Reflection and refraction of a plane wave at oblique incidence

Let us consider a plane wave that obliquely incidents at the boundary of two media that

are characterized by their permittivity and permeability (see Figure 1). The plane

containing both the normal to the surface and the direction of propagation of the incident

wave is known as the plane of incidence.

We consider two different cases:

Case A: The electric field of the incident wave is perpendicular to the plane of

incidence.

Case B: The electric field of the incident wave is in the plane of incidence.

Any other incident wave can be decomposed into linear combination of these two.

Fig. 1 shows a wave of either polarization incident on the boundary of two media. In this

Figure, the angle

i between the normal to the boundary and the propagation direction is

the angle of incidence. We choose the plane of incidence to be the

x z plane with the

axes directions shown in the Figure1, the y-axes is out of the page.

There may also be reflected and refracted (transmitted) waves, as shown in Fig. 1.

Directions of propagation of these waves have angles

r and

t with the normal to the

boundary. The unit wave vectors of the incident, reflected and transmitted waves can be

written as the following:

k i [sini ,0,cosi ] (1)

k r [sinr ,0,cosr ] (2)

k t [sint ,0,cost ] (3)

The phasors of the traveling incident, reflected, and refracted plane waves can be written

in the following form

)exp()( 0 rkiErE ii , )exp()( 1 rkiErE rr

, and )exp()( 2 rkiErE tt

correspondingly. If the wave fields depend on coordinate as

exp(ik r ), then Maxwell

equations for phasors

HiE

(4)

EiH

can be written in the form:

i(k E ) iH (4)

i(k H ) iE (5)

Now we can write the wave fields for the cases A and B,

Incident wave, )}cossin(exp{~, 1 iiii zxikHE

Case A (

E perpendicular to plane of incidence)

}exp{)sincos(

}exp{

1

10

0

rkikiEH

rkiEjE

iiii

ii

(6)

Case B (

E in plane of incidence)

}exp{

}exp{)sincos(

1

10

0

rkiEjH

rkikiEE

ii

iiii

(7)

Reflected wave, )}cossin(exp{~, 1 rrrr zxikHE

Case A (

E perpendicular to plane of incidence)

}exp{)sincos(

}exp{

1

11

1

rkikiEH

rkiEjE

rrrr

rr

(8)

Case B (

E in plane of incidence)

}exp{

}exp{)sincos(

1

11

1

rkiEjH

rkikiEE

rr

rrrr

(10)

Refracted (transmitted) wave, )}cossin(exp{~, 2 tttt zxikHE

Case A (

E perpendicular to plane of incidence)

}exp{)sincos(

}exp{

2

22

2

rkikiEH

rkiEjE

tttt

tit

(11)

Case B (

E in plane of incidence)

}exp{

}exp{)sincos(

2

20

2

rkiEjH

rkikiEE

tt

tttt

(12)

We introduced unit vectors kji ,, along zyx ,, axes in Eqs. (6)-(12).

Case B: Let us consider a plane wave with electric field E

in the plane of incidence

incident on the discontinuity between two dielectrics ( 11, ),( 22 , )

The boundary conditions at 0z are continuity of tangential components of electric and

magnetic fields tanE

and tanH

:

)sinexp()sinexp()sinexp(

)sinexp(cos)sinexp(cos)sinexp(cos

222

211

1

110

1

1

221110

tri

ttrrii

xikExikExikE

xikExikExikE

(13)

Eqs. (13) must hold for all values of x , which is possible only if

tri kkk sinsinsin 211 (14)

We see that

ri (angle of refraction equals angle of incidence) (15)

and

2

1)1(

)2(

22

11

2

1

sin

sin

n

n

v

v

k

k

ph

ph

i

t

(16)

Here n is index of refraction.

Eq. (16) is the familiar Snells law.

Canceling the exponential terms in (13), (14) by means of (15), we obtain

22

21

1

10

1

1

210 coscoscos

EEE

EEE tii

(17)

We can now solve Eqs. (17) for 1E and 2E with the result

it

i

it

it

E

E

E

E

coscos

cos2

coscos

coscos

2211

11

0

2

2211

2211

0

1

(18)