Electromagnetic Wave Properties to Know for AP Physics 2

What You Need to Know

Electromagnetic (EM) waves are self-propagating oscillating electric and magnetic fields. In AP Physics 2, you’re expected to connect wave behavior (optics) with field/energy ideas (Maxwell + photons), and to fluently move between $\lambda$, $f$, $v$, energy, intensity, and field amplitudes.

Core ideas (the “must-say” properties)

• EM waves are transverse: $\vec E \perp \vec B \perp \text{direction of travel}$.
• In vacuum, all EM waves travel at the same speed:
  $c = 3.00\times 10^8\ \text{m/s}$
• Speed in vacuum relates to constants:
  $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$
• Fields are linked (vacuum plane wave):
  $\frac{E}{B} = c\quad\Rightarrow\quad E = cB$
• EM waves carry energy and momentum (radiation pressure).
• Light can be treated as a wave and as photons:
  $E_{\text{photon}} = hf = \frac{hc}{\lambda}$

Big exam trigger: If you see intensity, think energy transport (Poynting vector) and/or field amplitudes. If you see $\lambda$ or $f$, think wave relation and photon energy.

Step-by-Step Breakdown

A) Converting between $\lambda$, $f$, $v$ (vacuum or medium)

1. Use the wave relation:
   $v = f\lambda$
2. If it’s vacuum/air (usually treated as vacuum), set $v=c$.
3. If it’s a material, use refractive index:
   $n = \frac{c}{v}\quad\Rightarrow\quad v = \frac{c}{n}$
4. At a boundary, remember:
   • Frequency stays the same: $f_1=f_2$
   • Speed and wavelength change: $v_2 = \frac{c}{n_2}$ and $\lambda_2 = \frac{v_2}{f}$

Mini-check: If $n$ increases, speed decreases, so wavelength decreases (frequency unchanged).

B) From intensity to electric/magnetic field amplitude

For a sinusoidal plane EM wave, average intensity is
$I = \left\langle S\right\rangle = \frac{1}{2}c\epsilon_0 E_0^2 = \frac{1}{2}\frac{c}{\mu_0}B_0^2$

1. Choose which form matches what you need (usually solve for $E_0$ first).
2. Solve for amplitude:
   $E_0 = \sqrt{\frac{2I}{c\epsilon_0}}$
3. Then get the other field using
   $B_0 = \frac{E_0}{c}$

Decision point: Use $E_0$ and $B_0$ (amplitudes), not instantaneous values, unless the question explicitly asks for instantaneous fields.

C) Photon energy/momentum (when it’s “quantum light”)

1. From wavelength or frequency:
   $E = hf = \frac{hc}{\lambda}$
2. Photon momentum:
   $p = \frac{E}{c} = \frac{h}{\lambda}$
3. If you have power and want photons per second:
   $\text{photons/s} = \frac{P}{E_{\text{photon}}}$

D) Radiation pressure (force from light)

1. If light with intensity $I$ hits area $A$, power on it is
   $P = IA$
2. Radiation pressure:
   • Absorbed: $p = \frac{I}{c}$
   • Reflected: $p = \frac{2I}{c}$
3. Force:
   $F = pA$

Key Formulas, Rules & Facts

Constants you should recognize

EM wave structure (vacuum plane wave)

Wave relations & media

Critical reminder: At refraction, students often (wrongly) change frequency. Don’t. Only $v$ and $\lambda$ change.

Intensity, energy, photons

Radiation pressure

Polarization (transverse wave property)

“Wave optics” conditions (often tested as EM-wave behavior)

Reflection phase shifts (thin film / reflection logic)

• Reflecting from lower $n$ to higher $n$ gives a phase shift of $\pi$.
• Reflecting from higher $n$ to lower $n$ gives no phase shift.

This shows up when deciding whether two reflected rays are in-phase or out-of-phase.

Examples & Applications

1) Photon energy from wavelength (spectrum + quantum)

Problem: What is the energy of a photon with $\lambda = 500\ \text{nm}$?

Setup:

• Convert: $500\ \text{nm} = 5.00\times 10^{-7}\ \text{m}$
• Use $E=\frac{hc}{\lambda}$

Key insight:
$E = \frac{\left(6.63\times 10^{-34}\right)\left(3.00\times 10^8\right)}{5.00\times 10^{-7}} \approx 3.98\times 10^{-19}\ \text{J}$
Higher frequency (shorter wavelength) means higher photon energy.

2) Field amplitude from intensity (link wave energy to E-field)

Problem: Sunlight intensity is about $I = 1000\ \text{W/m}^2$. Find $E_0$ and $B_0$.

Setup:
$E_0 = \sqrt{\frac{2I}{c\epsilon_0}}$

Key insight (numbers rough OK on AP):
$E_0 \approx \sqrt{\frac{2(1000)}{(3.00\times 10^8)(8.85\times 10^{-12})}} \approx 870\ \text{N/C}$
Then
$B_0 = \frac{E_0}{c} \approx \frac{870}{3.00\times 10^8} \approx 2.9\times 10^{-6}\ \text{T}$

3) Radiation pressure and force (momentum transfer)

Problem: A perfectly reflecting sail of area $A=2.0\ \text{m}^2$ receives $I=600\ \text{W/m}^2$ normally. Find force.

Setup:

• Reflecting: $p=\frac{2I}{c}$
• Force: $F=pA$

Key insight:
$F = \left(\frac{2I}{c}\right)A = \frac{2(600)(2.0)}{3.00\times 10^8} \approx 8.0\times 10^{-6}\ \text{N}$
Tiny force, but real.

4) Refraction: what changes and what doesn’t

Problem: Light enters glass with $n=1.50$ from air. If $\lambda_{\text{air}}=600\ \text{nm}$, find $\lambda_{\text{glass}}$.

Setup:

• Frequency unchanged.
• Speed reduces: $v=\frac{c}{n}$
• So wavelength scales the same way: $\lambda_{\text{glass}} = \frac{\lambda_{\text{air}}}{n}$

Key insight:
$\lambda_{\text{glass}} = \frac{600\ \text{nm}}{1.50} = 400\ \text{nm}$

Common Mistakes & Traps

1. Changing frequency when light enters a medium

   • Wrong move: Setting $f_2=\frac{c}{\lambda_2}$ using vacuum speed after refraction.
   • Why wrong: At boundaries, $f$ stays fixed; the source sets it.
   • Fix: Use $f_1=f_2$, then update $v$ and $\lambda$ with $n$.

2. Forgetting the factor $\frac{1}{2}$ in average intensity

   • Wrong move: Using $I=c\epsilon_0E_0^2$.
   • Why wrong: Intensity uses time-averaged Poynting vector for sinusoidal waves.
   • Fix: Memorize $I=\frac{1}{2}c\epsilon_0E_0^2$.

3. Mixing amplitude with instantaneous fields

   • Wrong move: Treating $E_0$ as the field at all times.
   • Why wrong: The fields oscillate: $E(t)$ varies from $-E_0$ to $+E_0$.
   • Fix: If intensity is given, you’re almost always solving for amplitudes $E_0,B_0$.

4. Using $E=cB$ in a material without thinking

   • Wrong move: Applying $E/B=c$ inside a medium automatically.
   • Why wrong: That exact ratio is for a plane wave in vacuum; AP problems usually keep it vacuum/air unless stated, but be cautious.
   • Fix: Unless the problem explicitly moves into a medium and asks about fields, use the vacuum relationship in vacuum/air contexts.

5. Confusing electric field units with energy units

   • Wrong move: Thinking bigger $E$ field means bigger photon energy.
   • Why wrong: Photon energy depends on $f$: $E_{\text{photon}}=hf$. Field amplitude affects intensity, not per-photon energy.
   • Fix: Separate: frequency => photon energy, amplitude => intensity.

6. Radiation pressure factor of 2 mistake

   • Wrong move: Using $p=\frac{I}{c}$ for a reflecting surface.
   • Why wrong: Reflection reverses momentum, doubling change in momentum.
   • Fix: Absorb: $I/c$. Reflect: $2I/c$.

7. Interference sign errors (path vs phase shift on reflection)

   • Wrong move: Using $\Delta L=m\lambda$ without accounting for a $\pi$ phase flip.
   • Why wrong: A reflection from higher $n$ can add an extra half-wavelength phase shift.
   • Fix: Track whether one (and only one) ray gets a $\pi$ shift; that swaps constructive/destructive conditions.

8. Spectrum ordering backwards (energy vs wavelength)

   • Wrong move: Saying radio has higher energy than gamma because “longer is bigger.”
   • Why wrong: $E\propto f\propto 1/\lambda$.
   • Fix: Shorter wavelength => higher frequency => higher energy.

Memory Aids & Quick Tricks

Quick Review Checklist

• You can state: EM waves are transverse with $\vec E \perp \vec B \perp \text{travel}$.
• You know in vacuum: $c=3.00\times 10^8\ \text{m/s}$ and $c=\frac{1}{\sqrt{\mu_0\epsilon_0}}$.
• You can use $v=f\lambda$ and in media $v=c/n$.
• You remember refraction rule: $f$ stays the same; $v$ and $\lambda$ change.
• You can convert photon energy: $E=hf=\frac{hc}{\lambda}$ and momentum: $p=\frac{h}{\lambda}$.
• You can get field amplitudes from intensity: $I=\frac{1}{2}c\epsilon_0E_0^2$ and $B_0=E_0/c$.
• You can compute radiation pressure: absorb $I/c$, reflect $2I/c$, then $F=pA$.
• You can apply polarization rules: $I=\frac{1}{2}I_0$ then $I=I_0\cos^2\theta$.
• You can rank spectrum correctly: radio => gamma increases $f$ and energy, decreases $\lambda$.

You’ve got this—if you keep frequency vs amplitude straight, most EM-wave questions collapse into a one-line equation.