Part 4 · Chapter 33

Electromagnetic Waves

Maxwell's crowning achievement was to show that a beam of light is a traveling wave of electric and magnetic fields — an electromagnetic wave — and so that optics is a branch of electromagnetism. This chapter is the hinge between the two: we finish our study of electric and magnetic fields and lay the foundation for optics. We watch the two fields create each other endlessly through induction as the wave races outward at the speed of light, find how it carries energy and momentum, and then meet the everyday optics of polarization, reflection, refraction, total internal reflection, and the bright geometry of a rainbow.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 75 min
i What you'll learn
  • Maxwell's rainbow: the continuous electromagnetic spectrum (radio to gamma), all traveling at one speed \(c\) in vacuum, with visible light a thin slice from about 430 to 690 nm.
  • The traveling wave \(E = E_m\sin(kx-\omega t)\), \(B = B_m\sin(kx-\omega t)\) — transverse, mutually perpendicular, in phase, with \(\vec{E}\times\vec{B}\) pointing the way it travels — and the dual induction that links \(c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}}\) and \(\dfrac{E_m}{B_m} = c\).
  • Energy transport via the Poynting vector \(\vec{S} = \dfrac{1}{\mu_0}\vec{E}\times\vec{B}\), the intensity \(I = \dfrac{1}{c\mu_0}E_{\text{rms}}^{2}\), the inverse-square law \(I = \dfrac{P_s}{4\pi r^{2}}\), and radiation pressure \(p_r = \dfrac{I}{c}\) (absorbed) or \(\dfrac{2I}{c}\) (reflected).
  • Polarization: the one-half rule \(I = \tfrac{1}{2}I_0\) for unpolarized light and the cosine-squared rule \(I = I_0\cos^{2}\theta\) for already-polarized light.
  • Reflection and refraction: \(\theta_1' = \theta_1\) and Snell's law \(n_2\sin\theta_2 = n_1\sin\theta_1\); total internal reflection above the critical angle \(\theta_c = \sin^{-1}\dfrac{n_2}{n_1}\); and full polarization on reflection at the Brewster angle \(\theta_B = \tan^{-1}\dfrac{n_2}{n_1}\).
Section 33-1

What Is Physics?

The information age rests almost entirely on the physics of electromagnetic waves. Television, telephones, and the Web connect the globe, and we are continuously immersed in the signals of countless transmitters. The starting point for understanding all of it — and for imagining what comes next — is the basic physics of electromagnetic waves, which span so vast a range of types that they are poetically called Maxwell's rainbow.

Section 33-2

Maxwell's Rainbow

In Maxwell's day only the visible, infrared, and ultraviolet forms of light were known. Spurred by his theory, Heinrich Hertz discovered radio waves and showed they travel at the same speed as visible light. We now know a continuous spectrum stretching from long radio waves through microwaves, infrared, visible, ultraviolet, x rays, and gamma rays. There are no gaps and no inherent upper or lower bound on wavelength — and every one of these waves travels through vacuum at the same speed \(c\).

The visible band is a thin slice of this spectrum. The eye's sensitivity peaks near 555 nm — the yellow-green we see most readily — and falls off smoothly toward the red and violet ends. Taking the limits where sensitivity drops to about 1% of its maximum gives roughly 430 to 690 nm, though intense light can be seen a little beyond.

One spectrum, one speed. The familiar labels — "x rays," "AM radio," "TV channels" — mark wavelength ranges defined by common sources, detectors, or legal assignment, not by any physical break. All electromagnetic waves, no matter where they sit in the spectrum, are the same kind of thing and travel at the same \(c\) in free space.
Section 33-3

The Traveling Electromagnetic Wave, Qualitatively

How is a radio-region wave made? At the heart of a transmitter is an LC oscillator running at \(\omega = 1/\sqrt{LC}\) (Chapter 31), coupled through a transformer and transmission line to an antenna of two conducting rods. The oscillating charge makes the antenna a tiny electric dipole whose moment swings sinusoidally; the changing dipole produces changing electric and magnetic fields. Those changes do not appear everywhere at once — they travel outward at speed \(c\) as an electromagnetic wave of the same angular frequency \(\omega\).

Far from the antenna the wavefronts are nearly flat — a plane wave — and four features stand out, true no matter how the wave is made: the fields \(\vec{E}\) and \(\vec{B}\) are both perpendicular to the travel direction (it is a transverse wave); \(\vec{E}\) is perpendicular to \(\vec{B}\); the cross product \(\vec{E}\times\vec{B}\) points the way the wave travels; and the fields vary sinusoidally, in phase, at the same frequency.

The plane-wave fields
\[ E = E_m\sin(kx - \omega t), \qquad B = B_m\sin(kx - \omega t) \]
With the wave traveling along \(x\), \(\vec{E}\) oscillating along \(y\), and \(\vec{B}\) along \(z\). Here \(k = 2\pi/\lambda\) is the angular wave number and \(\omega\) the angular frequency. Each field forms its own wave, but the two cannot exist independently.
Wave speed and the field ratios
\[ c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \approx 3.0\times10^{8}\,\mathrm{m/s}, \qquad \frac{E_m}{B_m} = c, \qquad \frac{E}{B} = c \]
All electromagnetic waves share this speed in vacuum. The fields are so tightly coupled — at every instant and place \(E = cB\) — that we usually track only one of them, almost always \(\vec{E}\), since detectors respond to the electric component.
📐
A most curious wave
An electromagnetic wave needs no medium — it propagates through the vacuum of space, and the two fields sustain each other by induction.

A changing \(\vec{B}\) induces a perpendicular \(\vec{E}\) (Faraday), and that changing \(\vec{E}\) induces a perpendicular \(\vec{B}\) (Maxwell), and so on without end — the fields continuously create each other and the pattern travels as light. Relativity later revealed its speed is special: every observer measures the same \(c\), whatever their motion. The meter is now defined so that \(c = 299\,792\,458\,\mathrm{m/s}\) exactly.

Section 33-4

The Traveling Wave, Quantitatively

The two field ratios are not assumptions — they follow from Faraday's and Maxwell's laws applied to a thin rectangle riding along with the wave. As the wave sweeps past, the changing magnetic flux through a rectangle in the \(xy\) plane induces the wave's electric field; the changing electric flux through a rectangle in the \(xz\) plane induces its magnetic field. Each law yields a partial-derivative relation between the fields.

The coupled field equations
\[ \frac{\partial E}{\partial x} = -\frac{\partial B}{\partial t}, \qquad -\frac{\partial B}{\partial x} = \mu_0\varepsilon_0\frac{\partial E}{\partial t} \]
The first comes from Faraday's law, the second from Maxwell's law of induction. The minus signs are necessary: where one field rises with position, the other falls with time. Substituting the sinusoidal forms turns these into algebraic relations between amplitudes.
What the two laws give
\[ \frac{E_m}{B_m} = c \quad(\text{from Faraday}), \qquad \frac{E_m}{B_m} = \frac{1}{\mu_0\varepsilon_0 c} \quad(\text{from Maxwell}) \]
Setting the two expressions for \(E_m/B_m\) equal forces \(c = 1/\sqrt{\mu_0\varepsilon_0}\). So the very same induction that makes the wave self-sustaining also fixes its speed — light's speed is built into the electric and magnetic constants of empty space.
Section 33-5

Energy Transport and the Poynting Vector

Every sunbather knows light carries energy. The rate of energy flow per unit area is given by the Poynting vector, whose direction is the wave's direction of travel.

Poynting vector and instantaneous flow rate
\[ \vec{S} = \frac{1}{\mu_0}\vec{E}\times\vec{B}, \qquad S = \frac{1}{\mu_0}EB = \frac{1}{c\mu_0}E^{2} \]
Because \(\vec{E}\perp\vec{B}\), the magnitude of \(\vec{E}\times\vec{B}\) is \(EB\); using \(B = E/c\) writes everything in terms of the electric field. The SI unit of \(S\) is the watt per square metre.
Intensity (time-averaged S) and rms field
\[ I = S_{\text{avg}} = \frac{1}{c\mu_0}E_{\text{rms}}^{2}, \qquad E_{\text{rms}} = \frac{E_m}{\sqrt{2}} \]
The intensity is the average flow rate, using \([\sin^{2}]_{\text{avg}} = \tfrac{1}{2}\). Although \(E = cB\) with \(c\) large, the electric and magnetic energy densities are exactly equal, \(u_E = u_B\) everywhere along the wave — the two components carry energy on an equal footing.
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The inverse-square law
A point source emitting power \(P_s\) isotropically spreads its energy over an expanding sphere, so the intensity falls as \(1/r^{2}\).

All the emitted power must cross any sphere centered on the source, of area \(4\pi r^{2}\), giving \(I = \dfrac{P_s}{4\pi r^{2}}\). Double the distance and the intensity drops to a quarter — the reason a distant star, however luminous, delivers only the faintest fields to your eye.

Section 33-6

Radiation Pressure

Light carries momentum as well as energy, so it pushes on whatever it strikes. The push is tiny — you feel no punch from a camera flash — but it is real. If a body absorbs energy \(\Delta U\) from a beam it also gains momentum; if it reflects the beam straight back, it gains twice as much, just as an elastic tennis ball delivers twice the impulse of a lump of putty.

Momentum delivered by radiation
\[ \Delta p = \frac{\Delta U}{c}\ \;(\text{total absorption}), \qquad \Delta p = \frac{2\Delta U}{c}\ \;(\text{total reflection}) \]
The momentum change points along the incident beam. Partial absorption gives a value in between.
Radiation pressure
\[ p_r = \frac{I}{c}\ \;(\text{total absorption}), \qquad p_r = \frac{2I}{c}\ \;(\text{total reflection}) \]
From \(F = \Delta p/\Delta t\) and intensity \(I = \text{power}/\text{area}\), the force on area \(A\) is \(IA/c\) (absorbed) or \(2IA/c\) (reflected); dividing by \(A\) gives the pressure, in pascals. Focused laser beams can deliver pressures vastly larger than a lamp's, enough to manipulate small particles.
Section 33-7

Polarization

The plane containing a wave's oscillating \(\vec{E}\) is its plane of oscillation, and the wave is plane-polarized in that direction. Common sources — the Sun, a bulb — emit unpolarized light: the electric field stays perpendicular to the travel direction but points randomly. A polarizing sheet transmits only the field component along its polarizing direction and absorbs the perpendicular component, so the emerging light is polarized along that direction.

The two intensity rules
\[ I = \tfrac{1}{2}I_0\ \;(\text{one-half rule, unpolarized in}), \qquad I = I_0\cos^{2}\theta\ \;(\text{cosine-squared rule, polarized in}) \]
For unpolarized light the random orientations average to half, so exactly half the intensity passes. For already-polarized light, \(\theta\) is the angle between the incoming polarization and the sheet's axis; the transmitted field is \(E\cos\theta\), and since intensity goes as the field squared, the transmitted intensity goes as \(\cos^{2}\theta\).
📐
Work through a stack sheet by sheet
Apply the one-half rule at the first sheet only if the incoming light is unpolarized; everywhere after, the light is polarized, so use the cosine-squared rule.

After each sheet the transmitted light is polarized along that sheet's axis, so the angle for the next sheet is measured from there. Crossed sheets (axes at \(90^\circ\)) pass nothing; parallel sheets pass everything the first one let through. Scattering and reflection also polarize light — which is why the sky's glow is partly polarized and why bees and Viking navigators could read the hidden Sun's direction from it.

Section 33-8

Reflection and Refraction

Treating light as straight-line rays is the realm of geometrical optics. When a ray meets a boundary between two transparent media, part reflects and part passes through, bending — refracting — as it crosses. All angles are measured from the normal, the line perpendicular to the surface. Two laws govern what happens.

Law of reflection and Snell's law of refraction
\[ \theta_1' = \theta_1, \qquad n_2\sin\theta_2 = n_1\sin\theta_1 \]
The reflected ray leaves at the incidence angle. The index of refraction \(n = c/v\) compares the speed of light in vacuum to its speed in the medium; \(n = 1\) exactly for vacuum, about 1.00 for air, 1.33 for water, ~1.5 for glass, 2.42 for diamond. Light entering a higher-index medium (\(n_2 > n_1\)) bends toward the normal; entering a lower-index medium, it bends away.
Chromatic dispersion and rainbows. A medium's index depends slightly on wavelength — generally larger for blue than for red — so white light spreads into colors on refraction, blue bending more than red. A prism exploits this twice. A raindrop refracts, reflects once inside, and refracts again, sending the separated colors back at about \(42^\circ\) from the antisolar point: the primary rainbow. A second internal reflection makes the fainter, color-reversed secondary rainbow at about \(52^\circ\).
Section 33-9

Total Internal Reflection

Send light from glass toward air. As the incidence angle grows, the refracted ray bends ever farther from the normal until, at the critical angle \(\theta_c\), it grazes along the surface (\(\theta_2 = 90^\circ\)). Beyond \(\theta_c\) no light escapes at all — every bit reflects back inside. This is total internal reflection.

Critical angle
\[ n_1\sin\theta_c = n_2\sin 90^\circ \;\Rightarrow\; \theta_c = \sin^{-1}\frac{n_2}{n_1} \]
Since a sine cannot exceed 1, this requires \(n_2 < n_1\): total internal reflection happens only when light tries to pass into a lower-index medium. It is the principle of the optical fiber, where light bounces along a curved glass core with almost no loss — the basis of fiber-optic communication and the medical endoscope.
Section 33-10

Polarization by Reflection

Light reflected from a surface is partly or fully polarized — which is why a polarizing sunglass lens cuts glare. Resolve the incident field into components perpendicular and parallel to the plane of incidence. In general both reflect, but at one special incidence angle, the Brewster angle \(\theta_B\), the reflected light contains only the perpendicular component — it is fully polarized perpendicular to the plane of incidence.

Brewster angle and Brewster's law
\[ \theta_B = \tan^{-1}\frac{n_2}{n_1}, \qquad \theta_B = \tan^{-1} n\ \;(\text{from air, } n_1 \approx 1) \]
At \(\theta_B\) the reflected and refracted rays are perpendicular, \(\theta_B + \theta_r = 90^\circ\). From a horizontal surface like water, the reflected glare is polarized horizontally — so polarizing sunglasses are mounted with their axis vertical to block it. Sir David Brewster found both relations experimentally in 1812.
Worked Examples

Putting It to Work

1 Field amplitudes of sunlight

Problem. Sunlight just outside Earth's atmosphere has intensity \(I = 1.40\,\mathrm{kW/m^2}\). Treating it as a plane wave, find the electric-field amplitude \(E_m\) and the magnetic-field amplitude \(B_m\).

Solution. Intensity relates to the amplitude through \(I = E_m^{2}/2c\mu_0\), so \(E_m = \sqrt{2c\mu_0 I}\); then \(B_m = E_m/c\).

E_m = √(2cμ₀I)
\[ E_m = \sqrt{2(3.00\times10^{8})(4\pi\times10^{-7})(1.40\times10^{3})} \approx 1.03\times10^{3}\,\mathrm{V/m} \]
B_m = E_m/c
\[ B_m = \frac{1.03\times10^{3}}{3.00\times10^{8}} \approx 3.4\times10^{-6}\,\mathrm{T} = 3.4\,\mu\mathrm{T} \]

Note how small \(B_m\) looks beside \(E_m\) — but only because they carry different units. Their energy densities are equal; you cannot say one component is "stronger."

2 Intensity and rms field of a point source

Problem. An isotropic point source radiates \(P_s = 250\,\mathrm{W}\). At \(r = 12\,\mathrm{m}\), find the intensity and the rms electric field.

Solution. Use the inverse-square law for \(I\), then \(I = E_{\text{rms}}^{2}/c\mu_0\) for the field.

I = P_s / 4πr²
\[ I = \frac{250}{4\pi(12)^{2}} \approx 0.138\,\mathrm{W/m^2} \]
E_rms = √(I c μ₀)
\[ E_{\text{rms}} = \sqrt{(0.138)(3.00\times10^{8})(4\pi\times10^{-7})} \approx 7.2\,\mathrm{V/m} \]

Move twice as far away and the intensity drops to one-quarter, while the rms field — going as \(\sqrt{I}\) — drops only to one-half.

3 Three polarizing sheets

Problem. Unpolarized light of intensity \(I_0\) passes through three sheets: the first with axis along \(y\), the second at \(60^\circ\) counterclockwise from \(y\), the third along \(x\). What fraction of \(I_0\) emerges?

Solution. One-half rule at sheet 1; cosine-squared rule thereafter. The angle into sheet 2 is \(60^\circ\), and into sheet 3 (along \(x\), i.e. \(90^\circ\) from \(y\)) it is \(90^\circ - 60^\circ = 30^\circ\).

I₃ = ½ I₀ cos²60° cos²30°
\[ I_3 = \tfrac{1}{2}I_0\cos^{2}60^\circ\cos^{2}30^\circ = \tfrac{1}{2}I_0\,(0.25)(0.75) \approx 0.094\,I_0 \]

About \(9.4\%\) emerges, polarized along \(x\). Curiously, removing the middle sheet would leave the first and third crossed at \(90^\circ\) — and nothing would get through. The middle sheet lets light past by rotating the polarization in two manageable steps.

4 Reflection and refraction at two interfaces

Problem. A monochromatic beam in material 1 (\(n_1 = 1.33\)) meets material 2 (\(n_2 = 1.77\)), making \(50^\circ\) with the interface. Find the reflection angle and the refraction angle. Then the refracted beam strikes a parallel interface with air (\(n_3 = 1.00\)); find the new refraction angle.

Solution. The incidence angle from the normal is \(90^\circ - 50^\circ = 40^\circ\), so the reflection angle is \(\theta_1' = 40^\circ\). Snell's law gives the refraction into material 2.

θ₂ = sin⁻¹[(n₁/n₂) sin θ₁]
\[ \theta_2 = \sin^{-1}\!\left(\frac{1.33}{1.77}\sin 40^\circ\right) \approx 28.9^\circ \]

The beam bends toward the normal, as it must on entering a higher-index medium. At the parallel interface the incidence angle equals \(\theta_2\), so

θ₃ = sin⁻¹[(n₂/n₃) sin θ₂]
\[ \theta_3 = \sin^{-1}\!\left(\frac{1.77}{1.00}\sin 28.9^\circ\right) \approx 58.8^\circ \]

Now the beam bends away from the normal, entering lower-index air — swinging from \(29^\circ\) back out to \(59^\circ\).

5 Critical angle and Brewster angle for water

Problem. For a water–air interface (\(n_{\text{water}} = 1.33\), \(n_{\text{air}} = 1.00\)), find (a) the critical angle for light in the water, and (b) the Brewster angle for light reflecting off the water surface from the air side.

Solution. Total internal reflection: \(\theta_c = \sin^{-1}(n_2/n_1)\) with \(n_1 = 1.33\), \(n_2 = 1.00\).

θ_c = sin⁻¹(1.00/1.33)
\[ \theta_c = \sin^{-1}\!\left(\frac{1.00}{1.33}\right) \approx 48.8^\circ \]

Light inside the water striking the surface beyond \(48.8^\circ\) is trapped. For glare from above, Brewster's law uses \(n = 1.33\):

θ_B = tan⁻¹(n)
\[ \theta_B = \tan^{-1}(1.33) \approx 53.1^\circ \]

Sunlight hitting the water at about \(53^\circ\) from the vertical reflects fully horizontally polarized — exactly the glare a vertically-oriented polarizing lens removes.

Review

Chapter Summary

✓ The wave

\(E = E_m\sin(kx-\omega t)\), \(B = B_m\sin(kx-\omega t)\); transverse, perpendicular, in phase; \(\vec{E}\times\vec{B}\) gives travel direction.

✓ Speed & field ratio

\(c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}}\); \(\dfrac{E}{B} = \dfrac{E_m}{B_m} = c\); the fields induce each other.

✓ Poynting & intensity

\(\vec{S} = \dfrac{1}{\mu_0}\vec{E}\times\vec{B}\); \(I = \dfrac{1}{c\mu_0}E_{\text{rms}}^{2}\), \(E_{\text{rms}} = E_m/\sqrt{2}\).

✓ Inverse square & pressure

\(I = \dfrac{P_s}{4\pi r^{2}}\); \(p_r = \dfrac{I}{c}\) (absorbed), \(\dfrac{2I}{c}\) (reflected).

✓ Polarization

One-half rule \(I = \tfrac{1}{2}I_0\) (unpolarized in); cosine-squared rule \(I = I_0\cos^{2}\theta\) (polarized in).

✓ Reflection & refraction

\(\theta_1' = \theta_1\); Snell's law \(n_2\sin\theta_2 = n_1\sin\theta_1\), with \(n = c/v\).

✓ Total internal reflection

\(\theta_c = \sin^{-1}\dfrac{n_2}{n_1}\), only when \(n_2 < n_1\); basis of optical fibers.

✓ Brewster angle

\(\theta_B = \tan^{-1}\dfrac{n_2}{n_1}\); reflected light fully polarized, \(\theta_B + \theta_r = 90^\circ\).

Practice

Problems

Take \(c = 3.00\times10^{8}\,\mathrm{m/s}\), \(\mu_0 = 4\pi\times10^{-7}\,\mathrm{T\cdot m/A}\), and \(\varepsilon_0 = 8.85\times10^{-12}\,\mathrm{C^2/N\cdot m^2}\). For polarizing stacks, apply the one-half rule only at the first sheet (and only if the light enters unpolarized), and measure each cosine-squared angle from the previous sheet's axis. For ray problems, measure every angle from the normal.

  1. A plane electromagnetic wave has magnetic-field amplitude \(B_m = 1.0\times10^{-4}\,\mathrm{T}\). (a) Find \(E_m\). (b) Find the intensity of the wave.
  2. A plane wave has maximum electric field \(3.20\times10^{-4}\,\mathrm{V/m}\). Find the magnetic-field amplitude.
  3. A radio wave of wavelength 3.0 m travels in vacuum along \(+x\) with electric-field amplitude 300 V/m oscillating along \(y\). Find (a) the frequency, (b) the angular frequency, (c) the angular wave number, and (d) the magnetic-field amplitude. (e) Parallel to which axis does \(\vec{B}\) oscillate?
  4. What inductance must be paired with a 17 pF capacitor in an LC oscillator to generate 550 nm (visible) waves? Comment on whether such a circuit is practical.
  5. The maximum electric field 10 m from an isotropic point source of light is 2.0 V/m. Find (a) the maximum magnetic field, (b) the average intensity there, and (c) the power of the source.
  6. Assume a TV station radiates isotropically at 1.0 MW. Find the intensity of the signal reaching a planet 4.3 ly away (1 ly \(= 9.46\times10^{15}\,\mathrm{m}\)).
  7. What is the radiation pressure 1.5 m from a 500 W bulb radiating uniformly, on a perfectly absorbing surface facing the bulb?
  8. Solar radiation just outside the atmosphere has intensity \(1.4\,\mathrm{kW/m^2}\). (a) Find the radiation pressure on a fully absorbing surface. (b) Find the ratio of this pressure to sea-level atmospheric pressure \(1.0\times10^{5}\,\mathrm{Pa}\).
  9. Unpolarized light passes through two sheets with polarizing directions at \(70^\circ\) and \(90^\circ\) to the \(y\) axis. If the incident intensity is \(43\,\mathrm{W/m^2}\), what intensity emerges?
  10. Unpolarized light enters three sheets with polarizing directions at \(\theta_1 = 40^\circ\), \(\theta_2 = 20^\circ\), \(\theta_3 = 40^\circ\) from the \(y\) axis. What percentage of the initial intensity is transmitted?
  11. Light in vacuum strikes a glass slab at \(32.0^\circ\) from the normal and refracts to \(21.0^\circ\) inside. Find the glass's index of refraction.
  12. A point source of light is 80.0 cm below the surface of water (\(n = 1.33\)). Find the diameter of the circle at the surface through which light escapes.
  13. Find the critical angle for a ray in benzene (\(n = 1.8\)) meeting a flat layer of air above it.
  14. (a) At what angle of incidence is sunlight reflected from water (\(n = 1.33\)) completely polarized? (b) Does this Brewster angle depend on the wavelength of the light?
Tip: three habits keep this chapter's problems clean. First, lean on the wave's tight internal links — \(c = f\lambda\), \(E_m = cB_m\), \(I = E_m^{2}/2c\mu_0 = E_{\text{rms}}^{2}/c\mu_0\) — so that one given quantity unlocks all the others, and remember the electric and magnetic energies are equal even though the numbers in volts and teslas look wildly different. Second, for any distant source reach for the inverse-square law \(I = P_s/4\pi r^{2}\) first and note that the field falls only as \(1/r\), not \(1/r^{2}\). Third, in geometrical optics measure every angle from the normal, never from the surface; apply the one-half rule once and only for unpolarized light entering the first sheet; and before invoking a critical angle, check that the light is heading into a lower-index medium — otherwise total internal reflection simply cannot occur.