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Electromagnetic Waves

2.1 Properties of EM Waves

From Maxwell’s equations, the following properties hold for plane EM waves:

  1. Transversality: E\mathbf{E} and B\mathbf{B} are perpendicular to k\mathbf{k} and to each other.
  2. Orthogonality: EB\mathbf{E} \perp \mathbf{B}And E=cB|\mathbf{E}| = c|\mathbf{B}|.
  3. In-phase: E\mathbf{E} and B\mathbf{B} oscillate in phase.
  4. Dispersion relation: ω=ck\omega = ck in vacuum.

Proof of transversality. For a plane wave E=E0ei(krωt)\mathbf{E} = \mathbf{E}_0 e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)} Gauss’s law gives ikE0=0i\mathbf{k}\cdot\mathbf{E}_0 = 0So kE0\mathbf{k} \perp \mathbf{E}_0. Similarly From B=0\nabla \cdot \mathbf{B} = 0: kB0\mathbf{k} \perp \mathbf{B}_0. \blacksquare

Proof of E=cB|\mathbf{E}| = c|\mathbf{B}|. From Faraday’s law for a plane wave: k×E0=ωB0\mathbf{k} \times \mathbf{E}_0 = \omega\mathbf{B}_0. Taking magnitudes: kE0=ωB0kE_0 = \omega B_0So E0/B0=ω/k=cE_0/B_0 = \omega/k = c. \blacksquare

Worked Example: Plane wave fields and intensity

Problem. A plane wave in vacuum has E=(20x^+30y^)cos(kzωt)\mathbf{E} = (20\hat{\mathbf{x}} + 30\hat{\mathbf{y}})\cos(kz - \omega t) V/m with λ=500\lambda = 500 nm. Find B\mathbf{B}The intensity, and describe the polarisation state.

Solution. E0=202+302=130036.1|\mathbf{E}_0| = \sqrt{20^2 + 30^2} = \sqrt{1300} \approx 36.1 V/m. B0=E0/c=36.1/(3×108)=1.20×107B_0 = E_0/c = 36.1/(3 \times 10^8) = 1.20 \times 10^{-7} T.

Since k=kz^\mathbf{k} = k\hat{\mathbf{z}} and B0=k^×E0/c\mathbf{B}_0 = \hat{\mathbf{k}} \times \mathbf{E}_0/c: B=(20y^+30x^)B0/E0cos(kzωt)/c\mathbf{B} = (-20\hat{\mathbf{y}} + 30\hat{\mathbf{x}})B_0/E_0 \cdot \cos(kz - \omega t)/c =(30x^20y^)(1/c)cos(kzωt)= (30\hat{\mathbf{x}} - 20\hat{\mathbf{y}})(1/c)\cos(kz - \omega t) T.

Intensity: I=12cε0E02=12(3×108)(8.854×1012)(1300)=1.73I = \frac{1}{2}c\varepsilon_0 E_0^2 = \frac{1}{2}(3 \times 10^8)(8.854 \times 10^{-12})(1300) = 1.73 W/m2^2.

Polarisation: E0\mathbf{E}_0 has components along x^\hat{\mathbf{x}} and y^\hat{\mathbf{y}} with a Constant phase relationship (δ=0\delta = 0), so the wave is linearly polarised at angle θ=arctan(30/20)=56.3°\theta = \arctan(30/20) = 56.3° from the xx-axis.

2.2 Energy and Intensity

The Poynting vector:

S=1μ0E×B\mathbf{S} = \frac{1}{\mu_0}\mathbf{E} \times \mathbf{B}

Represents the energy flux (W/m2^2). The time-averaged intensity for a plane wave:

I=S=12cε0E02I = \langle S \rangle = \frac{1}{2}c\varepsilon_0 E_0^2

The energy density of an EM field is:

u=12ε0E2+12μ0B2=ε0E2u = \frac{1}{2}\varepsilon_0 E^2 + \frac{1}{2\mu_0}B^2 = \varepsilon_0 E^2

(the electric and magnetic contributions are equal for a plane wave). The intensity is related to The energy density by I=ucI = uc.

Radiation pressure. For a perfectly absorbing surface: Prad=I/cP_{\mathrm{rad} = I/c}. For a perfectly Reflecting surface: Prad=2I/cP_{\mathrm{rad} = 2I/c}.

Worked Example: Radiation pressure from a laser

Problem. A 5 mW laser beam (λ=632.8\lambda = 632.8 nm) is normally incident on a perfectly reflecting Mirror. The beam has a diameter of 1 mm. Find the radiation pressure and the force on the mirror.

Solution. Beam area: A=π(0.5×103)2=7.85×107A = \pi(0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} m2^2. Intensity: I=P/A=5×103/(7.85×107)=6.37×103I = P/A = 5 \times 10^{-3}/(7.85 \times 10^{-7}) = 6.37 \times 10^3 W/m2^2.

Radiation pressure (reflecting): Prad=2I/c=2(6.37×103)/(3×108)=4.25×105P_{\mathrm{rad} = 2I/c = 2(6.37 \times 10^3)/(3 \times 10^8) = 4.25 \times 10^{-5}} Pa.

Force: F=PradA=(4.25×105)(7.85×107)=3.34×1011F = P_{\mathrm{rad} \cdot A = (4.25 \times 10^{-5})(7.85 \times 10^{-7}) = 3.34 \times 10^{-11}} N.

2.3 EM Waves in Matter

In a linear, isotropic, non-magnetic medium with refractive index nn:

v=cn,k=nωck^v = \frac{c}{n}, \quad \mathbf{k} = n\frac{\omega}{c}\hat{\mathbf{k}}

The index of refraction is related to the relative permittivity and permeability:

n=εrμrn = \sqrt{\varepsilon_r \mu_r}

For non-magnetic materials (μr1\mu_r \approx 1): nεrn \approx \sqrt{\varepsilon_r}.

The wavelength inside a medium of refractive index nn is λn=λ0/n\lambda_n = \lambda_0/nWhere λ0\lambda_0 Is the vacuum wavelength. The frequency remains unchanged across the boundary.

Worked Example: EM wave propagation in glass

Problem. A plane wave of wavelength λ0=600\lambda_0 = 600 nm in vacuum enters a glass slab (n=1.50n = 1.50) at normal incidence. Find (a) the wavelength and wave speed inside the glass, (b) the Frequency, and (c) the ratio of intensities inside and outside the glass, accounting for reflection At the front surface.

Solution.

(a) λn=λ0/n=600/1.50=400\lambda_n = \lambda_0/n = 600/1.50 = 400 nm. v=c/n=2.0×108v = c/n = 2.0 \times 10^8 m/s.

(b) f=c/λ0=(3×108)/(600×109)=5.0×1014f = c/\lambda_0 = (3 \times 10^8)/(600 \times 10^{-9}) = 5.0 \times 10^{14} Hz (unchanged).

(c) At normal incidence: R=[(n1n2)/(n1+n2)]2=[(11.5)/(1+1.5)]2=(0.5/2.5)2=0.04R = [(n_1 - n_2)/(n_1 + n_2)]^2 = [(1 - 1.5)/(1 + 1.5)]^2 = (0.5/2.5)^2 = 0.04. Transmittance: T=1R=0.96T = 1 - R = 0.96. The intensity inside the glass is Iinside=0.96I0I_{\mathrm{inside} = 0.96\,I_0}But the power per unit area Referenced to the vacuum intensity is Iinside=(n2/n1)TI0=1.5×0.96×I0=1.44I0I_{\mathrm{inside} = (n_2/n_1)\,T\,I_0 = 1.5 \times 0.96 \times I_0 = 1.44\,I_0} If we compare the electric field amplitudes squared times the respective impedances.

2.4 Boundary Conditions and Snell’s Law

At a planar interface between two linear, isotropic media, the tangential components of E\mathbf{E} and H\mathbf{H} and the normal components of D\mathbf{D} and B\mathbf{B} are Continuous across the boundary.

Consider a plane wave incident from medium 1 (n1n_1) onto medium 2 (n2n_2), with the interface at z=0z = 0 and the plane of incidence the xzxz-plane.

The phase matching condition requires the phases of all three waves (incident, reflected, Transmitted) to match at z=0z = 0 for all xx and tt. This gives:

k1sinθi=k1sinθr=k2sinθtk_1\sin\theta_i = k_1\sin\theta_r = k_2\sin\theta_t

From the first equality: θi=θr\theta_i = \theta_r (law of reflection). From the second equality: n1sinθi=n2sinθtn_1\sin\theta_i = n_2\sin\theta_t (Snell’s law).

Proof. The incident, reflected, and transmitted fields are:

Eiei(k1xx+k1zzωt),Erei(k1xx+k1zzωt),Etei(k2xx+k2zzωt)E_i \propto e^{i(k_{1x}x + k_{1z}z - \omega t)}, \quad E_r \propto e^{i(k_{1x}'x + k_{1z}'z - \omega t)}, \quad E_t \propto e^{i(k_{2x}x + k_{2z}z - \omega t)}

At z=0z = 0The tangential field must be continuous for all xx and tt: k1x=k1x=k2xk_{1x} = k_{1x}' = k_{2x}I.e., k1sinθi=k1sinθr=k2sinθtk_1\sin\theta_i = k_1\sin\theta_r = k_2\sin\theta_t. Since k=nω/ck = n\omega/cThis yields Snell’s law. \blacksquare

2.5 Fresnel Equations

Applying the boundary conditions for the tangential fields yields the Fresnel equations for the Amplitude reflection and transmission coefficients.

s-polarisation (E\mathbf{E} perpendicular to the plane of incidence, along y^\hat{\mathbf{y}}): The tangential components of E\mathbf{E} and H\mathbf{H} give:

rs=E0rE0i=n1cosθin2cosθtn1cosθi+n2cosθtr_s = \frac{E_{0r}}{E_{0i}} = \frac{n_1\cos\theta_i - n_2\cos\theta_t}{n_1\cos\theta_i + n_2\cos\theta_t}

ts=E0tE0i=2n1cosθin1cosθi+n2cosθtt_s = \frac{E_{0t}}{E_{0i}} = \frac{2n_1\cos\theta_i}{n_1\cos\theta_i + n_2\cos\theta_t}

p-polarisation (E\mathbf{E} parallel to the plane of incidence): The tangential components of E\mathbf{E} and H\mathbf{H} give:

rp=E0rE0i=n2cosθin1cosθtn2cosθi+n1cosθtr_p = \frac{E_{0r}}{E_{0i}} = \frac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t}

tp=E0tE0i=2n1cosθin2cosθi+n1cosθtt_p = \frac{E_{0t}}{E_{0i}} = \frac{2n_1\cos\theta_i}{n_2\cos\theta_i + n_1\cos\theta_t}

Reflectance and transmittance (energy fractions):

Rs=rs2,Ts=n2cosθtn1cosθits2,Rs+Ts=1R_s = |r_s|^2, \quad T_s = \frac{n_2\cos\theta_t}{n_1\cos\theta_i}|t_s|^2, \quad R_s + T_s = 1

Rp=rp2,Tp=n2cosθtn1cosθitp2,Rp+Tp=1R_p = |r_p|^2, \quad T_p = \frac{n_2\cos\theta_t}{n_1\cos\theta_i}|t_p|^2, \quad R_p + T_p = 1

At normal incidence (θi=0\theta_i = 0): rs=rp=(n1n2)/(n1+n2)r_s = r_p = (n_1 - n_2)/(n_1 + n_2) and R=[(n1n2)/(n1+n2)]2R = [(n_1 - n_2)/(n_1 + n_2)]^2.

Worked Example: Fresnel coefficients at a glass-air interface

Problem. Light is incident from air (n1=1.00n_1 = 1.00) onto glass (n2=1.50n_2 = 1.50) at θi=30°\theta_i = 30°. Calculate rsr_s, rpr_p, RsR_sAnd RpR_p.

Solution. From Snell’s law: sinθt=sin30°/1.50=0.333\sin\theta_t = \sin 30°/1.50 = 0.333So θt=19.47°\theta_t = 19.47°. cosθi=cos30°=0.866\cos\theta_i = \cos 30° = 0.866, cosθt=cos19.47°=0.943\cos\theta_t = \cos 19.47° = 0.943.

rs=1.00×0.8661.50×0.9431.00×0.866+1.50×0.943=0.8661.4140.866+1.414=0.5492.280=0.241r_s = \frac{1.00 \times 0.866 - 1.50 \times 0.943}{1.00 \times 0.866 + 1.50 \times 0.943} = \frac{0.866 - 1.414}{0.866 + 1.414} = \frac{-0.549}{2.280} = -0.241

Rs=rs2=0.0580R_s = r_s^2 = 0.0580

rp=1.50×0.8661.00×0.9431.50×0.866+1.00×0.943=1.2990.9431.299+0.943=0.3562.242=0.159r_p = \frac{1.50 \times 0.866 - 1.00 \times 0.943}{1.50 \times 0.866 + 1.00 \times 0.943} = \frac{1.299 - 0.943}{1.299 + 0.943} = \frac{0.356}{2.242} = 0.159

Rp=rp2=0.0252R_p = r_p^2 = 0.0252

At this angle, p-polarised light is reflected less efficiently than s-polarised light. The Negative sign of rsr_s indicates a phase shift of π\pi upon reflection.

:::caution Common Pitfall The Fresnel coefficients rsr_s and rpr_p have different forms. A common error is to swap the n1cosθin_1\cos\theta_i and n2cosθtn_2\cos\theta_t terms. Remember: for rsr_sThe numerator starts with n1cosθin_1\cos\theta_i; for rpr_pThe numerator starts with n2cosθin_2\cos\theta_i. Also, rr and tt are Amplitude coefficients, while RR and TT are energy coefficients — they are related but not Interchangeable.

2.6 Total Internal Reflection and Evanescent Waves

When light travels from a denser to a rarer medium (n1>n2n_1 \gt n_2) and the angle of incidence Exceeds the critical angle:

θc=arcsin ⁣(n2n1)\theta_c = \arcsin\!\left(\frac{n_2}{n_1}\right)

Snell’s law gives sinθt=(n1/n2)sinθi>1\sin\theta_t = (n_1/n_2)\sin\theta_i \gt 1So θt\theta_t becomes complex. Writing cosθt=isin2θt1\cos\theta_t = i\sqrt{\sin^2\theta_t - 1}The Fresnel coefficients become complex with rs2=rp2=1|r_s|^2 = |r_p|^2 = 1: all energy is reflected.

The transmitted field becomes an evanescent wave:

Eteκzei(kxxωt)E_t \propto e^{-\kappa z}\, e^{i(k_x x - \omega t)}

Where:

κ=k0n12sin2θin22,kx=k0n1sinθi\kappa = k_0\sqrt{n_1^2\sin^2\theta_i - n_2^2}, \quad k_x = k_0 n_1\sin\theta_i

The field decays exponentially with penetration depth δ=1/κ\delta = 1/\kappa into the second medium, But propagates without loss along the interface. No net energy is transported across the boundary (T=0T = 0).

Frustrated total internal reflection (FTIR). If a third medium (with n3n2n_3 \geq n_2) is brought Within a distance comparable to δ\delta of the interface, the evanescent wave can couple into it, Allowing energy transmission across the gap. This is the optical analogue of quantum mechanical Tunnelling.

Worked Example: Critical angle and evanescent wave penetration

Problem. Light travels from glass (n1=1.50n_1 = 1.50) to air (n2=1.00n_2 = 1.00) at θi=50°\theta_i = 50°. Find (a) the critical angle, (b) the penetration depth for λ=500\lambda = 500 nm, and (c) the Propagation constant along the interface.

Solution.

(a) θc=arcsin(n2/n1)=arcsin(1/1.50)=41.8°\theta_c = \arcsin(n_2/n_1) = \arcsin(1/1.50) = 41.8°. Since 50°>41.8°50° \gt 41.8°TIR occurs.

(b) κ=k0n12sin2θin22\kappa = k_0\sqrt{n_1^2\sin^2\theta_i - n_2^2} =2πλ(1.50)2sin250°1.002= \frac{2\pi}{\lambda}\sqrt{(1.50)^2\sin^2 50° - 1.00^2} =2π500×1092.25×0.5871.00= \frac{2\pi}{500 \times 10^{-9}}\sqrt{2.25 \times 0.587 - 1.00} =(1.257×107)0.320= (1.257 \times 10^7)\sqrt{0.320} =(1.257×107)(0.566)=7.11×106= (1.257 \times 10^7)(0.566) = 7.11 \times 10^6 m1^{-1}

Penetration depth: δ=1/κ=1.41×107\delta = 1/\kappa = 1.41 \times 10^{-7} m =141= 141 nm.

(c) kx=k0n1sinθi=(2π/(500×109))(1.50)(0.766)=1.44×107k_x = k_0 n_1\sin\theta_i = (2\pi/(500 \times 10^{-9}))(1.50)(0.766) = 1.44 \times 10^7 m1^{-1}.

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