Questions on longitudinal and transverse waves

Question 1

Explain the difference between a transverse and a longitudinal wave, in terms of the direction of particle oscillation.

Accepted Answer

In a transverse wave the particles oscillate at right angles to the direction of energy transfer (1); in a longitudinal wave they oscillate parallel to it (1).

Question 2

A sound wave of frequency 170 Hz travels through air at 340 m s⁻¹. Calculate its wavelength.

Accepted Answer

λ = c/f = 340/170 (1) λ = 2.0 m (1)

Question 3

Give one example of a transverse wave and one example of a longitudinal wave.

Accepted Answer

Transverse: light (or any electromagnetic wave, or a wave on a string) (1). Longitudinal: sound (or a compression wave on a spring) (1).

Question 4

Sound travels through air as a longitudinal wave. State what is meant by a compression and by a rarefaction.

Accepted Answer

A compression is a region where the air particles are bunched together, so the pressure is a maximum (1). A rarefaction is a region where the particles are spread apart, so the pressure is a minimum (1).

Question 5

All electromagnetic waves travel at the same speed in a vacuum. State this speed and give the approximate range of wavelengths of visible light.

Accepted Answer

3.00 × 10⁸ m s⁻¹ (1). Visible light runs from about 400 nm (violet) to about 700 nm (red) (1).

Question 6

State what is meant by plane-polarised light, and explain why only transverse waves can be polarised.

Accepted Answer

Plane-polarised light has its oscillations confined to a single plane (1). Only transverse waves can be polarised because their oscillations are perpendicular to the travel direction, so a plane can be selected (1); longitudinal oscillations are along the travel direction, so there is no such choice (1).

Question 7

Unpolarised light passes through a single polarising filter, which is then slowly rotated in its own plane. Describe what happens to the brightness of the transmitted light, and contrast this with rotating a filter placed in front of already-polarised light.

Accepted Answer

For unpolarised light the transmitted brightness stays roughly constant as the single filter rotates (1), because about half the light passes at any angle (1). For already-polarised light, the brightness varies from a maximum to zero and back as the filter rotates, being zero when its axis is perpendicular to the polarisation (1).

Question 8

Microwaves of frequency 3.0 × 10⁹ Hz travel at 3.0 × 10⁸ m s⁻¹. Calculate their wavelength, and state one way to show that microwaves are polarised.

Accepted Answer

λ = c/f = (3.0 × 10⁸)/(3.0 × 10⁹) (1)
λ = 0.10 m (1)
Rotating a metal grille (or receiver) between transmitter and detector changes the received signal, which shows the microwaves are polarised (1)

Question 9

Television signals are transmitted as plane-polarised radio waves. Explain why a receiving aerial must be mounted with its rods parallel to the plane of polarisation of the signal, and describe what happens to the received signal strength as the aerial is slowly rotated through 90° about the line to the transmitter.

Accepted Answer

The wave drives oscillations (a current) in the aerial only when the rods lie along the plane of oscillation of the wave (1). As the aerial is rotated away from this plane the received signal strength falls (1), reaching a minimum (near zero) when the rods are at 90° to the plane of polarisation (1).

Question 10

Visible light in a vacuum has wavelengths from about 400 nm to about 700 nm. The speed of light in a vacuum is 3.00 × 10⁸ m s⁻¹. Calculate the range of frequencies of visible light.

Accepted Answer

f = c/λ (1). At 400 nm: f = (3.00 × 10⁸)/(400 × 10⁻⁹) = 7.5 × 10¹⁴ Hz; at 700 nm: f = 4.3 × 10¹⁴ Hz (1). Visible frequencies run from about 4.3 × 10¹⁴ Hz up to 7.5 × 10¹⁴ Hz (1).

Question 11

A sound wave and a radio wave both travel due north. State the possible direction(s) of oscillation of (a) the air particles carrying the sound and (b) the oscillations of the radio wave.

Accepted Answer

(a) North–south only: sound is longitudinal, so the particles oscillate parallel to the direction of travel (1). (b) Any direction perpendicular to north (for example vertical, or east–west): radio waves are transverse (1).

Question 12

Plane-polarised light of intensity 24 W m⁻² is incident on a polarising filter whose transmission axis is at 30° to the plane of polarisation of the light. Calculate the intensity of the light transmitted by the filter.

Accepted Answer

Malus's law: I = I₀cos²θ (1). I = 24 × cos²30° = 24 × 0.750 (1) = 18 W m⁻² (1).

Question 13

Two polarising filters are arranged with their axes crossed (at 90°), so no light passes. A third filter is then inserted between them with its axis at 45°. Explain why light now emerges from the arrangement.

Accepted Answer

Each filter transmits the component of the oscillation along its axis (1). The first filter polarises the light; the 45° filter transmits a component at 45°, re-orienting the polarisation (1); that light then has a component along the final filter's axis, so some passes (1). Without the middle filter the light reaching the last filter is perpendicular to its axis and is blocked (1).

Question 14

Explain why polarising sunglasses reduce glare from reflective surfaces such as water or a wet road.

Accepted Answer

Light reflected from such surfaces is partially polarised in the horizontal plane (1). Polarising sunglasses have a vertical transmission axis (1), so they block much of this horizontally-polarised reflected glare while still letting other light through (1).

Question 15

Explain why sound waves cannot be polarised.

Accepted Answer

Sound is longitudinal: its oscillations are along the direction of travel, not perpendicular to it (1). There is no plane of oscillation to select (1), so a filter cannot restrict it and sound cannot be polarised (1).

Question 16

A student is given two beams of light of equal brightness. One beam is unpolarised; the other is plane polarised. The beams look identical to the eye. Deduce how the student can use a single polarising filter to identify which beam is which, stating the observation expected for each beam.

Accepted Answer

Place the filter in each beam in turn and rotate it in its own plane through at least 90° (1). For the unpolarised beam the transmitted brightness stays roughly constant at every angle (1). For the plane-polarised beam the brightness varies from a maximum to near zero, the minimum occurring when the transmission axis is at 90° to the plane of polarisation (1). The beam whose brightness varies on rotation is the plane-polarised one (1).

Question 17

Describe an experiment, using a microwave transmitter and a receiver connected to a meter, to show that the microwaves emitted by the transmitter are plane polarised. Explain how the results also show that microwaves are transverse waves.

Accepted Answer

Point the receiver at the transmitter and record the meter reading (1). Rotate the receiver (or a metal grille placed between them) about the line joining transmitter and receiver, recording the signal at intervals (1). The signal falls to a minimum (near zero) when the receiver has turned through 90° and returns to a maximum at 180° (1). The signal depends on orientation, so the oscillations are confined to a single plane: the microwaves are plane polarised (1). Only oscillations perpendicular to the direction of travel can be selected by orientation in this way, so microwaves must be transverse (1).

Question 18

The table gives the wavelengths in a vacuum of four electromagnetic waves. P: 3.0 cm; Q: 550 nm; R: 10 μm; S: 3.0 nm. Deduce which wave is visible light and which is a microwave, and calculate the frequency of the visible wave. The speed of electromagnetic waves in a vacuum is 3.00 × 10⁸ m s⁻¹.

Accepted Answer

Q is visible light: 550 nm lies in the visible range, about 400 nm to 700 nm (1). P is a microwave: 3.0 cm lies between about 10 cm and 1 mm (1). f = c/λ = (3.00 × 10⁸)/(550 × 10⁻⁹) = 5.5 × 10¹⁴ Hz (1).

Question 19

Light that is already plane polarised, of intensity I₀, passes through two polarising filters in turn. The transmission axis of the first filter is at 30° to the plane of polarisation of the light, and the transmission axis of the second is at 75° to that same plane. Calculate the intensity emerging from the second filter, as a fraction of I₀.

Accepted Answer

Take the filters one at a time, because the light leaving a filter is polarised along that filter's own axis (1). First filter: I₁ = I₀cos²30° = 0.750 I₀ (1). The angle at the second filter is the angle to the filter before it, 75° − 30° = 45°, and not the 75° back to the original plane (1). I₂ = 0.750 I₀ × cos²45° = 0.375 I₀ (1).