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## Reflection and Refraction of electromagnetic waves, Physics tutorial

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∫_{s}B.dsLet 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 = E

_{1}l - E_{2}l = 0 or E_{1}= E_{2}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 = ∫

_{s}jf.ds + d/dt∫_{s}D.dsFor 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 H

_{1}l - H_{2}l = 0 or H_{1 }= H_{2}across the interface. H is continuous across interface. From equation of Gauss law for electric field i.e.∇.E = Ρ/ε

_{0}or ∇.D = ΡfWhere D = ε

_{0}E - PWe can write ∇.D = 0 as there is no free charge inside cylinder placed perpendicularly to interface

We can write ∇.D = 0 in integration as ∫

_{1}D.ds + ∫_{2}D.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 writeD

_{⊥}ds_{1}= D_{⊥}ds_{2 }implying that D_{⊥}is continuousLast 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

∫

_{1}B.ds + ∫_{2}B.ds = 0B

_{⊥}ds_{1}+ B_{⊥}ds_{2}= 0 where B_{⊥}is continuousFrom the four boundary conditions i.e.

E

_{1along}= E_{2along}H

_{1}along = H_{2along}D

_{1across}= D_{2across}, andB

_{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}/μ_{0}= E_{i}/cμ_{0}As B = E/c and H

_{i}is in free space. Substituting forc = √1/εμ

_{0}Hi = (ε

_{0})^{1/2}E_{i}/(μ_{0})^{1/2}Likewise H

_{r}(also in free space) = (ε_{0})^{1/2}E_{r}/(μ_{0})^{1/2}H

_{t}is though in medium of relative permittivity, ε_{r}and relative permeability, μ_{r}.Therefore H

_{t}= (ε_{0}ε_{r})^{1/2}E_{t}/(μ_{0}μ_{r})^{1/2}Equation can be written as:

((ε

_{0})^{1/2}E_{i})/(μ_{0}^{1/2}) - ((ε_{0})^{1/2}E_{r})/(μ_{0}^{1/2}) = [((ε_{0}ε_{r})^{1/2}E_{t})/(μ_{0}μ_{r})^{1/2}]Or E

_{i }- E_{r}= (ε_{r})^{1/2}E_{t}/(μ_{r})^{1/2}μ

_{r}≈1 for most dielectrics so that equation becomesE

_{i}- E_{r}= (ε_{r})^{1/2}E_{t}Using amplitude of electric field of incident, reflected and transmitted waves and substituting for n = (ε

_{r})^{1/2}equation can be written asE

_{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 isE

_{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)(q^{2}/c) or w = 1/2CV^{2}For parallel plate capacitor C = ε

_{0}A/dThe energy density of the electric field, U

_{E}, i.e. energy per unit volume,U

_{E}= (ε_{0}A/Ad.d)V^{2}= 1/2ε_{0}(V/d)^{2}= 1/2ε_{0}E^{2}(volume = Ad, E = V/d and ε

_{0}E = D) = 1/2D.EEnergy 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 = 1/2Li

^{2}Energy density of magnetic field will then be, U

_{B}= (1/2)(Li^{2}/Al)Substituting for L = μ

_{0}n^{2}lA and i = B/μ_{0}nWhere n = number of turns of solenoid

U

_{B}= (1/2)(B/μ_{0})Substituting for H = (B/μ

_{0}), U = (1/2)B.H and total magnetic energy isU

_{B}= 1/2∫B.HdvEquations 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)dvFor 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}= 1/2ε_{0}E^{2}+ 1/2(B^{2}/μ_{0})Energy U

_{em}stored in box =(U_{E}+ U_{B})Adx i.e.dU

_{em}= (1/2ε_{0}E^{2}+ B^{2}/μ_{0})AdxSubstituting relation B = E/c in equation we have

dU

_{em}[(1/2)ε_{0}(E)(cB) + (B/μ_{0})(E/c)]Adx = [1/2EB(ε_{0}c + 1/μ_{0}c)]Adx = [1/2EB(ε_{0}μ_{0}c^{2}+ 1)/μ_{0}c]AdxSo that equation is

dU

_{em}= EBAdx/μ_{0}cRate of transfer of this energy per unit area, i.e. Poynting vector, N,

N = dU

_{em}/dtA = EBdx/μ_{0}cdtBut dx/dt = c and equation becomes N = EB/μ

_{0}= EHSince B/μ

_{0}= H.The average energy crossing unit area per second = E

_{0}H_{0}/2 i.e. N = E_{0}H_{0}/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_{or}^{2}.cμ_{0}/E_{oi}^{2}.cμ_{0 }= E_{or}^{2}/E_{oi}^{2}R = (1-n/1+n)

^{1/2}Where √ε

_{r}= n and √μ_{r}≈ 1Using equations T = 4n/(1+n)

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