Introduction:

Reflection and refraction of electromagnetic waves:

When the plane wave is incident at boundary between two different media, part of wave is reflected and other part is refracted or transmitted. We need to find out ratio of intensity of reflected part to intensity of incident wave and ratio of intensity of refracted (transmitted) part to intensity of incident wave. Intensity is proportional to rate of energy per unit area i.e. Poynting vector. These ratios are termed as coefficients of reflection and transmission respectively. These coefficients can be determined in terms of refractive indices of media.

From equation i.e.

∇xE =-dB/dt

In integral form this equation becomes

_r E.dl = -d/dt∫_sB.ds

Let r be rectangular loop placed along surface common to two media as

In limit that Δb→0 i.e. along interface ∫B.ds = 0, ds being surface area and product of Δl and Δb. Therefore LHS of equation equals zero and

E.dl = E₁l - E₂l = 0 or E₁ = E₂

This implies that electric field is continuous. This is boundary condition for electric field. Boundary condition for magnetic intensity can be determined from modified Ampere's law (i.e. Maxwell's 4th equation) by considering figure below

Equation can be rendered in integral form as

H.dl = ∫_sjf.ds + d/dt∫_sD.ds

For other medium (i.e. second medium) of finite conductivity, first term in RHS of equation becomes zero. Second term is zero in limit that Δb→0 i.e. at interface as ds is product of dl and db. Therefore equation is

H.dl = 0 or H₁l - H₂l = 0 or H₁ = H₂ across the interface. H is continuous across interface. From equation of Gauss law for electric field i.e.

∇.E = Ρ/ε₀ or ∇.D = Ρf

Where D = ε₀E - P

We can write ∇.D = 0 as there is no free charge inside cylinder placed perpendicularly to interface

We can write ∇.D = 0 in integration as ∫₁D.ds + ∫₂D.ds = 0 as total flux of electric displacement has contribution from top and bottom surfaces of cylinder as dh → 0 at interface. Flux of D is normal to surfaces 1 and 2 of cylinder. Therefore we can write

D_⊥ds₁ = D_⊥ds₂ implying that D_⊥ is continuous

Last boundary condition is determined from Gauss law for magnetic field i.e. ∇.B = 0 which in integral form becomes ∫B.ds = 0. Using same cylinder in Figure given above, contributions of flux of magnetic field is from top and bottom surfaces of cylinder so that we can write

∫₁B.ds + ∫₂B.ds = 0

B_⊥ds₁ + B_⊥ds₂ = 0 where B_⊥ is continuous

From the four boundary conditions i.e.

E_1along = E_2along

H₁along = H_2along

D_1across = D_2across, and

B_1across = B_2across.

Reflection and refraction (transmission) coefficients of electromagnetic waves:

Noting that in medium 1 we have only the incident and reflected waves i.e. two waves  while in medium 2 we have only refracted or transmitted wave i.e. one wave, applying the 1st boundary condition we can write:

E_i + E_r = E_t

The second boundary condition gives

H_i - H_r = H_t

Where subscripts i, r and t represent incident, refracted and transmitted. Note that reflected magnetic intensity, H_r, points in direction opposite direction of H_i. This can be visualized by making middle finger of left hand point in direction of propagation (z-direction) while forefinger and thumb represents E and B respectively (i.e. in x and y directions). By pointing middle finger in negative z-direction, thumb representing magnetic field is observed to be pointing downward i.e. opposite its former direction.

Using H_i = B_i/μ₀ = E_i/cμ₀

As B = E/c and H_i is in free space. Substituting for

c = √1/εμ₀

Hi = (ε₀)^1/2E_i/(μ₀)^1/2

Likewise H_r (also in free space) = (ε₀)^1/2E_r/(μ₀)^1/2

H_t is though in medium of relative permittivity, ε_r and relative permeability, μ_r.

Therefore H_t = (ε₀ε_r)^1/2E_t/(μ₀μ_r)^1/2

Equation can be written as:

((ε₀)^1/2E_i)/(μ₀^1/2) - ((ε₀)^1/2E_r)/(μ₀^1/2) = [((ε₀ε_r)^1/2E_t)/(μ₀μ_r)^1/2]

Or E_i - E_r = (ε_r)^1/2E_t/(μ_r)^1/2

μ_r≈1 for most dielectrics so that equation becomes

E_i - E_r = (ε_r)^1/2E_t

Using amplitude of electric field of incident, reflected and transmitted waves and substituting for n = (ε_r)^1/2 equation can be written as

E_oi - E_or = nE_ot

Also using amplitude of electric fields in equation we have

E_oi + E_or = E_ot

By combining equations to eliminate E_ot provides:

E_or/E_ot = (1-n)(1+n)

Equations can also be combined to eliminate E_or. The result is

E_ot/E_oi = 2/(1+n)

Ratio of reflected energy of wave to that of incident energy is referred to as reflection coefficient, R. Also ratio of transmitted energy of wave to that of incident energy is referred to as transmitted coefficient, T. Such ratios or coefficients are determined from time average of Poynting vector over one cycle for incident wave, reflected wave and transmitted wave as Poynting vector is energy per unit time per unit area. We can use combination of energy of static electric and magnetic fields to approximate energy of electromagnetic wave. Using the capacitor, energy of static electric field is achieved as follows:

Work done in transferring charge from one plate of capacitor to other plate is

dW = Vdq = (q/c)dq since V = q/c

Integrating we have

W =₀∫^q(q/c)dq = (1/2)(q²/c) or w = 1/2CV²

For parallel plate capacitor C = ε₀A/d

The energy density of the electric field, U_E, i.e. energy per unit volume,

U_E = (ε₀A/Ad.d)V² = 1/2ε₀(V/d)² = 1/2ε₀E²

(volume = Ad, E = V/d and ε₀E = D) = 1/2D.E

Energy density in magnetic field is obtained as follows: let solenoid have cross sectional area. Volume Al of solenoid would store magnetic energy. Energy density is energy stored divided by Al where A is cross sectional area of solenoid and l its length. Magnetic energy dU, stored = Vidt or dU/dt = Vi or dU/dt = Vi or dU/dt = Vi or dU/dt = Li(di/dt)

As V = Ldi/dt where L is self inductance integrating,

∫dU = ∫Lidi = 1/2Li²

Energy density of magnetic field will then be, U_B = (1/2)(Li²/Al)

Substituting for L = μ₀n²lA and i = B/μ₀n

Where n = number of turns of solenoid

U_B = (1/2)(B/μ₀)

Substituting for H = (B/μ₀), U = (1/2)B.H and total magnetic energy is

U_B = 1/2∫B.Hdv

Equations can be combined to provide energy of electromagnetic wave assuming that it does not change for varying fields. Therefore

U_em = 1/2∫(E.D + B.H)dv

For the plane wave travelling along z-axis at speed c in box of cross sectional area, A and thickness dx, sum of equations give energy density of electromagnetic wave i.e.

U_em = U_E + U_B = 1/2ε₀E² + 1/2(B²/μ₀)

Energy U_em stored in box =(U_E + U_B)Adx i.e.

dU_em = (1/2ε₀E² + B²/μ₀)Adx

Substituting relation B = E/c in equation we have

dU_em[(1/2)ε₀(E)(cB) + (B/μ₀)(E/c)]Adx = [1/2EB(ε₀c + 1/μ₀c)]Adx = [1/2EB(ε₀μ₀c² + 1)/μ₀c]Adx

So that equation is

dU_em = EBAdx/μ₀c

Rate of transfer of this energy per unit area, i.e. Poynting vector, N,

N = dU_em/dtA = EBdx/μ₀cdt

But dx/dt = c and equation becomes N = EB/μ₀ = EH

Since B/μ₀ = H.

The average energy crossing unit area per second = E₀H₀/2 i.e. N = E₀H₀/2 as N is in direction of propagation and directions of E and H are orthogonal we can write,

N = E x H

By definitions of reflection coefficient, R and transmission coefficient, T given earlier,

R = (E_r x H_r)ave(E_i x H_i)ave = E_or².cμ₀/E_oi².cμ₀ = E_or²/E_oi²

R = (1-n/1+n)^1/2

Where √ε_r = n and √μ_r ≈ 1

Using equations T = 4n/(1+n)²

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