EECS 117 Lecture 24: Oblique Incidence on Dielectrics University of California, Berkeley Oblique Incidence (TM waves) Ei Er Hi Hr ki kr θi θr n1,ε1,µ1 n2,ε2,µ2 x z θt kt Ht Et TE Waves (cont.) A transverse magnetic (TM) wave, also so called the Ppolarized wave, has the electric field oscillating in the plane of incidence and the magnetic field out of the plane. The plane of incidence is defined to be the one made by the incident wave vector (propagation direction vector) and the line normal to the surface of discontinuity. In general, a field (either electric or magnetic) of a wave can be described by r r r jk • r ˆ F ( x, z ) = F e f r r ˆ is a unit vector along the field F , k where is the wave vector, f r r and r is a general space vector, i.e., r = xxˆ + yyˆ + zzˆ Incident, Reflected and Transmitted Waves For the incident electric field, r r r r jk i • r jn1 kkˆi • r Ei ( x, z ) = Ei e eˆi = Ei e eˆi = Ei e jn1 k sin θ i x e jn1 k cos θ i z (cosθ i xˆ − sin θ i zˆ ) = Ei x xˆ + Ei z zˆ k is the wave number of the field in vacuum, and k1 = For the reflected and transmitted fields, ω v = ω c / n1 r Er ( x, z ) = Er e jn1k sin θ r x e − jn1k cosθ r z (− cos θ r xˆ − sin θ r zˆ ) r Et ( x, z ) = Et e jn2 k sin θt x e jn2 k cosθt z (cos θ t xˆ − sin θ t zˆ ) = n1k Boundary Conditions Applying the boundary condition for tangential electric field at z = 0, we have Eix + Erx = Et x Ei cos θ i e jn1k sin θi x − Er cos θ r e jn1k sin θ r x = Et cos θ t e jn2 k sin θt x Just like in the case of oblique incidence on a perfect conductor, this condition needs to be satisfied independent of x. This is possible by equating the exponents and coefficients separately: (1) n1k sin θ i = n1k sin θ r = n2 k sin θ t Ei cos θ i − Er cos θ r = Et cos θ t (2) Law of Reflection and Refraction The first equality in (1) gives the law of reflection, i.e.,θ i = θ r and the second equality in (2) gives the Snell’s law, namely, n1 sin θ i = n1 sin θ r = n2 sin θ t For most of the materials, which are non-magnetic, µ1= µ2= µ0. Then, sin θ t n1 ε 1µ1 ε1 ε 1µ 2 η 2 = = ≈ ≈ = sin θ i n2 ε 2 µ2 ε2 ε 2 µ1 η1 Total Reflection If n1 > n2 or η1 < η 2 , we have total reflection for incident angle larger than the critical angle θc , where n2 sin π / 2 ε2 −1 −1 η1 ≈ sin θ c = sin ≈ sin n1 ε1 η2 −1 For θi > θc ε1 sin θ t = sin θ i > 1 ε2 ε1 2 cos θ t = 1 − sin θ t = ± j sin θ i − 1 ε2 2 Total Reflection (cont.) The transmitted electric and magnetic fields are then equal to: r jn 2 k sin θ t x jn 2 k cos θ t z (cosθt xˆ − sin θt zˆ ) Et ( x, z ) = Et e e = Et e jβ tx x e jα tz z (cosθ t xˆ − sin θ t zˆ ) r H t ( x, z ) = H t e jn 2 k sin θ t x e jn 2 k cos θ t z yˆ = H t e jβ tx x e jα tz z yˆ where ε1 ε1 2 β tx = n2 k sin θ i , α tz = n2 k sin θ i − 1 ε2 ε2 The wave propagates in x direction only!! It does penetrate into medium 2, with its amplitude decaying exponentially inside the medium. This surface wave is called an evanescent wave. Reflection and Transmission Coefficients The equation (2) obtained from applying the boundary condition to the tangential electric field can be used to find the reflection and transmission coefficients. But we need one more piece of information, which we can get from applying the tangential magnetic field boundary condition: Hi e jn1k sin θ i x + Hr e jn1k sin θ r x = Ht e jn2 k sin θ t x Hi + Hr = Ht We know that H and E are related to each other through impedance of the medium. Thus, we have H i = − Ei / η1 , H r = − Er / η1 , H t = − Et / η 2 Coefficients (cont.) We need the negative signs r in r the equations because with our definition of direction of Er , Hrmust be negative according to the right hand rule so that E × H is pointing the same direction as does. Now we have two simultaneous equations: Ei cos θ i − Er cos θ r = Et cos θ t Ei / η1 + Er / η1 = Et / η 2 Solving these two equations, we can expressions for the reflection and transmission coefficients. ΓTM Er η1 cosθ i − η 2 cosθ t = = Ei η1 cosθ i + η 2 cosθ t Coefficients (cont.) τ TM Et η 2 2η 2 cos θ i = = (1 + ΓTM ) = Ei η 2 η1 cos θ i + η 2 cos θ t We can also define the coefficients in terms of the tangential components of the electric fields. They have expressions: ΓZ TM = τZ TM = E rx Ei x Et x Ei x − Er cosθ r η 2 cosθ t − η1 cosθ i = = η1 cosθi + η 2 cosθt Ei cosθ i Et cosθ t 2η 2 cosθ t = = Ei cosθ i η1 cosθ i + η 2 cosθ t Brewster Angle (TM Waves) There is a situation that we don’t have reflection off the boundary, when the numerator of ΓTM is equal to 0, or η1 cos θ i = η 2 cos θ t The incident angle that this happens is called Brewster angle. It can be shown that sin θ BTM = 2 1 1 + (η 2 /η1 ) 2 1 − µ 2ε1 / µ1ε 2 1 = 2 ≈ 1 + ε1 / ε 2 1 − (ε1 / ε 2 ) or θB TM ≈ tan −1 ε1 / ε 2 ≈ tan −1 n1 / n2 Oblique Incidence (TE Waves) For a TE wave, which has the electric field oscillating out of the plane of incidence and magnetic field parallel to the plane, we can obtain the same set of quantities as in TM waves quickly by swapping E and H: E → H , H → E with some cautions. In the TM case, we applied the boundary condition to the tangential electric field. Here we need to apply the condition to the tangential magnetic field. Because both conditions require the field to be continuous across the boundary, we have H i cosθi e jn1k sin θ i x − H r cosθ r e jn1k sin θ r x = H t cosθt e jn 2 k sin θ t x which also yield the laws of reflection and refraction. If the incident angle is larger than the critical angle (of course n1 > n2), the transmitted wave is evanescent. Boundary Conditions for TE Waves Analogous to the TM case, we obtain this from the boundary condition for tangential magnetic fields: H i cosθ i − H r cosθ r = H t cosθ t Just like what we had in TM case, we also have this equality due to the boundary condition for tangential electric fields: Ei e jn1 k sin θ i x + Er e jn1 k sin θ r x = Et e jn 2 k sin θ t x Ei + Er = Et The electric and magnetic fields are related: H i = − Ei / η1 , H r = − Er / η1 , H t = − Et / η 2 Boundary Conditions (cont.) So we have these two simultaneous equations in the TE case: (E i cosθ i − Er cosθ r ) /η1 = Et /η 2 cosθ t Ei + Er = Et These two equations yield the expressions of reflection and transmission coefficients: τ TE Er η 2 cosθ i − η1 cosθ t ΓTE = = Ei η 2 cosθ i + η1 cosθ t Et 2η 2 cosθ i = = 1 + ΓTE = Ei η 2 cosθ i + η1 cosθ t Since the electric fields are tangential to the boundary, ΓZ TE = ΓTE , τ Z TE = τ TE Brewster Angle (TE Waves) We can also define Brewster angle just like before. In this case, we have sin θ BTM 2 1 − µ1ε 2 / µ 2ε1 = 2 1 − (µ1 / µ 2 ) For most of the materials which have µ1= µ2= µ0, this angle usually does not exist. Impedance for Waves at Oblique Incidence Since in many cases the tangential components of the fields are continuous across the boundary, it is useful to define a characteristic wave impedance referred to the z direction (as in our setup) in terms of the components in planes transverse to that direction. as the ratio of electric to magnetic field components in planes parallel to the boundary. In other words, in the TM case, ηz TM Ex + Ex − = =− = η cosθ H y+ H y− ηz = − TE Ey + H x+ = Ey − H x− = η secθ + and – refer, respectively, to incident and reflected wave. The sign of the ratio is chosen for each wave to yield a positive impedance. Reflection and Transmission Revisit If we substitute these new definitions into the reflection and transmission coefficient expressions, we can see that these expressions are just what we learned in the transmission line. ΓZ TM τZ η 2 cosθt − η1 cosθi η 2, Z − η1, Z = = η1 cosθi + η 2 cosθt η1, Z + η 2, Z 2η 2, Z 2η 2 cosθ t = = η1 cosθ i + η 2 cosθ t η1, Z + η 2, Z TM TM TM TM TM TM TM ΓZ TE TM η 2 cosθ i − η1 cosθ t η 21, Z − η1, Z = ΓTE = = η 2 cosθi + η1 cosθt η1, Z + η 2, Z TE TE τ Z = τ TE TE 2η 2, Z TE 2η 2 cosθ i = = η 2 cosθ i + η1 cosθ t η 2, Z TE + η1, Z TE TE TE Reflection and Transmission Revisit (cont.) In general, η 2, Z − η1, Z ΓZ = η1, Z + η 2, Z τZ 2η 2, Z = η1, Z + η 2, Z The impedance defined by the fields’ tangential components is analogous to the impedance in transmission lines.