result. Ans: NOTE: R+T=1 => conservation of energy E 0R E 0T 0.268; 0.732 E 0I E 0I H 0R H 0.268; 0T 1.268 H 0I H 0I R 0.072;T 0.928 Chapter 24, Optics by GHATAK 4th Ed. Oblique Incidence Case 1: E parallel to the plane of incidence The reflection of a plane wave with its electric vector parallel to the plane of incidence. The parallel polarization (or the p polarization) is also called the transverse magnetic (or the TM) polarization as the magnetic field is perpendicular to the plane of incidence. Case 1: E parallel to the plane of incidence 1 D B1 E1 H1 x 0 D2 x 0 x 0 B2 // x 0 E2 x 0 // x 0 H2 x 0 // // x 0 The z component of the electric field represents a tangential component which should be continuous across the surface. Thus at x=0 E1z+E3z=E2z Further, the normal component of D must also be continuous, and since D = E, 1E1x+ 1E3x= 2E2x The z component of the electric field represents a tangential component which should be continuous across the surface. Thus at x=0 Further, the normal component of D must also be continuous, and since D = E, Substituting for E20 from Eq. (15), The amplitude reflection coefficient , r||=E30/E10 The amplitude refraction coefficient , t||=E20/E10 For nonmagnetic media, 120=4x10–7 N A–2, and the expression for the amplitude reflection coefficient simplifies to Case 2. E perpendicular to the plane of incidence 1 D B1 E1 H1 x 0 D2 x 0 x 0 B2 // x 0 E2 x 0 // x 0 H2 x 0 // // x 0 The reflection and refraction of a plane wave with the electric vector lying perpendicular to the plane of incidence. (s-polarized) Since, the y axis is tangential to the interface, the y component of E must be continuous across the interface; consequently E10 + E30 = E20 the z -component of the magnetic field to be continuous, H10 cos1 H 30 cos1 H 20 cos 2 or k1 1 ( E10 E30 )cos1 k2 2 E20 cos 2 We summarize the amplitude reflection and transmission coefficients for the two cases; the results are valid for nonmagnetic media: Equations are known as the Fresnel equations. 2 2 2 Ex 1: Light is incident at 60o on a boundary separated by media of refractive index n1 = 1 and n2 = sqrt(3) . Find the amplitude of reflection coefficient r when electric field is perpendicular to the plane of incidence. n1 cosi n2 cost r n1 cosi n2 cost n1 sin i n2 sin t Ex 2: Consider a linearly polarized electromagnetic wave (with its electric field vector along y-direction) of magnitude 5 V/m propagating in vacuum. It is incident on a dielectric interface at x = 0 at an angle of incidence of 30o. The frequency associated with the wave is 6×1014 Hz and refractive index of the dielectric is 1.5. Show that R+ T=1. n1 cosi n2 cost r 0.2404 n1 cosi n2 cost 2cosi sin t n2 cos 2 2 t 0.7596 ; T t sin(i t ) n1 cos1 R rr ' 0.057796 T 4n1n2 cos1 cos 2 n1 cos1 n2 cos2 Hence Proved, 2 R T 1 0.942204