College Physics Quiz: Electromagnetic Waves
7 questions · exam conditions
0:00
Electromagnetic WavesQuestion 1 of 7

An electromagnetic wave traveling in vacuum has a wavelength of 500500 nm. What is the energy of a single photon of this wave?

3.97×10−193.97 \times 10^{-19} J
6.63×10−346.63 \times 10^{-34} J
1.33×10−271.33 \times 10^{-27} J
2.48×10−192.48 \times 10^{-19} J
5.97×10145.97 \times 10^{14} J
← Back to quizzes

College Physics Quiz

College Physics Quiz: Electromagnetic Waves

Practice Electromagnetic Waves in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Electromagnetic Waves, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

An electromagnetic wave traveling in vacuum has a wavelength of 500500 nm. What is the energy of a single photon of this wave?

  1. 3.97×10−193.97 \times 10^{-19} J (correct answer)
  2. 6.63×10−346.63 \times 10^{-34} J
  3. 1.33×10−271.33 \times 10^{-27} J
  4. 2.48×10−192.48 \times 10^{-19} J
  5. 5.97×10145.97 \times 10^{14} J
Explanation: This question tests your understanding of photon energy and the relationship between electromagnetic wave properties and quantum mechanics. When you encounter wavelength and need to find photon energy, you're dealing with the particle nature of light. To find photon energy, use Planck's equation: E=hfE = hf, where hh is Planck's constant (6.63×10−346.63 \times 10^{-34} J·s) and ff is frequency. Since you're given wavelength instead of frequency, use c=λfc = \lambda f to find frequency: f=cλf = \frac{c}{\lambda}. Combining these gives E=hcλE = \frac{hc}{\lambda}. Substituting the values: E=(6.63×10−34)(3.00×108)500×10−9E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{500 \times 10^{-9}}. Converting 500 nm to meters gives 500×10−9500 \times 10^{-9} m. Calculating: E=1.989×10−255.00×10−7=3.97×10−19E = \frac{1.989 \times 10^{-25}}{5.00 \times 10^{-7}} = 3.97 \times 10^{-19} J. Answer B (6.63×10−346.63 \times 10^{-34} J) is just Planck's constant itself—a common trap for students who forget to complete the calculation. Answer C (1.33×10−271.33 \times 10^{-27} J) results from incorrectly using wavelength in the numerator instead of the denominator. Answer D (2.48×10−192.48 \times 10^{-19} J) comes from unit conversion errors, likely using wavelength in micrometers instead of nanometers. Remember the formula E=hcλE = \frac{hc}{\lambda} for photon energy problems. Always double-check your unit conversions—wavelength is often given in nm but needs to be in meters for the calculation.

Question 2

A linearly polarized electromagnetic wave has its electric field vector making a 30°30° angle with respect to a polarizing filter's transmission axis. What fraction of the incident intensity is transmitted?

  1. 0.750.75 (correct answer)
  2. 0.500.50
  3. 0.870.87
  4. 0.300.30
  5. 0.250.25
Explanation: When electromagnetic radiation encounters a polarizing filter, you're dealing with Malus's Law, which governs how much light passes through based on the angle between the incident polarization and the filter's transmission axis. Malus's Law states that the transmitted intensity equals the incident intensity multiplied by cos⁡2θ\cos^2\theta, where θ\theta is the angle between the electric field vector and the transmission axis. Here, θ=30°\theta = 30°, so the fraction transmitted is: ItransmittedIincident=cos⁡2(30°)=(32)2=34=0.75\frac{I_{transmitted}}{I_{incident}} = \cos^2(30°) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} = 0.75 This confirms answer A is correct. Looking at the wrong answers: B (0.50) would result from using cos⁡(60°)\cos(60°) squared, suggesting confusion about which angle to use—perhaps mixing up the complementary angle. C (0.87) comes from taking just cos⁡(30°)=0.866...\cos(30°) = 0.866... without squaring it, which misses the crucial point that intensity depends on the square of the electric field amplitude. D (0.30) doesn't correspond to any reasonable trigonometric calculation and might represent a random guess or misremembering the cosine value. Remember this key pattern: polarization problems always involve the cosine squared of the angle between the polarization direction and the transmission axis. The intensity relationship is quadratic, not linear, because light intensity is proportional to the square of the electric field amplitude. When you see polarization angles, immediately think "Malus's Law" and cos⁡2θ\cos^2\theta.

Question 3

Two electromagnetic waves of the same frequency travel in opposite directions and interfere. If both waves have the same amplitude E0E_0, what is the amplitude of the electric field at a point where the waves interfere constructively?

  1. 2E02E_0 (correct answer)
  2. E0E_0
  3. 2E0\sqrt{2}E_0
  4. E02E_0^2
  5. 00
Explanation: When electromagnetic waves interfere, you need to understand how wave amplitudes combine based on their phase relationship. This question tests the fundamental principle of wave superposition. At points of constructive interference, the waves arrive in phase, meaning their electric field oscillations are perfectly synchronized. When two waves with the same amplitude E0E_0 interfere constructively, their amplitudes add directly: Etotal=E0+E0=2E0E_{total} = E_0 + E_0 = 2E_0. This is because the electric field is a vector quantity, and when the vectors point in the same direction (in phase), you simply add their magnitudes. Let's examine why the other answers are incorrect. Answer B (E0E_0) would represent no change in amplitude, which would only occur if one wave had zero amplitude or if there were no interference at all. Answer C (2E0\sqrt{2}E_0) might seem appealing because it suggests some kind of root-mean-square relationship, but this doesn't apply to constructive interference of equal-amplitude waves. Answer D (E02E_0^2) incorrectly suggests that amplitudes multiply during interference, which confuses amplitude with intensity (intensity is proportional to the square of amplitude, but amplitudes themselves add linearly). The correct answer is A: 2E02E_0. Remember this key principle: for wave interference, amplitudes add algebraically. In constructive interference, they add positively; in destructive interference, they subtract. Don't confuse this with intensity calculations, where you deal with squares of amplitudes. Focus on the phase relationship to determine whether interference is constructive or destructive.

Question 4

A radio antenna emits electromagnetic waves uniformly in all directions with a total power of 1000 W. What is the intensity of the electromagnetic radiation at a distance of 100 m from the antenna?

  1. 7.96×10−37.96 \times 10^{-3} W/m² (correct answer)
  2. 3.18×10−33.18 \times 10^{-3} W/m²
  3. 1.59×10−21.59 \times 10^{-2} W/m²
  4. 2.65×10−32.65 \times 10^{-3} W/m²
Explanation: For uniform emission in all directions, the power is distributed over a spherical surface of area 4πr24\pi r^2. The intensity is I=P/(4πr2)=1000/(4π×1002)=1000/(4π×104)=7.96×10−3I = P/(4\pi r^2) = 1000/(4\pi \times 100^2) = 1000/(4\pi \times 10^4) = 7.96 \times 10^{-3} W/m². Choice B uses 2πr22\pi r^2 instead of 4πr24\pi r^2. Choice C uses 2πr22\pi r^2 and makes an arithmetic error. Choice D uses an incorrect geometric factor.

Question 5

Two coherent electromagnetic waves with equal amplitudes interfere. The phase difference between them is π/3\pi/3 radians. What fraction of the maximum possible intensity is observed?

  1. 0.25
  2. 0.50
  3. 0.75 (correct answer)
  4. 0.87
Explanation: For two coherent waves with equal amplitudes E0E_0, the resultant amplitude is Eresult=2E0∣cos⁡(ϕ/2)∣E_{result} = 2E_0|\cos(\phi/2)|, where ϕ\phi is the phase difference. With ϕ=π/3\phi = \pi/3, we get Eresult=2E0∣cos⁡(π/6)∣=2E0(3/2)=E03E_{result} = 2E_0|\cos(\pi/6)| = 2E_0(\sqrt{3}/2) = E_0\sqrt{3}. Since intensity ∝E2\propto E^2, the intensity ratio is (E03)2/(2E0)2=3/4=0.75(E_0\sqrt{3})^2/(2E_0)^2 = 3/4 = 0.75. Choice A would correspond to ϕ=2π/3\phi = 2\pi/3. Choice B would be ϕ=π/2\phi = \pi/2. Choice D uses an incorrect formula.

Question 6

A linearly polarized electromagnetic wave is incident on a polarizing filter. The transmitted intensity is 75% of the incident intensity. At what angle is the polarization direction of the wave relative to the transmission axis of the filter?

  1. 30° (correct answer)
  2. 45°
  3. 60°
  4. 37°
Explanation: According to Malus's law, I=I0cos⁡2θI = I_0 \cos^2\theta, where θ\theta is the angle between the incident polarization and the transmission axis. Given I/I0=0.75I/I_0 = 0.75, we have cos⁡2θ=0.75\cos^2\theta = 0.75, so cos⁡θ=0.75=3/2\cos\theta = \sqrt{0.75} = \sqrt{3}/2, giving θ=30°\theta = 30°. Choice B would give 50% transmission. Choice C would give 25% transmission. Choice D would give approximately 64% transmission.

Question 7

Unpolarized light of intensity I0I_0 passes through three ideal polarizing filters. The first and third filters have their transmission axes perpendicular to each other, while the second filter's axis makes a 45° angle with both. What is the transmitted intensity?

  1. I0/8I_0/8 (correct answer)
  2. I0/4I_0/4
  3. I0/2I_0/2
  4. 0
Explanation: Unpolarized light through the first filter gives intensity I1=I0/2I_1 = I_0/2. The second filter at 45° to the first: I2=I1cos⁡2(45°)=(I0/2)(1/2)=I0/4I_2 = I_1\cos^2(45°) = (I_0/2)(1/2) = I_0/4. The third filter at 45° to the second (90° to first): I3=I2cos⁡2(45°)=(I0/4)(1/2)=I0/8I_3 = I_2\cos^2(45°) = (I_0/4)(1/2) = I_0/8. Choice B stops after the second filter. Choice C stops after the first filter. Choice D assumes the first and third filters block all light (ignoring the middle filter).