MIT 2.71/2.710 Optics10/19/05 wk7-b-1Today’s summary• Energy / Poynting’s vector• Reflection and refraction at a dielectric interface:– wave approach to derive Snell’s law– reflection and transmission coefficients– total internal reflection (TIR) revisited• Two-beam interferenceMIT 2.71/2.710 Optics10/19/05 wk7-b-2EnergyMIT 2.71/2.710 Optics10/19/05 wk7-b-3The Poynting vectorEBSBEBES ×=×=0201εµcso in free spacekS ||S has units of W/m2so it representsenergy flux (energy perunit time & unit area)MIT 2.71/2.710 Optics10/19/05 wk7-b-4Poynting vector and phasors (I)20021ESEEBEkBBESεωωεcckc=⇒==⇒×=×=For example, sinusoidal field propagating along z()()tkzEctkzEωεω−=⇒−=22000coscosˆSxERecall: for visible light, ω~1014-1015HzMIT 2.71/2.710 Optics10/19/05 wk7-b-5Poynting vector and phasors (II)Recall: for visible light, ω~1014-1015HzSo any instrument will record the averageaverage incident energy flux∫+=TtttTd1SSwhere T is the period (T=λ/c) Sis called the irradianceirradiance, akaintensityintensityof the optical field (units: W/m2)MIT 2.71/2.710 Optics10/19/05 wk7-b-6Poynting vector and phasors (III)21d)(cos)(cos22=−=−∫+ttkztkzTttωω20021Ecε=SFor example: sinusoidal electric field,Then, at constant z:()()tkzEctkzEωεω−=⇒−=22000coscosˆSxEMIT 2.71/2.710 Optics10/19/05 wk7-b-7Relationship between E and BEBk()EkBkxEBErk×=⇒−≡∂∂×≡∇×⇒=∂∂−=×∇−⋅ωωω1 and eˆ where0itiEttiVectors k, E, B form aright-handed triad.Note:free space or isotropic media onlyMIT 2.71/2.710 Optics10/19/05 wk7-b-8Poynting vector and phasors (IV)Recall phasor representation:()()() ( ) ( ) e : phasor""or amplitudecomplex sincos,ˆ cos,φφωφωφωiAtkziAtkzAtzftkzAtzf−−−+−−=−−=Can we use phasors to compute intensity?MIT 2.71/2.710 Optics10/19/05 wk7-b-9Poynting vector and phasors (V)Consider the superposition oftwo two fields of the samesame frequency:()()() ( )φωω−−=−=tkzEtzEtkzEtzEcos,cos,202101()()φεεcos22...d 201022021002210EEEEctEETcTtt++==+=∫+SNow consider the two corresponding phasorsphasors:φiEE−e2010()φεεφcos22...e220102202100220100EEEEcEEci++==+−and the quantityMIT 2.71/2.710 Optics10/19/05 wk7-b-10Poynting vector and phasors (V)Consider the superposition oftwo two fields of the samesame frequency:()()() ( )φωω−−=−=tkzEtzEtkzEtzEcos,cos,202101()()φεεcos22...d 201022021002210EEEEctEETcTtt++==+=∫+SNow consider the two corresponding phasorsphasors:φiEE−e2010()φεεφcos22...e220102202100220100EEEEcEEci++==+−and the quantity= !!MIT 2.71/2.710 Optics10/19/05 wk7-b-11Poynting vector and irradianceSummarySummary200 ;1ESBESεµc=×=∫+=TtttTd1SSPoynting vectorIrradiance (or intensity)220phasor or phasor2∝= SSεc(free space or isotropic media)MIT 2.71/2.710 Optics10/19/05 wk7-b-12Reflection / RefractionFresnel coefficientsMIT 2.71/2.710 Optics10/19/05 wk7-b-13Reflection & transmission @ dielectric interfaceMIT 2.71/2.710 Optics10/19/05 wk7-b-14EikikiEiReflection & transmission @ dielectric interfaceMIT 2.71/2.710 Optics10/19/05 wk7-b-15Reflection & transmission @ dielectric interfaceI. Polarization normal to plane of incidenceI. Polarization normal to plane of incidence()[]()[]txykiEtiEiiiiiiiωθθω−+−==−⋅=)sincos(expˆ expˆ00zrkzEIncident electric field:()[]()[]txykiEtiErrirrrrωθθω−++==−⋅=)sincos(expˆ expˆ00zrkzEReflected electric field:()[]()[]txykiEtiEtttttttωθθω−+−==−⋅=)sincos(expˆ expˆ00zrkzETransmitted electric field:where:vacuumvacuum2 ,2λπλπttiinknk ==MIT 2.71/2.710 Optics10/19/05 wk7-b-16Reflection & transmission @ dielectric interfaceI. Polarization normal to plane of incidenceI. Polarization normal to plane of incidence()[]()[]()[]txykiEtxykiEtxykiEEEEttttrririiiitriωθθωθθωθθ−+−=−+++−+−⇒==+)sincos(exp )sincos(exp )sincos(expl)(tangentia l)(tangential)(tangentia000Continuity of tangential electric fieldat the interface:But at the interface y=0 so()[]()[]()[]txkiEtxkiEtxkiEtttririiiωθωθωθ−=−+−sinexp sinexp sinexp000MIT 2.71/2.710 Optics10/19/05 wk7-b-17Reflection & transmission @ dielectric interfaceI. Polarization normal to plane of incidenceI. Polarization normal to plane of incidenceContinuity of tangential electric fieldat the interface:()[]()[]()[]txkiEtxkiEtxkiEtttririiiωθωθωθ−=−+−sinexp sinexp sinexp000Since the exponents must be equalfor all x, we obtainttiittiirinnkkθλπθλπθθθθsin2sin2sinsinand vacuumvacuum=⇔==MIT 2.71/2.710 Optics10/19/05 wk7-b-18Reflection & transmission @ dielectric interfaceI. Polarization normal to plane of incidenceI. Polarization normal to plane of incidenceContinuity of tangential electric fieldat the interface: riθθ=ttiinnθθsinsin =law of reflectionSnell’s law of refractionso wave description is equivalentto Fermat’s principle!! ☺MIT 2.71/2.710 Optics10/19/05 wk7-b-19Reflection & transmission @ dielectric interfaceI. Polarization normal to plane of incidenceI. Polarization normal to plane of incidence()[]txykiEiiiiiωθθ−+−=)sincos(expˆ0zEIncident electric field:()[]txykiErrirrωθθ−++=)sincos(expˆ0zEReflected electric field:()[]txykiEtttttωθθ−+−=)sincos(exp0zETransmitted electric field:Need to calculate the reflected and transmitted amplitudes E0r, E0t i.e. need twotwo equationsMIT 2.71/2.710 Optics10/19/05 wk7-b-20Reflection & transmission @ dielectric interfaceI. Polarization normal to plane of incidenceI. Polarization normal to plane of incidenceContinuity of tangential electric fieldat the interface gives us one equation:()[]()[]()[]txkiEtxkiEtxkiEtttririiiωθωθωθ−=−+−sinexp sinexp sinexp000which after satisfying Snell’s law becomestriEEE000 =+MIT 2.71/2.710 Optics10/19/05 wk7-b-21Reflection & transmission @ dielectric interfaceI. Polarization normal to plane of incidenceI. Polarization normal to plane of incidencel)(tangentia l)(tangential)(tangentiatriBBB==+The second equation comes from continuity of tangential magnetic fieldat the interface:Recall0000cossinˆˆˆ11EkkθθωωzyxEkB=×=MIT 2.71/2.710 Optics10/19/05 wk7-b-22Reflection & transmission @ dielectric interfaceI. Polarization normal to plane of incidenceI. Polarization normal to plane of incidenceSo continuity of tangential magnetic field Bxat the interface y=0 becomes:tttrriiiitttrriiiiEnEnEnEkEkEkθθθθθθcoscoscoscoscoscos000000=−⇔=−MIT 2.71/2.710 Optics10/19/05 wk7-b-23Reflection & transmission @