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TM (transverse magnetic) Mode

Figure . Vector of incident wave reflected and transmitted under TM modeOn this case : Polarization components of E at each position parallel to the plane of incidenceWhile B is perpendicular to the plane of incidence and parallel to the boundary of the two medium, the direction of vector ErBr have to be opposite with the vector EiBiThe incident wave :

The reflected wave :

The transmitted wave:

At x = 0 the tangential component of E are continue, which yields Eoi cos i - Eor cos r = Eot cos t

by the 4th boundary condition that the tangential components of H are continue yields to:n1 Eoi + n1 Eor = n2 Eot

By defining that r = E0r/E0i and t = E0t/E0i, we get

And by applying the Snells law :

TE (transverse Electric) Mode

Reflection and transmission of em wave for TE modeThe electric field on incident wave is perpendicular to the plane of incidence thus the magnetic field is on the plane of incidence On this system the magnetic field of the reflected wave and transmitted wave undergoes a change in polarization direction.It satisfy the continuity of tangetial components of E and H at z = 0, therefore

The 2nd boundary condition gives

So the coefficient of reflection and transmission respectively given by:

7The Reflection and Transmission Related to EnergyHow the energy of the incident wave divided into energy of reflected wave and energy of transmitted wave

Remember that

thusTherefore the coefficient of reflection and the coefficient of transmission of each refelected mode can be expressed by:

By the same method for The transmitted energy:

by the approximation of1= 2 dan (2/1)1/2 = (n2/n1)1/2

Recall that

The equation becomesAnd finally we get that:

By using the same method, it is obtained that:

R + T = 1It is approved thatThe case that incident wave is perpendicular to the boundary medium

For the reflection coefficient of the two modes