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.

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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 reflection𝑟𝑟⃗ 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 ∥ ∥ 𝐼𝐼 𝑅𝑅 𝑇𝑇 𝑇𝑇 This𝐸𝐸 𝐼𝐼 equation�𝑖𝑖 �𝑘𝑘��⃗ ⋅ 𝑟𝑟must⃗ − 𝑖𝑖𝑖𝑖𝑖𝑖 remain� 𝐸𝐸 valid𝑅𝑅 at� 𝑖𝑖all�𝑘𝑘���⃗ points⋅ 𝑟𝑟⃗ − 𝑖𝑖𝑖𝑖𝑖𝑖 in the� interface𝐸𝐸 � and𝑖𝑖 𝑘𝑘����⃗ at⋅ 𝑟𝑟⃗all− times.𝑖𝑖𝑖𝑖𝑖𝑖� 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 = �𝑘𝑘��⃗𝐼𝐼 ⋅ 𝑟𝑟⃗ �𝑘𝑘���𝑅𝑅1⃗ ⋅ 𝑟𝑟⃗ 𝜙𝜙 𝐼𝐼 𝑅𝑅 We are familiar with the vector equation�𝑘𝑘��� ⃗to− a𝑘𝑘� ��plane�⃗� ⋅ 𝑟𝑟⃗ and𝜙𝜙 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 𝑟𝑟⃗ 𝑛𝑛� 𝑛𝑛� ⋅ 𝑟𝑟⃗ plane containing the and in the x-z plane. �𝑘𝑘��⃗𝐼𝐼 − �𝑘𝑘���𝑅𝑅⃗ ���⃗𝐼𝐼 ����𝑅𝑅⃗ Since and the𝑘𝑘 normal𝑘𝑘 to the plane is normal to the interface are in the same plane, we have, 𝑘𝑘���⃗𝐼𝐼 − 𝑘𝑘����𝑅𝑅⃗ 𝑛𝑛� 𝑛𝑛� × = 0 ���⃗𝐼𝐼 ����𝑅𝑅⃗ Thus we have, sin = sin . Since�𝑘𝑘 − and𝑘𝑘 � 𝑛𝑛are� in the same medium, though their directions = = and hence, we have, are different, their��𝑘𝑘��⃗𝐼𝐼� magnitudes𝜃𝜃𝐼𝐼 �𝑘𝑘����𝑅𝑅⃗ �are the𝜃𝜃𝑅𝑅 same �𝑘𝑘��⃗𝐼𝐼 �𝑘𝑘���𝑅𝑅⃗ 𝜔𝜔 ���⃗𝐼𝐼 ����𝑅𝑅⃗ �sin𝑘𝑘 � =�𝑘𝑘sin� 𝑐𝑐 = 𝜃𝜃𝐼𝐼 𝜃𝜃𝑅𝑅

𝜃𝜃𝐼𝐼 𝜃𝜃𝑅𝑅 which is the law of reflection.

We will now prove the Snell’s law.

Let us look into the equation

= 2

𝐼𝐼 𝑇𝑇 In a fashion similar to the above, we�𝑘𝑘� �can�⃗ − show𝑘𝑘����⃗� ⋅ 𝑟𝑟that⃗ 𝜙𝜙 × = 0

�𝑘𝑘���⃗𝐼𝐼 − 𝑘𝑘����𝑇𝑇⃗� 𝑛𝑛� 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, ��𝑘𝑘��⃗𝐼𝐼� 𝜃𝜃𝐼𝐼 �𝑘𝑘����𝑇𝑇⃗� 𝜃𝜃𝑇𝑇 �𝑘𝑘��⃗𝐼𝐼 𝑘𝑘����𝑇𝑇⃗ = , = 𝜔𝜔 𝜔𝜔 where is the velocity of the wave� �𝑘𝑘in��⃗𝐼𝐼 �the transmitted�𝑘𝑘�⃗𝑇𝑇� medium. 𝑐𝑐 𝑣𝑣𝑇𝑇 We, therefore, have, 𝑣𝑣𝑇𝑇 sin = = = sin 𝜃𝜃𝐼𝐼 �𝑘𝑘�⃗𝑇𝑇� 𝑐𝑐 𝑛𝑛𝑇𝑇 𝑇𝑇 𝐼𝐼 ≡ 𝑛𝑛 𝜃𝜃 ��𝑘𝑘��⃗𝐼𝐼� 𝑣𝑣 𝑛𝑛 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 𝜇𝜇 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 ).

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Taking the tangential components of the electric field (parallel to the interface), we have, cos cos = cos (1) = = 𝐸𝐸𝐼𝐼 𝜃𝜃𝐼𝐼 − 𝐸𝐸𝑅𝑅 𝜃𝜃𝑅𝑅 𝐸𝐸𝑇𝑇 𝜃𝜃𝑇𝑇

𝐻𝐻𝐼𝐼 𝐻𝐻𝑅𝑅 𝐻𝐻𝑇𝑇 As our medium is linear, we have the following relationship between the electric and the magnetic field magnitudes,

0 = = = = 𝐵𝐵 𝐸𝐸 𝐸𝐸�𝜇𝜇 𝜖𝜖 𝜖𝜖 𝐻𝐻 � 𝐸𝐸 𝜇𝜇 𝑣𝑣𝑣𝑣 𝜇𝜇 𝜇𝜇 Thus the continuity of the tangential component of the magnetic field H can be expressed in terms of electric field amplitudes

( + ) = (2) 𝜖𝜖𝐼𝐼 𝜖𝜖𝑇𝑇 � 𝐸𝐸𝐼𝐼 𝐸𝐸𝑅𝑅 � 𝐸𝐸𝑇𝑇 𝜇𝜇𝐼𝐼 𝜇𝜇𝑇𝑇 Equations (1) and (2) can be simplified (we use = )

𝜃𝜃𝑅𝑅 𝜃𝜃𝐼𝐼 + cos =

𝐸𝐸𝐼𝐼 𝐸𝐸𝑅𝑅 𝜇𝜇𝐼𝐼𝜖𝜖𝑇𝑇 𝜃𝜃𝐼𝐼 � 𝐸𝐸𝐼𝐼 − 𝐸𝐸𝑅𝑅 𝜇𝜇𝑇𝑇𝜖𝜖𝐼𝐼 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑇𝑇

= = = 𝜇𝜇𝐼𝐼𝜖𝜖𝑇𝑇 √𝜇𝜇𝑇𝑇𝜖𝜖𝑇𝑇 𝜇𝜇𝐼𝐼 𝑣𝑣𝐼𝐼 𝜇𝜇𝐼𝐼 𝑛𝑛𝑇𝑇 𝜇𝜇𝐼𝐼 � 𝑇𝑇 𝐼𝐼 𝑇𝑇 𝑇𝑇 𝑇𝑇 𝐼𝐼 𝑇𝑇 where, and are refractive𝜇𝜇 indices𝜖𝜖 of√𝜇𝜇 the𝐼𝐼𝜖𝜖𝐼𝐼 transmitted𝜇𝜇 𝑣𝑣 𝜇𝜇 medium𝑛𝑛 𝜇𝜇 and incident medium, respectively, with respect to free space. 𝑛𝑛𝑇𝑇 𝑛𝑛𝐼𝐼

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Thus, we have,

+ cos =

𝐸𝐸𝐼𝐼 𝐸𝐸𝑅𝑅 𝑛𝑛𝑇𝑇 𝜇𝜇𝐼𝐼 𝜃𝜃𝐼𝐼

𝐸𝐸𝐼𝐼 − 𝐸𝐸𝑅𝑅 𝑛𝑛𝐼𝐼 𝜇𝜇𝑇𝑇 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃𝑇𝑇 which gives,

( )cos ( ) cos = (3) ( )cos + ( ) cos 𝐸𝐸𝑅𝑅 𝑛𝑛𝑇𝑇⁄𝜇𝜇𝑇𝑇 𝜃𝜃𝐼𝐼 − 𝑛𝑛𝐼𝐼⁄𝜇𝜇𝐼𝐼 𝜃𝜃𝑇𝑇

𝐸𝐸𝐼𝐼 𝑛𝑛𝑇𝑇⁄𝜇𝜇𝑇𝑇 𝜃𝜃𝐼𝐼 𝑛𝑛𝐼𝐼⁄𝜇𝜇𝐼𝐼 𝜃𝜃𝑇𝑇 Substituting (3) in (2), 2 ( )cos = (4) ( )cos + ( ) cos 𝐸𝐸𝑇𝑇 𝑛𝑛𝐼𝐼⁄𝜇𝜇𝐼𝐼 𝜃𝜃𝐼𝐼

𝐸𝐸𝐼𝐼 𝑛𝑛𝑇𝑇⁄𝜇𝜇𝑇𝑇 𝜃𝜃𝐼𝐼 𝑛𝑛𝐼𝐼⁄𝜇𝜇𝐼𝐼 𝜃𝜃𝑇𝑇

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 𝜇𝜇𝐼𝐼 𝜇𝜇𝑇𝑇

n cos cos = T nTcos + cos 𝐸𝐸𝑅𝑅 𝜃𝜃𝐼𝐼 − 𝑛𝑛𝐼𝐼 𝜃𝜃𝑇𝑇 2 nIcos ≡ 𝑟𝑟𝑝𝑝 𝐸𝐸𝐼𝐼 = 𝜃𝜃𝐼𝐼 𝑛𝑛𝐼𝐼 𝜃𝜃𝑇𝑇 nTcos + cos 𝐸𝐸𝑇𝑇 𝜃𝜃𝐼𝐼 where, and are, respectively, the reflection and transmission≡ 𝑡𝑡𝑝𝑝 coefficients for field 𝐸𝐸𝐼𝐼 𝜃𝜃𝐼𝐼 𝑛𝑛𝐼𝐼 𝜃𝜃𝑇𝑇 amplitudes. (Caution : the phrases “reflection/transmission” coefficients are also used to denote 𝑟𝑟𝑝𝑝 𝑡𝑡𝑝𝑝 fraction of transmitted intensities.)

Brewster’s Angle sin = Using sin , we can express , as follows. 𝑛𝑛𝑇𝑇 𝜃𝜃𝐼𝐼 , 𝑛𝑛𝐼𝐼 𝜃𝜃𝑇𝑇 𝑟𝑟𝑝𝑝 sin cos sin cos = I sin Icos + sin cos θ 𝜃𝜃𝐼𝐼 − 𝜃𝜃𝑇𝑇 𝜃𝜃𝑇𝑇 𝑟𝑟𝑝𝑝 sin 2 sin 2 = θ 𝜃𝜃𝐼𝐼 𝜃𝜃𝑇𝑇 𝜃𝜃𝑇𝑇 sin 2 𝐼𝐼 sin 2 𝑇𝑇 2 sin(𝜃𝜃 − )𝜃𝜃cos( + ) = 𝐼𝐼 𝑇𝑇 (𝜃𝜃 − )𝜃𝜃 ( ) 2 sin 𝐼𝐼 + 𝑇𝑇 cos 𝐼𝐼 𝑇𝑇 tan( 𝜃𝜃 − 𝜃𝜃 ) 𝜃𝜃 𝜃𝜃 = 𝐼𝐼 𝑇𝑇 𝐼𝐼 𝑇𝑇 tan( 𝜃𝜃 + 𝜃𝜃 ) 𝜃𝜃 − 𝜃𝜃 𝜃𝜃𝐼𝐼 − 𝜃𝜃𝑇𝑇 + = 0 If 2, i.e. if the angle between the𝐼𝐼 reflected𝑇𝑇 ray and the transmitted ray is 90 , the 𝜋𝜋 𝜃𝜃 𝜃𝜃 reflection coefficient for the parallel polarization becomes zero, because tan( + ) . If 𝜃𝜃𝐼𝐼 𝜃𝜃𝑇𝑇 we had started with an equal mixture of p polarized and s polarized waves (i.e. an unpolarized 𝜃𝜃𝐼𝐼 𝜃𝜃𝑇𝑇 → ∞ 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”.

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For this angle, we have cos sin tan = cot = = = sin sin 𝜃𝜃𝑇𝑇 𝜃𝜃𝐼𝐼 The following figure (right) , the 𝜃𝜃red𝐼𝐼 curve𝜃𝜃 𝑇𝑇shows the variation in the𝑛𝑛 reflected amplitude with 𝜃𝜃𝑇𝑇 𝜃𝜃𝑇𝑇 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,

𝐼𝐼 𝑇𝑇 𝜃𝜃 𝜃𝜃 1 = = + + 1 𝑛𝑛𝑇𝑇 − 𝑛𝑛𝐼𝐼 𝑛𝑛 − 𝑟𝑟𝑝𝑝 𝑛𝑛𝑇𝑇 𝑛𝑛𝐼𝐼 𝑛𝑛

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.

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The corresponding directions of the magnetic field is shown in the figure. The boundary conditions give + = ( ) cos = cos 𝐸𝐸𝐼𝐼 𝐸𝐸𝑅𝑅 𝐸𝐸𝑇𝑇

𝐻𝐻𝐼𝐼 − 𝐻𝐻𝑅𝑅 𝜃𝜃𝐼𝐼 𝐻𝐻𝑇𝑇 𝜃𝜃𝑇𝑇 Substituting = , we get for the reflection and transmission coefficient for the amplitudes 𝜖𝜖 of the electric𝐻𝐻 field� 𝜇𝜇 𝐸𝐸 n cos cos = I nIcos + cos 𝐸𝐸𝑅𝑅 𝜃𝜃𝐼𝐼 − 𝑛𝑛𝑇𝑇 𝜃𝜃𝑇𝑇 2 nTcos ≡ 𝑟𝑟𝑠𝑠 𝐸𝐸𝐼𝐼 = 𝜃𝜃𝐼𝐼 𝑛𝑛𝑇𝑇 𝜃𝜃𝑇𝑇 nIcos + cos 𝐸𝐸𝑇𝑇 𝜃𝜃𝐼𝐼 ≡ 𝑡𝑡𝑠𝑠 𝐸𝐸𝐼𝐼 𝜃𝜃𝐼𝐼 𝑛𝑛𝑇𝑇 𝜃𝜃𝑇𝑇 Using Snell’s law, we can simplify these expressions to get,

sin( ) = ( ) sin 𝐼𝐼 + 𝑇𝑇 𝑠𝑠 2 sin 𝜃𝜃 cos− 𝜃𝜃 𝑟𝑟 = − 𝐼𝐼 𝑇𝑇 sin( 𝜃𝜃+ 𝜃𝜃) 𝜃𝜃𝑇𝑇 𝜃𝜃𝐼𝐼 𝑡𝑡𝑠𝑠 𝜃𝜃𝐼𝐼 𝜃𝜃𝑇𝑇

For normal incidence, = = 0, we have,

𝐼𝐼 𝑇𝑇 𝜃𝜃 𝜃𝜃 1 = = + 1 + 𝑛𝑛𝐼𝐼 − 𝑛𝑛𝑇𝑇 − 𝑛𝑛 𝑟𝑟𝑠𝑠 𝑛𝑛𝑇𝑇 𝑛𝑛𝐼𝐼 𝑛𝑛 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 𝑟𝑟𝑝𝑝 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 𝜃𝜃𝐼𝐼 − 𝑛𝑛𝐼𝐼 𝜃𝜃𝑇𝑇 𝑟𝑟𝑝𝑝 ncos 1 sin2 = 𝜃𝜃𝐼𝐼 𝑛𝑛𝐼𝐼 𝜃𝜃𝑇𝑇 ncos + 1 sin2 𝜃𝜃𝐼𝐼 − � − 𝜃𝜃𝑇𝑇 where we have used the refractive index of the𝐼𝐼 second medium𝑇𝑇 as = < 1. Substituting 𝜃𝜃 � − 𝜃𝜃 𝑛𝑛𝑇𝑇 Snell’s law into the above, we have, sin = sin , we can write the above as 𝑛𝑛 𝑛𝑛

𝜃𝜃𝑇𝑇 𝜃𝜃𝐼𝐼⁄𝑛𝑛 n2cos 2 sin2 = n2cos + 2 sin2 𝜃𝜃𝐼𝐼 − �𝑛𝑛 − 𝜃𝜃𝐼𝐼 The quantity under the square root𝑟𝑟𝑝𝑝 could become negative for some values of since n <1 . we 𝜃𝜃𝐼𝐼 �𝑛𝑛 − 𝜃𝜃𝐼𝐼 therefore write, 𝜃𝜃𝐼𝐼 n2cos sin2 2 = n2cos + sin2 2 𝜃𝜃𝐼𝐼 − 𝑖𝑖� 𝜃𝜃𝐼𝐼 − 𝑛𝑛 In a very similar way, we can show𝑟𝑟𝑝𝑝 that the reflection coefficient for s polarization can be 𝜃𝜃𝐼𝐼 𝑖𝑖� 𝜃𝜃𝐼𝐼 − 𝑛𝑛 expressed as follows: n cos cos = I nIcos + cos 𝜃𝜃𝐼𝐼 − 𝑛𝑛𝑇𝑇 𝜃𝜃𝑇𝑇 𝑟𝑟𝑠𝑠 cos sin2 2 = 𝜃𝜃𝐼𝐼 𝑛𝑛𝑇𝑇 𝜃𝜃𝑇𝑇 cos + sin2 2 𝜃𝜃𝐼𝐼 − 𝑖𝑖� 𝜃𝜃𝐼𝐼 − 𝑛𝑛 It may be seen that for sin > , the magnitudes𝜃𝜃 𝐼𝐼of both𝑖𝑖� and𝜃𝜃𝐼𝐼 − 𝑛𝑛 are both equal to unity because for both these, the numerator and the denominator are complex conjugate of each other. Thus it implies 𝜃𝜃𝐼𝐼 𝑛𝑛 𝑟𝑟𝑝𝑝 𝑟𝑟𝑠𝑠 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 the𝜃𝜃 (rarer)𝑐𝑐 𝑛𝑛 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). 𝜃𝜃𝑐𝑐 ⁄𝑛𝑛

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In case of total internal reflection, is there a wave in the transmitted medium? The answer is yes, it does as the following analysis shows.

Let us take the incident plane to be xz plane and the interface to be the xy plane so that the normal to the plane is along the z direction. The transmitted wave can be written as

= 0 exp

𝑇𝑇 𝑇𝑇 𝑇𝑇 The space part of the wave can be expressed𝐸𝐸�⃗ 𝐸𝐸�⃗ as �𝑖𝑖𝑘𝑘�⃗ ⋅ 𝑟𝑟⃗ − 𝑖𝑖𝑖𝑖𝑖𝑖�

= ( sin + )

Writing this in terms of angle of incidence𝑖𝑖𝑘𝑘�⃗𝑇𝑇 ⋅ 𝑟𝑟⃗ 𝑖𝑖 𝑘𝑘𝑇𝑇𝑥𝑥 𝜃𝜃𝑇𝑇 𝑘𝑘𝑇𝑇𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧 𝜃𝜃𝑇𝑇

𝜃𝜃𝐼𝐼 sin2 = sin + 1 2 𝑘𝑘𝑇𝑇 𝜃𝜃𝐼𝐼 𝑖𝑖𝑘𝑘�⃗𝑇𝑇 ⋅ 𝑟𝑟⃗ 𝑖𝑖 � 𝑥𝑥 𝜃𝜃𝐼𝐼 𝑘𝑘𝑇𝑇𝑧𝑧� − � 𝑛𝑛 𝑛𝑛 For angles greater than the critical angle the quantity within the square root is negative and we rewrite it as

sin2 = sin + 2 1 𝑘𝑘𝑇𝑇 𝜃𝜃𝐼𝐼 𝑖𝑖𝑘𝑘�⃗𝑇𝑇 ⋅ 𝑟𝑟⃗ 𝑖𝑖 � 𝑥𝑥 𝜃𝜃𝐼𝐼 𝑖𝑖𝑘𝑘𝑇𝑇𝑧𝑧� − � = 𝑛𝑛 𝑛𝑛 where, we define “propagation vector “𝑖𝑖𝑖𝑖 𝑖𝑖 as− 𝛼𝛼𝛼𝛼

𝛽𝛽 = sin 𝑘𝑘𝑇𝑇 𝛽𝛽 𝜃𝜃𝐼𝐼 and the “attenuation factor” as 𝑛𝑛

𝛼𝛼 sin2 = 2 1 𝜃𝜃𝐼𝐼 𝛼𝛼 𝑘𝑘𝑇𝑇� − sin𝑛𝑛2 2 2 2 = 2 1 = sin 𝜔𝜔𝑛𝑛𝑇𝑇 𝜃𝜃𝐼𝐼 𝜔𝜔 � − �𝑛𝑛𝐼𝐼 𝜃𝜃𝐼𝐼 − 𝑛𝑛𝑇𝑇 𝑐𝑐 𝑛𝑛 𝑐𝑐

With this, the wave in the transmitted medium becomes,

= 0 exp( ) −𝛼𝛼𝛼𝛼 𝐸𝐸�⃗𝑇𝑇 𝐸𝐸�⃗𝑇𝑇 𝑒𝑒 𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑖𝑖𝑖𝑖𝑖𝑖

9 which shows that the wave in the second medium propagates along the interface. It penetrates into the medium but its amplitude attenuates exponentially. This is known as “evanescent wave”.

The amplitude variation with angle of incidence is shown in the following figure.

What is the propagation vector?

2 Recall that the magnitude of the propagation vector is defined as , where the “wavelength” is the 𝜋𝜋 distance between two successive crests or troughs of the wave measured along the direction of 𝜆𝜆 𝜆𝜆 propagation. However, consider, for instance, a water wave which moves towards the shore. Along the shore, one would be more inclined to conclude that the wavelength is as measured by the distance between successive crests or troughs along the shore. This is the wavelength with which the attenuating surface wave propagates in the second medium.

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No transfer of energy into the transmitted medium:

Though there is a wave in the transmitted medium, one can show that on an average, there is no transfer of energy into the medium from the incident medium.

Consider, p polarization, for which the transmitted electric field, being parallel to the incident (xz) plane is along the x direction.

= 0 exp( ) −𝛼𝛼𝛼𝛼 From this we obtain the H- field using𝐸𝐸�⃗ 𝑇𝑇Faraday’s𝐸𝐸𝑇𝑇 𝑒𝑒 law, Since𝑖𝑖𝑖𝑖 the𝑖𝑖 − H𝑖𝑖𝑖𝑖𝑖𝑖 field𝑖𝑖̂ is taken perpendicular to the electric field , it would be in the y-z plane . Taking the components of × , we have,

∇ 𝐸𝐸�⃗ = 0 = × = = 0 exp( ) 𝜕𝜕𝐵𝐵𝑦𝑦 𝜕𝜕𝐸𝐸𝑥𝑥 −𝛼𝛼𝛼𝛼 − 𝑖𝑖𝜇𝜇 𝜔𝜔𝐻𝐻𝑦𝑦 �∇ 𝐸𝐸�⃗�𝑦𝑦 −𝛼𝛼𝛼𝛼𝑇𝑇 𝑒𝑒 𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑖𝑖𝑖𝑖𝑖𝑖 and 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

= 0 = × = = + 0 exp( ) 𝜕𝜕𝐵𝐵𝑧𝑧 𝜕𝜕𝐸𝐸𝑦𝑦 −𝛼𝛼𝛼𝛼 − 𝑖𝑖𝜇𝜇 𝜔𝜔𝐻𝐻𝑧𝑧 �∇ 𝐸𝐸�⃗�𝑧𝑧 𝑖𝑖𝑖𝑖𝑖𝑖𝑇𝑇 𝑒𝑒 𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑖𝑖𝑖𝑖𝑖𝑖 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

so that,

0 = + exp( ) 0 𝐸𝐸𝑇𝑇 −𝛼𝛼𝛼𝛼 𝐻𝐻��⃗ 𝑒𝑒 �𝑖𝑖𝑖𝑖𝑗𝑗̂ 𝛽𝛽𝑘𝑘�� 𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑖𝑖𝑖𝑖𝑖𝑖 Thus the average energy transfer Poynting𝜇𝜇 𝜔𝜔 vector. The complex Poynting vector can be written as

1 = × 2 ∗ 𝑆𝑆⃗ 𝐸𝐸�⃗ 𝐻𝐻��⃗ so that the average power transferred into the second medium is

1 = Re × 2 2 ∗ ⃗ 0 �⃗ ��⃗ 〈=𝑆𝑆〉 e 2�𝐸𝐸zRe𝐻𝐻 � 2 0 𝑇𝑇 − α 𝐸𝐸 2 0 �−𝑖𝑖𝑖𝑖𝑘𝑘� − 𝛽𝛽𝑗𝑗̂� = 𝜇𝜇 𝜔𝜔 e 2 z 2 0 𝐸𝐸𝑇𝑇 − α − 𝛽𝛽𝑗𝑗̂ Since the normal to the surface is along the 𝜇𝜇z direction,𝜔𝜔 the average energy transferred to the second medium is zero.

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Electromagnetic Waves

Lecture34: Electromagnetic Theory

Professor D. K. Ghosh, Physics Department, I.I.T., Bombay

Tutorial Assignment

1. A plane electromagnetic wave described by its magnetic field is given by the expression

= 0 sin( ) Determine the corresponding electric field and the time average Poynting vector. 𝐻𝐻��⃗ 𝐻𝐻 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑦𝑦� If it is incident normally on a perfect conductor and is totally reflected what would be the pressure exerted on the surface? Determine the surface current generated at the interface. 2. A circularly polarized electromagnetic wave is given by

= 0 sin( ) + 0 cos( ) Show that the average value of the Poynting vector for the wave is equal to the sum of the 𝐸𝐸�⃗ 𝐸𝐸 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑥𝑥� 𝐸𝐸 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑦𝑦� Poynting vectors of its components. 3.

Solutions to Tutorial Assignments

1. The wave is propagating along the + z direction (before reflection). The electric field is given by Maxwell- Ampere law,

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× = 0 𝜕𝜕𝐸𝐸�⃗ ∇ 𝐻𝐻��⃗ 𝜖𝜖 Since is along direction, and its y-component depends𝜕𝜕𝜕𝜕 on z only, the curl is given by

𝐻𝐻��⃗ 𝑦𝑦� = 0 cos( ) = 0 𝜕𝜕𝐻𝐻𝑦𝑦 𝜕𝜕𝐸𝐸�⃗ −𝑥𝑥� −𝑥𝑥� 𝐻𝐻 𝑘𝑘 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝜖𝜖 This gives, 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

0 0 = sin( ) = sin( ) 0 0 𝐻𝐻 𝑘𝑘 𝐻𝐻 𝐸𝐸�⃗ 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑥𝑥� 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑥𝑥� The Poynting vector is given by 𝜔𝜔𝜖𝜖 𝑐𝑐𝜖𝜖

2 0 = × = sin2( ) 0 𝐻𝐻 ⃗ �⃗ ��⃗ 𝑆𝑆 𝐸𝐸 𝐻𝐻2 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑧𝑧̂ = 0 𝑐𝑐𝜖𝜖 The time average Poynting vector is 2 . 𝐻𝐻 0 〈𝑆𝑆⃗〉 𝑐𝑐𝜖𝜖 𝑧𝑧̂ Since the wave is totally reflected, the change in momentum is twice the initial momentum carried. Thus the pressure is given by

2 2| | 0 = = 2 0 〈𝑆𝑆〉 𝐻𝐻 𝑃𝑃 At the metallic interface (z=0), the tangential component𝑐𝑐 𝑐𝑐of𝜖𝜖 the electric field is zero. Since the wave is totally reflected, the reflected wave must have oppositely directed electric field, i.e. in direction. The direction of propagation having been reversed, the magnetic the field is given by −𝑥𝑥�

= 0 sin( + ) At the interface the total magnetic field is 2 0 sin( ) . 𝐻𝐻��⃗𝑟𝑟 −𝐻𝐻 𝑘𝑘𝑘𝑘 𝜔𝜔𝜔𝜔 𝑦𝑦� The surface current can be determined by taking an Amperian loop perpendicular to the − 𝐻𝐻 𝜔𝜔𝜔𝜔 𝑦𝑦� interface, Taking the direction of the loop parallel to the magnetic field, the line integral is seen

to be 2 0 sin( ) = , where K is the surface current density. The direction of the surface current is along the direction. Thus 𝐻𝐻 𝜔𝜔𝜔𝜔 𝑙𝑙 𝐾𝐾𝐾𝐾 = 2 sin( ) 𝑥𝑥� 0 2. The electric field is given by 𝐾𝐾��⃗ 𝐻𝐻 𝜔𝜔𝜔𝜔 𝑖𝑖̂ = 0 sin( ) + 0 cos( ) As the wave is propagating in the z direction, the corresponding magnetic field is given by �⃗ 𝐸𝐸 1 𝐸𝐸 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑥𝑥� 𝐸𝐸 1 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑦𝑦� = 0 cos( ) + 0 sin( ) 0 0 Poynting vector is 𝐻𝐻��⃗ − 𝐸𝐸 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑥𝑥� 𝐸𝐸 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑦𝑦� 𝜇𝜇 𝑐𝑐 𝜇𝜇 𝑐𝑐 1 2 = × = 0 0 𝑆𝑆⃗ 𝐸𝐸�⃗ 𝐻𝐻��⃗ 𝐸𝐸 𝑧𝑧̂ 𝜇𝜇 𝑐𝑐 13

The individual components have average Poynting vectors given by

1 2 2 1 2 1 = 0 sin ( ) = 0 0 2 0 〈𝑆𝑆 〉 1 𝐸𝐸2〈 2 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 〉 1 𝐸𝐸 2 2 = 𝜇𝜇 𝑐𝑐 0 cos ( ) = 𝜇𝜇 𝑐𝑐 0 0 2 0 〈𝑆𝑆 〉 𝐸𝐸 〈 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 〉 𝐸𝐸 Thus = 1 + 2 . 𝜇𝜇 𝑐𝑐 𝜇𝜇 𝑐𝑐

〈𝑆𝑆〉 〈𝑆𝑆 〉 〈𝑆𝑆 〉

Electromagnetic Waves

Lecture34: Electromagnetic Theory

Professor D. K. Ghosh, Physics Department, I.I.T., Bombay

Self Assessment Questions

1. The electric field of a plane electromagnetic wave propagating in free space is described by

= 0 sin( ) Determine the corresponding magnetic field and the time average Poynting vector for the 𝐸𝐸�⃗ 𝐸𝐸 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑦𝑦� wave. 2. Show that an s –polarized wave cannot be totally transmitted to another medium. 3. An electromagnetic wave given by

= 0 sin( ) is incident on the surface of a perfect conductor and is totally reflected. The incident and 𝐸𝐸�⃗ 𝐸𝐸 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑥𝑥� the reflected waves combine and form a pattern. What is the average Poynting vector?

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Solutions to Self Assessment Questions

1. The wave is propagating in +x direction so that the propagation vector is = . Using Faraday’s law, we have, × = which gives magnetic field directed along the z direction, 𝑘𝑘�⃗ 𝑘𝑘𝑥𝑥� 0 𝑘𝑘�⃗ 𝐸𝐸�⃗ 𝜔𝜔𝐵𝐵�⃗ = sin( ) 𝐸𝐸 The Poynting vector is given by 𝐵𝐵�⃗ 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑧𝑧̂ 𝑐𝑐 2 1 0 = × = sin2( ) 0 0 𝐸𝐸 2 The time average of the Poynting𝑆𝑆⃗ vector𝐸𝐸�⃗ is𝐵𝐵� ⃗ = 0 . 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑥𝑥� 𝜇𝜇 𝑐𝑐𝜇𝜇2 𝐸𝐸 0 2. Considering Fresnel equations for s polarization,〈𝑆𝑆⃗〉 𝑐𝑐𝜇𝜇

nIcos cos = nIcos 𝐼𝐼 + 𝑇𝑇 cos 𝑇𝑇 𝑠𝑠 𝜃𝜃 − 𝑛𝑛 𝜃𝜃 𝑟𝑟 𝐼𝐼 𝑇𝑇 𝑇𝑇 For total transmission = 0, so that nIcos = 𝜃𝜃cos𝑛𝑛 . From𝜃𝜃 Snell’s law, we have, nIsin = sin . Squaring and adding, we get 2 = 2 , which is not correct. 𝑟𝑟𝑠𝑠 𝜃𝜃𝐼𝐼 𝑛𝑛𝑇𝑇 𝜃𝜃𝑇𝑇 𝜃𝜃𝐼𝐼 𝑇𝑇 𝑇𝑇 𝑛𝑛 3. 𝜃𝜃The incident wave is 𝑛𝑛𝐼𝐼 𝑛𝑛𝑇𝑇

= 0 sin( ) Since the tangential component of the electric field must vanish at the interface (z=0), the 𝐸𝐸���⃗𝑖𝑖 𝐸𝐸 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑥𝑥� reflected wave us given by

= 0 sin( + ) The corresponding magnetic fields are given by �𝐸𝐸���𝑟𝑟⃗ 𝐸𝐸 𝑘𝑘𝑘𝑘 𝜔𝜔𝜔𝜔 𝑥𝑥� = 0 sin( ) 0 𝐸𝐸 �𝐻𝐻���⃗𝑖𝑖 0 𝑘𝑘𝑘𝑘 − 𝜔𝜔𝜔𝜔 𝑦𝑦� =𝑐𝑐𝜇𝜇 sin( + ) 0 𝐸𝐸 The individual Poynting vectors can�𝐻𝐻� ��be𝑟𝑟⃗ calculated− and𝑘𝑘𝑘𝑘 on𝜔𝜔𝜔𝜔 adding𝑦𝑦� it will turn out that the average 𝑐𝑐𝜇𝜇 Poynting vector is zero. The waves of the type obtained by superposition of the two waves have the structure,

= + = 2 0 sin cos where the space and time parts are separated. These are known as “standing waves” and they 𝐸𝐸�⃗ 𝐸𝐸���⃗𝑖𝑖 �𝐸𝐸���𝑟𝑟⃗ 𝐸𝐸 𝑘𝑘𝑘𝑘 𝜔𝜔𝜔𝜔 do not transport energy.

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