# Electromagnetic Waves

> AP Physics 2 · Unit 6: Geometric and Physical Optics
> Source: https://www.owlsprep.com/study/ap-physics-2-u6-electromagnetic-waves/

This module covers core properties of electromagnetic waves, the $c = f\lambda$ relation, EM spectrum classification, photon energy, polarization, and Malus’s law, with AP-style worked examples and exam-focused tips.

**Prerequisites:** Basic wave properties (frequency, wavelength, intensity) for mechanical waves; Electric and magnetic field vector definitions and units; Energy quantization fundamentals from modern physics

## Learning objectives

- Describe fundamental properties of electromagnetic waves
- Apply $c = f\lambda$ to solve wavelength/frequency problems across media
- Use Malus's law to calculate intensity after polarizers
- Calculate photon energy from wavelength/frequency
- Classify regions of the electromagnetic spectrum

## Fundamental Nature of Electromagnetic Waves

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.

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

**Check your understanding**

Test your basic understanding:

1. 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

   *Why:* Correct! This is the most fundamental difference between the two wave types.

## Speed, Wavelength Relations and the EM Spectrum

All EM waves in vacuum travel at the constant speed $c = 3.00 \times 10^8 \text{ m/s}$, 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 $\lambda_n = \lambda_0 /n$, where $\lambda_0$ 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 $c = f\lambda$ 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.

## Polarization and Malus's Law

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 $\theta$ is the angle between the polarization direction of the incident light and the transmission axis of the analyzer.

**Worked example:** Unpolarized light of intensity $160 \text{ W/m}^2$ 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 $\theta = 60^\circ$.
4. Apply Malus's law to find the final intensity:
5. $$I_2 = I_1 \cos^2(60^\circ)$$
6. We know $\cos(60^\circ) = 0.5$, so $\cos^2(60^\circ) = 0.25$. Calculate the final result:
7. $$I_2 = 80 \text{ W/m}^2 \times 0.25 = 20 \text{ W/m}^2$$

**Check your understanding**

Check your understanding of multiple polarizers:

1. Unpolarized light of intensity $I_0$ passes through three polarizers: first horizontal, middle at 45°, third vertical. What is the final transmitted intensity?

   - 0
   - $I_0/8$
   - $I_0/4$
   - $I_0/2$

   *Why:* Correct! After first polarizer: $I_0/2$. After middle: $(I_0/2)\cos^2(45^\circ) = I_0/4$. After third: $(I_0/4)\cos^2(45^\circ) = I_0/8$. 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.

## Photon Energy

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 $h = 6.626 \times 10^{-34} J \cdot s$ is Planck's constant. For most AP problems involving photon energy in electron-volts (eV), the shortcut $hc \approx 1240 eV \cdot nm$ 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: $E = \frac{hc}{\lambda} = \frac{1240 eV \cdot nm}{\lambda (nm)}$
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 $hc \approx 1240 eV \cdot nm$ shortcut — it saves 1-2 minutes of unit conversion on every photon energy problem on the exam.

## Common pitfalls

- **Wrong:** Applying Malus's law directly to unpolarized incident light on a polarizer, calculating $I = I_0 \cos^2 \theta$ instead of halving the intensity.
  - Why it fails: Students memorize Malus's law and forget it only applies to incident polarized light, not unpolarized.
  - Correct: 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:** Assuming wavelength is constant when an EM wave enters a new medium, solving for frequency using the old vacuum wavelength.
  - Why it fails: Students confuse frequency and wavelength dependence on medium, mixing up which quantity is set by the source.
  - Correct: When an EM wave crosses between media, always hold frequency constant, and scale speed and wavelength by $1/n$ for the new medium.
- **Wrong:** Calculating photon energy by using $hc = 1240 eV \cdot nm$ with wavelength given in meters, getting an answer 9 orders of magnitude off.
  - Why it fails: Students memorize the shortcut value but forget it is only valid when wavelength is in nanometers.
  - Correct: Always check units when using the $hc = 1240 eV \cdot nm$ shortcut; convert wavelength to nanometers before plugging in, or use SI units for $h$ and $c$ if working in joules.
- **Wrong:** Ordering the EM spectrum by energy and reversing the ranking, claiming red light has higher energy than violet light.
  - Why it fails: Students forget the inverse relationship between wavelength and energy.
  - Correct: Remember energy is directly proportional to frequency and inversely proportional to wavelength — shorter wavelength = higher frequency = higher energy per photon.
- **Wrong:** Claiming EM waves require a medium to travel, just like mechanical sound waves.
  - Why it fails: Students carry over properties of mechanical waves learned earlier to EM waves.
  - Correct: 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.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Speed of EM waves in vacuum | $c = f\lambda = 3.00 \times 10^8 \text{ m/s}$ | Applies to all EM waves in vacuum, regardless of frequency |
| Speed/wavelength in medium | $v = c/n$, $\lambda_n = \lambda_0 /n$ | $n$ = refractive index; $f$ is always constant across media |
| Photon Energy (SI units) | $E = hf = \frac{hc}{\lambda}$ | $h = 6.626 \times 10^{-34} J \cdot s$, $c = 3.00 \times 10^8 m/s$ |
| Photon Energy (eV shortcut) | $E (eV) = \frac{1240 eV \cdot nm}{\lambda (nm)}$ | Only valid when $\lambda$ 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 | $I_2 = I_1 \cos^2 \theta$ | Only applies to polarized incident light; $\theta$ = angle between polarization and transmission axis |
| EM Spectrum (low $\rightarrow$ high $f$) | Radio < Microwave < IR < Visible < UV < X-ray < Gamma | Energy per photon increases with frequency, wavelength decreases |
| Visible Spectrum (long $\rightarrow$ short $\lambda$) | Red < Orange < Yellow < Green < Blue < Violet | Wavelength range: 700 nm (red) to 400 nm (violet) |

## 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.

- [Geometric Optics: Refraction and Reflection](https://www.owlsprep.com/study/ap-physics-2-u6-geometric-optics-refraction-and-reflection/)
- [Images](https://www.owlsprep.com/study/ap-physics-2-u6-images/)
- [Interference and Diffraction](https://www.owlsprep.com/study/ap-physics-2-u6-interference-and-diffraction/)

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