Electromagnetic Waves

AP Physics 2· AP Physics 2 CED — Geometric and Physical Optics· 14 min read

1. Fundamental Nature of Electromagnetic Waves★★☆☆☆⏱ 2 min

Electromagnetic (EM) waves are transverse waves produced by the acceleration of charged particles, consisting of coupled oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation. Unlike mechanical waves, EM waves can travel through vacuum as well as material media. This topic accounts for ~3-4% of your total AP Physics 2 exam score, appearing in both multiple-choice and free-response sections.

📘 Definition

Electromagnetic Wave

Transverse wave of coupled oscillating electric and magnetic fields that propagates through space, requiring no material medium.

✓ Quick check

Test your basic understanding:

Which of the following is a key difference between electromagnetic and mechanical waves?
- EM waves have a constant wavelength, mechanical waves do not
- EM waves do not require a medium to propagate, mechanical waves do
- EM waves are longitudinal, mechanical waves are transverse
- EM waves cannot travel through vacuum, mechanical waves can
Reveal answer
EM waves do not require a medium to propagate, mechanical waves do —
Correct! This is the most fundamental difference between the two wave types.

2. Speed, Wavelength Relations and the EM Spectrum★★☆☆☆⏱ 5 min

All EM waves in vacuum travel at the constant speed $c = 3.00 \times 1 0^{8} m/s$ , regardless of their frequency or wavelength. For any wave, the fundamental relation between speed, frequency, and wavelength holds:

c = f \lambda

When traveling through a medium with refractive index $n$ , the speed becomes $v = c / n$ . Frequency is determined by the source and stays constant, so wavelength scales as $λ_{n} = λ_{0} / n$ , where $λ_{0}$ is the vacuum wavelength. The full range of EM waves is the electromagnetic spectrum, ordered from lowest frequency (longest wavelength) to highest frequency (shortest wavelength): radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. Visible light spans only ~400 nm (violet) to 700 nm (red).

📐 Worked Example

A FM radio station broadcasts at a frequency of 92.5 MHz. What is the wavelength of this EM wave in vacuum? If the signal travels through window glass with refractive index $n = 1.5$ , what is its wavelength in glass?

1
Convert frequency to SI units and list known values:
2
$f = 92.5 \text{ MHz} = 92.5 \times 10^6 \text{ Hz}, \quad c = 3.00 \times 10^8 \text{ m/s}, \quad n=1.5$
3
Rearrange $c = f λ$ to solve for vacuum wavelength:
4
$\lambda_0 = \frac{c}{f} = \frac{3.00 \times 10^8 \text{ m/s}}{92.5 \times 10^6 \text{ Hz}} \approx 3.24 \text{ m}$
5
Frequency does not change when entering a new medium — only speed and wavelength change. Wavelength in glass scales inversely with refractive index.
6
Calculate final wavelength in glass:
7
$\lambda_n = \frac{\lambda_0}{n} = \frac{3.24 \text{ m}}{1.5} = 2.16 \text{ m}$

Exam tip:

When an EM wave moves from one medium to another, frequency is set by the source and never changes — only speed and wavelength change. This is one of the most commonly tested conceptual points on AP multiple choice.

3. Polarization and Malus's Law★★★☆☆⏱ 5 min

EM waves are transverse, meaning the electric field oscillates perpendicular to the direction of propagation. In unpolarized light, the electric field oscillates in all possible planes perpendicular to propagation. Polarized light has electric field oscillation restricted to a single plane. A polarizer only transmits the component of the electric field parallel to its transmission axis, absorbing the perpendicular component.

When unpolarized light passes through a single polarizer, intensity is always cut in half, regardless of polarizer orientation: $I_{1} = I_{0} /2$ . When polarized light of intensity $I_{1}$ is incident on a second polarizer (analyzer), the transmitted intensity follows Malus's Law:

I_2 = I_1 \cos^2 \theta

Where $θ$ is the angle between the polarization direction of the incident light and the transmission axis of the analyzer.

📐 Worked Example

Unpolarized light of intensity $160 W/m^{2}$ passes through two polarizers. The first polarizer has a horizontal transmission axis, and the second has a transmission axis at 60 degrees from horizontal. What is the intensity of light transmitted through both polarizers?

1
The first polarizer receives unpolarized light, so intensity after the first polarizer is:
2
$I_1 = \frac{I_0}{2} = \frac{160 \text{ W/m}^2}{2} = 80 \text{ W/m}^2$
3
The light is now horizontally polarized, and the angle between incident polarization and the second polarizer's axis is $θ = 6 0^{\circ}$ .
4
Apply Malus's law to find the final intensity:
5
$I_2 = I_1 \cos^2(60^\circ)$
6
We know $cos (6 0^{\circ}) = 0.5$ , so $cos^{2} (6 0^{\circ}) = 0.25$ . Calculate the final result:
7
$I_2 = 80 \text{ W/m}^2 \times 0.25 = 20 \text{ W/m}^2$

✓ Quick check

Check your understanding of multiple polarizers:

Unpolarized light of intensity $I_{0}$ passes through three polarizers: first horizontal, middle at 45°, third vertical. What is the final transmitted intensity?
- 0
- $I_{0} /8$
- $I_{0} /4$
- $I_{0} /2$
Reveal answer
$I_0/8$ —
Correct! After first polarizer: $I_{0} /2$ . After middle: $(I_{0} /2) cos^{2} (4 5^{\circ}) = I_{0} /4$ . After third: $(I_{0} /4) cos^{2} (4 5^{\circ}) = I_{0} /8$ . The middle polarizer rotates the polarization, allowing transmission through the third polarizer.

Exam tip:

Always remember to halve the intensity first if the incident light on the first polarizer is unpolarized. Malus's law only applies to already polarized incident light, which is a frequent AP exam error.

4. Photon Energy★★★☆☆⏱ 4 min

AP Physics 2 connects EM wave properties to quantum behavior: EM radiation is quantized into discrete packets called photons, where each photon's energy depends only on the frequency (or wavelength) of the EM wave. The photon energy relation is:

E = hf = \frac{hc}{\lambda}

Where $h = 6.626 \times 1 0^{- 34} J \cdot s$ is Planck's constant. For most AP problems involving photon energy in electron-volts (eV), the shortcut $h c \approx 1240 e V \cdot nm$ is extremely useful, as it avoids unit conversions between joules and eV when wavelength is given in nanometers. Higher frequency (shorter wavelength) EM radiation has higher energy per photon.

📐 Worked Example

What is the energy, in electron-volts, of a photon of green light with vacuum wavelength 500 nm? How does this compare to the energy of a 100 MHz radio photon?

1
Use the unit shortcut for photon energy: $E = \frac{h c}{λ} = \frac{1240 e V \cdot nm}{λ ( nm )}$
2
Calculate green light photon energy:
3
$E_{\text{green}} = \frac{1240 eV \cdot nm}{500 nm} = 2.48 eV \approx 2.5 eV$
4
For the radio photon, first calculate wavelength and convert to nanometers:
5
$\lambda = \frac{c}{f} = \frac{3.0 \times 10^8 m/s}{100 \times 10^6 Hz} = 3 m = 3 \times 10^9 nm$
6
Calculate radio photon energy and compare:
7
$E_{\text{radio}} = \frac{1240 eV \cdot nm}{3 \times 10^9 nm} \approx 4.1 \times 10^{-7} eV$
8
The green photon is approximately 6 million times more energetic than the radio photon.

Exam tip:

Memorize the $h c \approx 1240 e V \cdot nm$ shortcut — it saves 1-2 minutes of unit conversion on every photon energy problem on the exam.

5. Common Pitfalls

Wrong move:

Applying Malus's law directly to unpolarized incident light on a polarizer, calculating $I = I_{0} cos^{2} θ$ instead of halving the intensity.

Why:

Students memorize Malus's law and forget it only applies to incident polarized light, not unpolarized.

Correct move:

Always first check if incident light is unpolarized; if it is, immediately divide initial intensity by 2 after the first polarizer before applying Malus's law to any subsequent polarizers.

Wrong move:

Assuming wavelength is constant when an EM wave enters a new medium, solving for frequency using the old vacuum wavelength.

Why:

Students confuse frequency and wavelength dependence on medium, mixing up which quantity is set by the source.

Correct move:

When an EM wave crosses between media, always hold frequency constant, and scale speed and wavelength by $1/ n$ for the new medium.

Wrong move:

Calculating photon energy by using $h c = 1240 e V \cdot nm$ with wavelength given in meters, getting an answer 9 orders of magnitude off.

Why:

Students memorize the shortcut value but forget it is only valid when wavelength is in nanometers.

Correct move:

Always check units when using the $h c = 1240 e V \cdot nm$ shortcut; convert wavelength to nanometers before plugging in, or use SI units for $h$ and $c$ if working in joules.

Wrong move:

Ordering the EM spectrum by energy and reversing the ranking, claiming red light has higher energy than violet light.

Why:

Students forget the inverse relationship between wavelength and energy.

Correct move:

Remember energy is directly proportional to frequency and inversely proportional to wavelength — shorter wavelength = higher frequency = higher energy per photon.

Wrong move:

Claiming EM waves require a medium to travel, just like mechanical sound waves.

Why:

Students carry over properties of mechanical waves learned earlier to EM waves.

Correct move:

When asked for a key difference between EM and mechanical waves, always note that EM waves do not require a medium and can propagate through vacuum.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Speed of EM waves in vacuum	$c = f λ = 3.00 \times 1 0^{8} m/s$	Applies to all EM waves in vacuum, regardless of frequency
Speed/wavelength in medium	$v = c / n$ , $λ_{n} = λ_{0} / n$	$n$ = refractive index; $f$ is always constant across media
Photon Energy (SI units)	$E = h f = \frac{h c}{λ}$	$h = 6.626 \times 1 0^{- 34} J \cdot s$ , $c = 3.00 \times 1 0^{8} m / s$
Photon Energy (eV shortcut)	$E (e V) = \frac{1240 e V \cdot nm}{λ ( nm )}$	Only valid when $λ$ is in nanometers; eliminates unit conversions
Unpolarized light through 1 polarizer	$I_{1} = I_{0} /2$	Intensity halved regardless of polarizer orientation
Malus's Law	$I_{2} = I_{1} cos^{2} θ$	Only applies to polarized incident light; $θ$ = angle between polarization and transmission axis
EM Spectrum (low $\to$ high $f$ )	Radio < Microwave < IR < Visible < UV < X-ray < Gamma	Energy per photon increases with frequency, wavelength decreases
Visible Spectrum (long $\to$ short $λ$ )	Red < Orange < Yellow < Green < Blue < Violet	Wavelength range: 700 nm (red) to 400 nm (violet)

7. Frequently Asked

Do electromagnetic waves need a medium to travel through?

No, unlike mechanical waves (such as sound), electromagnetic waves do not require a material medium to propagate. They can travel through vacuum, which is how sunlight reaches Earth from space.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
EM spectrum energy ranking
2022 · FRQ
Malus's law intensity calculation
2021 · MCQ
Frequency across medium change

Going deeper

Official SyllabusAP Physics 2 Course and Exam Description

What's Next

This module lays the foundational understanding of light as an electromagnetic wave, which is required for all subsequent topics in Unit 6: Geometric and Physical Optics. Next, you will apply EM wave properties to the behavior of light at interfaces (reflection and refraction), where the wavelength change we discussed here explains why refraction occurs at a medium boundary. Without mastering the relation between speed, frequency, and wavelength across media, you will not be able to correctly solve problems involving thin film interference, a high-weight AP exam topic. Polarization concepts also carry over to conceptual questions about real-world applications like polarized sunglasses, while photon energy connects to modern physics topics later in the course.