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Electromagnetic Waves Electromagnetic Waves Lecture34: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Electromagnetic wave at the interface between two dielectric media We have so far discussed the propagation of electromagnetic wave in an isotropic, homogeneous, dielectric medium, such as in air or vacuum. In this lecture, we woulddiscuss what happens when a plane electromagnetic wave is incident at the interface between two dielectric media. For being specific, we will take one of the medium to be air or vacuum and the other to be a dielectric such as glass. We have come across this in school in connection with the reflection and transmission of light waves at such an interface. In this lecture, we would investigate this problem from the point of view of electromagnetic theory. Let us choose the interface to be the xy plane (z=0). The angles of incidence, reflection and refraction are the angles made by the respective propagation vectors with the common normal at the interface. 1 We have indicated the propation vectors in the apprpriate medium by capital letters I, R and T so as not to confuse with the notation for the position vector and time t. The principle that we use to establish the laws of reflection⃗ and refraction is the continuity of the tangential components of the electric field at the interface, as discussed extensively during the course of these lectures. Let us represent the component of the electric field parallel to the interface by the superscript . We then have, ∥ 0 exp + 0 exp = 0 exp ∥ ∥ This equation� ���⃗ ⋅ must⃗ − remain� valid at� all����⃗ points⋅ ⃗ − in the� interface � and ����⃗ at⋅ ⃗all− times.� That is obviously possible if the exponential factors is the same for all the three terms or if they differe at best by a constant phase factor. Considering, the incident and the reflection terms, we have, = + 1 = ���⃗ ⋅ ⃗ ����1⃗ ⋅ ⃗ We are familiar with the vector equation���� ⃗to− a� ��plane�⃗� ⋅ ⃗ and we know that if the position vector of a plane is and the normal to the plane is represented by , the equation to the plane is given by = constant. Thus is normal to the interface plane. Since the interface is x-y plane, we take the ⃗ � � ⋅ ⃗ plane containing the and in the x-z plane. ���⃗ − ����⃗ ���⃗ ����⃗ Since and the normal to the plane is normal to the interface are in the same plane, we have, ���⃗ − ����⃗ � � × = 0 ���⃗ ����⃗ Thus we have, sin = sin . Since� − and � are� in the same medium, though their directions = = and hence, we have, are different, their����⃗� magnitudes �����⃗ �are the same ���⃗ ����⃗ ���⃗ ����⃗ �sin � =�sin� = which is the law of reflection. We will now prove the Snell’s law. Let us look into the equation = 2 In a fashion similar to the above, we�� �can�⃗ − show����⃗� ⋅ that⃗ × = 0 ����⃗ − ����⃗� � 2 which gives sin = sin . Since and are in different media, we have, recognizing that as the wave goes from one medium to another, its frequency does not change, ����⃗� �����⃗� ���⃗ ����⃗ = , = where is the velocity of the wave� �in��⃗ �the transmitted��⃗� medium. We, therefore, have, sin = = = sin ��⃗� ≡ ����⃗� Here n is the refractive index of the second medium with respect to the incident medium. Fresnel’s Equations What happens to the amplitudes of the wave on reflection and transmission? Let us summarize the boundary conditions that we have derived in these lectures. We will assume that both the media are non-magnetic so that the permeability of both media are the same, viz., 0. The two media differ by their dielectric constant, the incident medium is taken to be air as above. We further assume that there are no free charges or currents on the surface so that both the normal and tangential components of the fields are continuous. We will consider two cases, the first case where the electric fields are parallel to the incident plane. This case is known as p- polarization, p standing for “parallel”. The second case is where the electric field is perpendicular to the incident plane. This is referred to as s- polarization, s standing for a German word “senkrecht “ meaning perpendicular. p- polarization In this case, since the magnetic fields are perpendicular to the plane of incidence, we take the directions of the H fields to come out of the plane of the page (the incident plane). Since the electric field, the magnetic field and the direction of propagation are mutually perpendicular being a right handed triad, we have indicated the directions of the electric fields accordingly. It may be noted that in this picture, for the case of normal incidence, the incident and the reflected electric fields are directed oppositely. (The assumption is not restrictive because, if it is not true, a negative sign should appear in the equations ). 3 Taking the tangential components of the electric field (parallel to the interface), we have, cos cos = cos (1) = = − As our medium is linear, we have the following relationship between the electric and the magnetic field magnitudes, 0 = = = = � � Thus the continuity of the tangential component of the magnetic field H can be expressed in terms of electric field amplitudes ( + ) = (2) � � Equations (1) and (2) can be simplified (we use = ) + cos = � − = = = √ � where, and are refractive indices of√ the transmitted medium and incident medium, respectively, with respect to free space. 4 Thus, we have, + cos = − which gives, ( )cos ( ) cos = (3) ( )cos + ( ) cos ⁄ − ⁄ ⁄ ⁄ Substituting (3) in (2), 2 ( )cos = (4) ( )cos + ( ) cos ⁄ ⁄ ⁄ Let us take = . The Fresnel’s equations are now expressible in terms of refractive indices of the two media and the angles of incidence and transmission n cos cos = T nTcos + cos − 2 nIcos ≡ = nTcos + cos where, and are, respectively, the reflection and transmission≡ coefficients for field amplitudes. (Caution : the phrases “reflection/transmission” coefficients are also used to denote fraction of transmitted intensities.) Brewster’s Angle sin = Using sin , we can express , as follows. , sin cos sin cos = I sin Icos + sin cos θ − sin 2 sin 2 = θ sin 2 sin 2 2 sin( − )cos( + ) = ( − ) ( ) 2 sin + cos tan( − ) = tan( + ) − − + = 0 If 2, i.e. if the angle between the reflected ray and the transmitted ray is 90 , the reflection coefficient for the parallel polarization becomes zero, because tan( + ) . If we had started with an equal mixture of p polarized and s polarized waves (i.e. an unpolarized → ∞ wave), the reflected ray will have no p component, so that it will be plane polarized. The angle of incidence at which this happens is called the Brewster’s angle”. 5 For this angle, we have cos sin tan = cot = = = sin sin The following figure (right) , the red curve shows the variation in the reflected amplitude with the angle of incidence for p-polarization. The blue curve is the corresponding variation for s – polarization to be discussed below. rs rp For normal incidence, = = 0, we have, 1 = = + + 1 − − s –polarization We next consider s polarization where the electric field is perpendicular to the incident plane. As we have taken the plane of incidence to be the plane of the paper, the electric field will be taken to come out of the plane of the paper. 6 The corresponding directions of the magnetic field is shown in the figure. The boundary conditions give + = ( ) cos = cos − Substituting = , we get for the reflection and transmission coefficient for the amplitudes of the electric field� n cos cos = I nIcos + cos − 2 nTcos ≡ = nIcos + cos ≡ Using Snell’s law, we can simplify these expressions to get, sin( ) = ( ) sin + 2 sin cos− = − sin( + ) For normal incidence, = = 0, we have, 1 = = + 1 + − − Notice that the expression differs from the expression for for normal incidence, while both the results should have been the same. This is because of different conventions we took in fixing 7 the directions in the two cases; in the p –polarization case, for normal incidence, the electric field directions are opposite for the incident and reflected rays while for the s-polarization, they have been taken to be along the same direction. Total Internal Reflection and Evanescent Wave Let us return back to the case of p polarization and consider the case where in the incident medium has a higher refractive index than the transmitted medium. In this case, we can write the amplitude reflection coefficient as n cos cos = T nTcos + cos − ncos 1 sin2 = ncos + 1 sin2 − � − where we have used the refractive index of the second medium as = < 1. Substituting � − Snell’s law into the above, we have, sin = sin , we can write the above as ⁄ n2cos 2 sin2 = n2cos + 2 sin2 − � − The quantity under the square root could become negative for some values of since n <1 . we � − therefore write, n2cos sin2 2 = n2cos + sin2 2 − � − In a very similar way, we can show that the reflection coefficient for s polarization can be � − expressed as follows: n cos cos = I nIcos + cos − cos sin2 2 = cos + sin2 2 − � − It may be seen that for sin > , the magnitudes of both� and − are both equal to unity because for both these, the numerator and the denominator are complex conjugate of each other. Thus it implies that when electromagnetic wave is incident at an angle of incidence greater than a “critical angle” defined by sin = where n here represents the refractive index of the (rarer) transmitted medium with respect to the(denser) incident medium , the wave is totally reflected. (In text books on optics, the critical angle is defined by the relation sin = 1 , because the refractive index there is conventionally defined as that of the denser medium with respect to the rarer one).
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