Polarisation: E0 has components along x^ and y^ with a Constant phase relationship (δ=0), so the wave is linearly polarised at angle θ=arctan(30/20)=56.3° from the x-axis.
2.2 Energy and Intensity
The Poynting vector:
S=μ01E×B
Represents the energy flux (W/m2). The time-averaged intensity for a plane wave:
I=⟨S⟩=21cε0E02
The energy density of an EM field is:
u=21ε0E2+2μ01B2=ε0E2
(the electric and magnetic contributions are equal for a plane wave). The intensity is related to The energy density by I=uc.
Radiation pressure. For a perfectly absorbing surface: Prad=I/c. For a perfectly Reflecting surface: Prad=2I/c.
Worked Example: Radiation pressure from a laser
Problem. A 5 mW laser beam (λ=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.
Force: F=Prad⋅A=(4.25×10−5)(7.85×10−7)=3.34×10−11 N.
2.3 EM Waves in Matter
In a linear, isotropic, non-magnetic medium with refractive index n:
v=nc,k=ncωk^
The index of refraction is related to the relative permittivity and permeability:
n=εrμr
For non-magnetic materials (μr≈1): n≈εr.
The wavelength inside a medium of refractive index n is λn=λ0/nWhere λ0 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 nm in vacuum enters a glass slab (n=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.
(c) At normal incidence: R=[(n1−n2)/(n1+n2)]2=[(1−1.5)/(1+1.5)]2=(0.5/2.5)2=0.04. Transmittance: T=1−R=0.96. The intensity inside the glass is Iinside=0.96I0But the power per unit area Referenced to the vacuum intensity is Iinside=(n2/n1)TI0=1.5×0.96×I0=1.44I0 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 and H and the normal components of D and B are Continuous across the boundary.
Consider a plane wave incident from medium 1 (n1) onto medium 2 (n2), with the interface at z=0 and the plane of incidence the xz-plane.
The phase matching condition requires the phases of all three waves (incident, reflected, Transmitted) to match at z=0 for all x and t. This gives:
k1sinθi=k1sinθr=k2sinθt
From the first equality: θi=θr (law of reflection). From the second equality: n1sinθi=n2sinθt (Snell’s law).
Proof. The incident, reflected, and transmitted fields are:
At z=0The tangential field must be continuous for all x and t: k1x=k1x′=k2xI.e., k1sinθi=k1sinθr=k2sinθt. Since k=nω/cThis yields Snell’s law. ■
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 perpendicular to the plane of incidence, along y^): The tangential components of E and H give:
At this angle, p-polarised light is reflected less efficiently than s-polarised light. The Negative sign of rs indicates a phase shift of π upon reflection.
:::caution Common Pitfall The Fresnel coefficients rs and rp have different forms. A common error is to swap the n1cosθi and n2cosθt terms. Remember: for rsThe numerator starts with n1cosθi; for rpThe numerator starts with n2cosθi. Also, r and t are Amplitude coefficients, while R and T 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>n2) and the angle of incidence Exceeds the critical angle:
θc=arcsin(n1n2)
Snell’s law gives sinθt=(n1/n2)sinθi>1So θt becomes complex. Writing cosθt=isin2θt−1The Fresnel coefficients become complex with ∣rs∣2=∣rp∣2=1: all energy is reflected.
The transmitted field becomes an evanescent wave:
Et∝e−κzei(kxx−ωt)
Where:
κ=k0n12sin2θi−n22,kx=k0n1sinθi
The field decays exponentially with penetration depth δ=1/κ into the second medium, But propagates without loss along the interface. No net energy is transported across the boundary (T=0).
Frustrated total internal reflection (FTIR). If a third medium (with n3≥n2) is brought Within a distance comparable to δ 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=1.50) to air (n2=1.00) at θi=50°. Find (a) the critical angle, (b) the penetration depth for λ=500 nm, and (c) the Propagation constant along the interface.
Solution.
(a) θc=arcsin(n2/n1)=arcsin(1/1.50)=41.8°. Since 50°>41.8°TIR occurs.