![]() 5 0° ° ° ° R T Parallel polarization Incidence angle, qi 1.0. Perpendicular polarization Incidence angle, qi 1.0. It’s always true that and in terms of irradiance (I, W/m2) using laws of reflection and refraction, you can deduce andģ0 Summary: Reflectance and Transmittance for an Air-to-Glass Interface N = n2/n1 = 1.5 total internal reflection - incident angle where RTM = 0 is: both and reach values of unity before q=90° total internal reflection ![]() N = n2/n1 = 1.5 RTM = 0 (here, reflected light TE polarized RTE = 15%) at normal : 4% normal grazing - at normal and grazing incidence, coefficients have same magnitude - negative values of r indicate phase change fraction of power in reflected wave = reflectance = fraction of power transmitted wave = transmittance = Note: R+T = 1 Occur when external reflection: internal reflection: TE waves TM waves n1 n2 amplitudes are related:ġ7 Fresnel equations TE waves TM waves Get all in terms of E and apply law of reflection (qi = qr): For reflection: eliminate Et, separate Ei and Er, and take ratio: Apply law of refraction and let :ġ8 Fresnel equations TE waves TM waves For transmission: eliminate Er, separate Ei and Et, take ratio… And together: Reflected and transmitted beams travel in different media (same frequencies different wavelengths!): which leads to the law of refraction:ġ4 Boundary conditions from Maxwell’s eqnsįor both electric and magnetic fields, components parallel to boundary plane must be continuous as boundary is passed TE waves electric fields: parallel to boundary plane complex field amplitudes continuity requires:ġ5 Boundary conditions from Maxwell’s eqnsįor both electric and magnetic fields, components parallel to boundary plane must be continuous as boundary is passed TE waves magnetic fields: continuity requires: same analysis can be performed for TM waves Incident and reflected beams travel in same medium same l hence we arrive at the law of reflection: True for any boundary point and time, so let’s take or hence, the frequencies are equal and if we now consider which means all three propagation vectors lie in the same plane Using diagram from Pedrotti3 boundary pointġ1 At the boundary point: phases of the three waves must be equal: Oscillating electric field y y x x Any polarization state can be described as linear combination of these two: “complex amplitude” contains all polarization infoġ0 Derivation of laws of reflection and refraction Plane of incidence: formed by and k and the normal of the interface plane k normalĨ Polarization modes (= confusing nomenclature!)Īlways relative to plane of incidence TE: transverse electric s: senkrecht polarized (E-field sticks in and out of the plane) TM: transverse magnetic p: plane polarized (E-field in the plane) M E E M perpendicular ( ), horizontal parallel ( || ), vertical E Y light is a 3-D vector field z x linear polarization circular polarizationħ …and consider it relative to a plane interface 5 T qi R T 0° ° ° ° Incidence angle, qi and the change in the phase upon reflectionĦ Let’s start with polarization… light is a 3-D vector field transmitted external reflection, R 1.0. We’ll also determine the fraction of the light reflected vs. ![]() Where c is the velocity of the propagating wave,ĥ and the change in the phase upon reflection Normal Law of reflection: qi qr n1 n2 Law of refraction “Snell’s Law”: qt Incident, reflected, refracted, and normal in same plane Easy to derive on the basis of: Huygens’ principle: every point on a wavefront may be regarded as a secondary source of wavelets Fermat’s principle: the path a beam of light takes between two points is the one which is traversed in the least timeģ Today, we’ll show how they can be derived when we consider light to be an electromagnetic waveĤ E and B are harmonic Also, at any specified point in time and space, 1 Chapter 23: Fresnel equations Chapter 23: Fresnel equations
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