Electromagnetic Waves — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Electromagnetic wave fundamental properties, the $c=f\lambda$ frequency-wavelength relation, EM spectrum classification, photon energy, polarization, and Malus’s law for transmitted intensity, including AP-style problem solving.

You should already know: Basic wave properties (frequency, wavelength, intensity) for mechanical waves. Electric and magnetic field vector definitions and units. Energy quantization fundamentals from modern physics.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is 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 through material media, with their speed decreasing in a medium relative to vacuum. In the AP Physics 2 Course and Exam Description (CED), this topic is the foundational opening of Unit 6: Geometric and Physical Optics, accounting for approximately 12% of the unit’s exam weight and 3-4% of the total AP exam score. EM wave concepts appear in both multiple-choice (MCQ) and free-response (FRQ) sections: most often as conceptual MCQs testing spectrum ordering or polarization, or as short calculation-based parts of longer FRQs connected to interference or photon interaction topics. Common notation conventions are standardized for AP: $c$ for speed of light in vacuum, $f$ for frequency, $\lambda$ for wavelength, $E$ for both electric field magnitude and photon energy, $I$ for wave intensity, and $\theta$ for polarization angle relative to the transmission axis.

2. Fundamental Speed and Wavelength Relations

The defining property of all EM waves in vacuum is that they travel at the constant speed $c=3.00\times 10^{8}m/s$, regardless of their frequency or wavelength. For any wave, the fundamental relation between speed, frequency, and wavelength holds: $$c=f\lambda$$ This relation applies to all EM waves in vacuum; when traveling through a medium with refractive index $n$, the speed becomes $v=c/n$, and frequency stays constant (since frequency is determined by the source), so wavelength scales as ${\lambda }_n={\lambda }_0/n$ where ${\lambda }_0$ is the vacuum wavelength. The entire range of possible frequencies/wavelengths of EM waves is called 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 is only a tiny sliver of the spectrum, with wavelengths from ~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: $f=92.5MHz=92.5\times 10^{6}Hz$. List known values: $c=3.00\times 10^{8}m/s$, $n=1.5$.
2. Rearrange $c=f\lambda$ to solve for vacuum wavelength: ${\lambda }_0=c/f=(3.00\times 10^{8}m/s)/(92.5\times 10^{6}Hz)\approx 3.24m$.
3. Frequency does not change when the wave enters a new medium — only speed and wavelength change. Wavelength in glass scales inversely with refractive index.
4. Calculate wavelength in glass: ${\lambda }_n={\lambda }_0/n=3.24m/1.5=2.16m$.

Exam tip: When an EM wave moves from one medium to another, always remember frequency is set by the source and never changes — only speed and wavelength change. This is the most commonly tested conceptual point for this topic on MCQs.

3. 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 the direction of travel. Polarized light has the electric field oscillation restricted to a single plane. A polarizer is a material that 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, the intensity is always cut in half, regardless of the polarizer orientation: $I_1=I_0/2$, where $I_0$ is the incident unpolarized intensity. When polarized light of intensity $I_1$ is incident on a second polarizer (called an analyzer), the intensity transmitted through the analyzer is given by 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 $160W/m^2$ 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 $I_1=I_0/2=160W/m^2/2=80W/m^2$. The light is now horizontally polarized.
2. The angle between incident polarization (horizontal) and the second polarizer's axis is $\theta =60^{\circ }$.
3. Apply Malus's law to find final intensity: $I_2=I_1{\cos }^2(60^{\circ })$.
4. We know $\cos (60^{\circ })=0.5$, so ${\cos }^2(60^{\circ })=0.25$. Calculate: $I_2=80W/m^2\times 0.25=20W/m^2$.

Exam tip: Always remember to halve the intensity first if the incident light on the first polarizer is unpolarized — don't apply Malus's law directly to unpolarized light, that's a frequent exam error.

4. Photon Energy of EM Radiation

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 $hc\approx 1240eV\cdot nm$ 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: for example, UV photons carry more energy than visible photons, which is why UV can cause sunburn but visible light cannot.

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: $E=\frac{hc}{\lambda }=\frac{1240eV\cdot nm}{\lambda (nm)}$.
2. Plug in the green light wavelength: $E_{green}=1240eV\cdot nm/500nm=2.48eV\approx 2.5eV$.
3. For the radio photon, first calculate wavelength: $\lambda =c/f=3.0\times 10^{8}m/s/100\times 10^{6}Hz=3m=3\times 10^{9}nm$.
4. Calculate radio photon energy: $E_{radio}=1240eV\cdot nm/3\times 10^{9}nm\approx 4.1\times 10^{-7}eV$. The green photon is ~6 million times more energetic than the radio photon.

Exam tip: Memorize the $hc\approx 1240eV\cdot nm$ shortcut — it saves 1-2 minutes of unit conversion on every photon energy problem on the exam.

5. Common Pitfalls (and how to avoid them)

• Wrong move: 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: 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 $hc=1240eV\cdot nm$ 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 $hc=1240eV\cdot nm$ 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. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

Unpolarized light of intensity $I_0$ is incident on three polarizers. The first polarizer is aligned horizontally, the third is aligned vertically, and the middle polarizer is aligned at 45 degrees between the first and third. What is the final transmitted intensity in terms of $I_0$? A) $0$ B) $I_0/8$ C) $I_0/4$ D) $I_0/2$

Worked Solution: First, incident unpolarized light passes through the first horizontal polarizer, so intensity becomes $I_1=I_0/2$, and the light is horizontally polarized. Next, the middle polarizer is at 45 degrees to the first, so apply Malus's law: $I_2=I_1{\cos }^2(45^{\circ })=(I_0/2)(1/2)=I_0/4$. The light is now polarized at 45 degrees, incident on the third vertical polarizer, which is also 45 degrees relative to the middle polarizer's axis. Apply Malus's law again: $I_3=I_2{\cos }^2(45^{\circ })=(I_0/4)(1/2)=I_0/8$. The common wrong answer A comes from forgetting the middle polarizer rotates the polarization, allowing transmission. The correct answer is B.

Question 2 (Free Response)

Newer Wi-Fi 6E technology uses EM waves in the 6 GHz band for faster data transmission. (a) Calculate the wavelength of 6.0 GHz EM waves in vacuum. (b) If these waves travel through a plastic waveguide with refractive index $n=1.4$, calculate their speed, frequency, and wavelength inside the waveguide. (c) Explain why frequency does not change when the wave enters the plastic, but speed and wavelength do.

Worked Solution: (a) Use the fundamental relation $c=f\lambda$, rearranged for wavelength: $\lambda =\frac{c}{f}=\frac{3.0\times 10^{8}m/s}{6.0\times 10^{9}Hz}=0.050m=5.0cm$. This is the vacuum wavelength. (b) Frequency is determined by the transmitting source and does not change across media, so $f=6.0GHz$ remains unchanged. Speed in plastic is $v=\frac{c}{n}=\frac{3.0\times 10^{8}m/s}{1.4}\approx 2.1\times 10^{8}m/s$. Wavelength in plastic is ${\lambda }_n=\frac{v}{f}=\frac{2.1\times 10^{8}m/s}{6.0\times 10^{9}Hz}=0.036m=3.6cm$, or equivalently ${\lambda }_n=\frac{{\lambda }_0}{n}=\frac{5.0cm}{1.4}\approx 3.6cm$. (c) Frequency is set by the oscillation frequency of the source (the Wi-Fi antenna), which does not depend on the medium the wave travels through, so frequency is constant. The speed of EM waves depends on the electric and magnetic properties of the medium, which reduce the wave speed relative to vacuum. Since $v=f\lambda$, if $v$ decreases and $f$ is constant, wavelength must decrease proportionally to the speed change.

Question 3 (Application / Real-World Style)

Silicon, the material used in most commercial solar cells, has a band gap of 1.1 eV. Only photons with energy greater than or equal to the band gap can be absorbed to generate electricity. What is the maximum wavelength of a photon that can be absorbed by silicon? Explain what this means for silicon's ability to absorb visible light.

Worked Solution: Use the photon energy relation to solve for wavelength at the minimum energy of 1.1 eV: $\lambda =\frac{hc}{E}=\frac{1240eV\cdot nm}{1.1eV}\approx 1130nm$. All visible light photons have wavelengths between 400 nm (violet) and 700 nm (red), which are all shorter than 1130 nm. Shorter wavelength corresponds to higher energy, so all visible photons have energy greater than 1.1 eV. This means silicon can absorb the entire visible spectrum, which makes it an efficient material for solar panels that use sunlight (which is mostly visible and infrared) to generate electricity.

7. Quick Reference Cheatsheet

8. What's Next

This chapter 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 exam questions about real-world applications like polarized sunglasses, while photon energy connects to modern physics topics later in the course.

• Reflection and Refraction
• Thin Film Interference
• Double Slit Interference
• Photoelectric Effect