These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data and assumptions for every question: c = 3.00 × 10⁸ m/s in a vacuum and in air; c = fλ; all EM waves are treated as plane waves; polarizing filters are ideal. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
Which list puts these electromagnetic waves in order of decreasing wavelength?
- (A) radio, microwave, infrared, visible, ultraviolet, X-ray, gamma
- (B) gamma, X-ray, ultraviolet, visible, infrared, microwave, radio
- (C) radio, infrared, microwave, visible, X-ray, ultraviolet, gamma
- (D) radio, microwave, visible, infrared, ultraviolet, X-ray, gamma
Answer and explanation
Answer: (A). Radio waves have the longest wavelengths and gamma rays the shortest, with the other categories in between in the order shown.
- (B) is the correct sequence written backwards. It is the order of increasing wavelength, which is the same as decreasing frequency.
- (C) swaps microwave with infrared, and ultraviolet with X-ray.
- (D) swaps infrared with visible. Infrared (“below red”) has longer wavelengths than red light.
Question 2 (multiple choice · core)
A communications link uses electromagnetic waves of frequency 2.0 × 10¹⁰ Hz. What is their wavelength?
- (A) 0.015 m
- (B) 1.5 m
- (C) 6.0 × 10¹⁸ m
- (D) 67 m
Answer and explanation
Answer: (A). λ = c/f = (3.00 × 10⁸ m/s) ÷ (2.0 × 10¹⁰ Hz) = 1.5 × 10⁻² m = 0.015 m (1.5 cm).
- (B) is 100 times too large: a power-of-ten slip when dividing 10⁸ by 10¹⁰.
- (C) multiplies c by f. The units would be m/s², not m.
- (D) divides f by c (2.0 × 10¹⁰ ÷ 3.00 × 10⁸ = 66.7), which has units of m⁻¹.
Question 3 (multiple choice · core)
A plane electromagnetic wave travels in the +x direction. At one instant, at a certain point, its electric field points in the +z direction. Along which axis does the magnetic field at that point oscillate?
- (A) the x-axis
- (B) the y-axis
- (C) the z-axis
- (D) it has no direction, because B is zero wherever E is largest
Answer and explanation
Answer: (B). B is perpendicular to the direction of travel (x) and to E (z), so it lies along y.
- (A) puts B along the direction of travel. EM waves are transverse, so neither field oscillates along x.
- (C) makes B parallel to E. The two fields are always perpendicular.
- (D) In a plane EM wave, E and B are in step: B is largest where E is largest, not zero.
Question 4 (multiple choice · foundation)
A bright flare erupts on the surface of a distant star. Astronomers on Earth can see the flare but can never hear any sound from it. What is the best explanation?
- (A) Light is an electromagnetic wave, which needs no medium; sound is a mechanical wave, which cannot cross the near-vacuum of space.
- (B) Sound travels too slowly to reach the Earth within a human lifetime, although it would arrive eventually.
- (C) Light is a longitudinal wave and sound is a transverse wave, so only light can travel through space.
- (D) The sound is absorbed by the Earth’s atmosphere just before it arrives.
Answer and explanation
Answer: (A). Sound needs particles of a medium to pass the disturbance along. Space between the star and the Earth is almost empty, so the sound cannot travel through it at all. EM waves consist of fields, which can exist in a vacuum.
- (B) is wrong because sound does not travel through a vacuum at any speed. Slowness is not the issue.
- (C) reverses the wave types. Light is transverse; sound is longitudinal.
- (D) assumes the sound reaches the atmosphere. It never leaves the star’s surroundings.
Question 5 (calculation · core)
A green laser emits light of wavelength 5.2 × 10⁻⁷ m. A nearby radio transmitter works at 1.2 × 10⁶ Hz.
(a) Calculate the frequency of the green light. (b) Calculate the wavelength of the radio wave. (c) Both signals are sent to a receiver 1.8 × 10⁴ m away through air. Calculate the travel time for each, and compare them. (d) Red light and blue light are also used. Rank red, green and blue light in order of increasing frequency.
Worked solution
- (a) f = c/λ = (3.00 × 10⁸) ÷ (5.2 × 10⁻⁷) = 5.77 × 10¹⁴ Hz (5.8 × 10¹⁴ Hz to 2 s.f.).
- (b) λ = c/f = (3.00 × 10⁸) ÷ (1.2 × 10⁶) = 250 m.
- (c) t = d/c = (1.8 × 10⁴ m) ÷ (3.00 × 10⁸ m/s) = 6.0 × 10⁻⁵ s for both. All EM waves travel at the same speed, so the very different frequencies make no difference.
- (d) Increasing frequency means decreasing wavelength: red, green, blue.
Suggested mark points (4): 1 for the light frequency; 1 for the radio wavelength; 1 for equal times of 6.0 × 10⁻⁵ s with the reason (same speed c); 1 for the ranking red < green < blue.
Common error: giving the radio signal a longer travel time “because its wavelength is longer”. Speed, not wavelength, sets the travel time.
Question 6 (constructed response · core)
A plane electromagnetic wave travels in the +y direction. Its electric field oscillates along the x-axis.
(a) Sketch a snapshot of the wave over two wavelengths. Show and label the electric field, the magnetic field, the direction of travel and one wavelength. (b) Describe the shape and orientation of the wavefronts. (c) The wave meets an ideal polarizing filter. State how much of the intensity is transmitted when the filter’s axis is along x, and when it is along z. Explain using the direction of the fields.
Worked solution and suggested Marlbridge rubric
(a) Draw the y-axis as the direction of travel. The E field is a sine curve oscillating along x; the B field is a sine curve oscillating along z, at right angles to E. The two curves cross the axis at the same points and peak at the same positions. One wavelength is marked between two neighbouring crests (or any two corresponding points) along y.
(b) The wavefronts are flat planes perpendicular to the direction of travel, so they are parallel to the xz-plane. Neighbouring crest wavefronts are one wavelength apart.
(c) The polarization direction is the direction of E, which is x. With the filter axis along x (parallel), all the intensity is transmitted. With the axis along z (perpendicular to E), none is transmitted. The fact that B lies along z does not mean a z-axis filter passes the wave.
| Point | What earns it |
|---|---|
| 1 | E drawn oscillating along x and B along z, perpendicular to each other and to y |
| 1 | E and B in step (crossing zero together and peaking together), travel direction labelled +y |
| 1 | One wavelength correctly marked between corresponding points |
| 1 | Wavefronts are planes perpendicular to y (parallel to the xz-plane) |
| 1 | All transmitted along x, none along z, because polarization follows E |
Do not award the last point for the right outcome justified by the direction of B.
Question 7 (constructed response · stretch)
Light of wavelength 6.0 × 10⁻⁷ m in air enters a block of glass at right angles to its surface. In this glass, light travels at 2.0 × 10⁸ m/s.
(a) Calculate the frequency of the light in air and in the glass. (b) Calculate the wavelength of the light inside the glass. (c) The block is 0.030 m thick. Calculate how much longer the light takes to cross the block than to cross the same distance of air. (d) A student says: “Inside the glass the wavelength is shorter, so the light moves towards the violet end of the spectrum and turns bluer.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) In air: f = c/λ = (3.00 × 10⁸) ÷ (6.0 × 10⁻⁷) = 5.0 × 10¹⁴ Hz. The frequency does not change at a boundary, so it is 5.0 × 10¹⁴ Hz in the glass too.
(b) λ_glass = v/f = (2.0 × 10⁸) ÷ (5.0 × 10¹⁴) = 4.0 × 10⁻⁷ m.
(c) In glass: t = 0.030 ÷ 2.0 × 10⁸ = 1.5 × 10⁻¹⁰ s. In air: t = 0.030 ÷ 3.00 × 10⁸ = 1.0 × 10⁻¹⁰ s. Difference = 5.0 × 10⁻¹¹ s.
(d) The claim is wrong. The position in the spectrum and the colour we see are set by the frequency, which stays at 5.0 × 10¹⁴ Hz inside the glass. Only the speed and the wavelength change, and both go back to their original values when the light leaves the glass. The ordering of the spectrum by wavelength assumes the waves are compared in the same medium (vacuum or air).
| Point | What earns it |
|---|---|
| 1 | f = 5.0 × 10¹⁴ Hz in air |
| 1 | Same frequency in the glass, with the reason (frequency fixed at a boundary) |
| 1 | λ in glass = 4.0 × 10⁻⁷ m, using the glass speed |
| 1 | Time difference of 5.0 × 10⁻¹¹ s |
| 1 | Rejects the claim because the colour depends on frequency, which is unchanged |
| 1 | Notes that the spectrum order compares wavelengths in the same medium (vacuum or air), or that the wavelength returns on leaving the glass |
Carry forward a wrong frequency from (a) into (b) once.
How did you do?
- Q1 or Q5(d) wrong: re-learn the order in “The electromagnetic spectrum” and Figure 2 of the study guide.
- Q2 or Q5 wrong: redo Worked examples 1 and 3, watching the unit prefixes.
- Q3 or Q6 wrong: revisit “What is waving?”, Figure 1 and Worked example 2.
- Q4 wrong: re-read “No medium needed”.
- Q7 incomplete: link back to the fixed-frequency rule in Topic 14.3.
Then tick off the topic checklist.