Waves at interfaces and pulse-echoEdexcel A-Level Physics: Revision notes

An interface is the boundary between two media. When a wave meets an interface, part of its energy is reflected and part is transmitted into the second medium, where it may be refracted.

The frequency of the wave is set by the source and does not change. The wave speed and wavelength change in the new medium. The more different the two media, the larger the fraction reflected.

Ultrasound is sound with a frequency above the range of human hearing, above about 20 kHz. Medical scanners use frequencies of several megahertz.

Air and skin are very different media, so an air gap between the transducer and the skin would reflect nearly all the ultrasound. Gel is applied to remove the air gap so that most of the ultrasound enters the body.

Wavelength follows from v = fλ. For 3.5 MHz in tissue (v = 1540 m s⁻¹), λ = 1540 ÷ 3.5 × 10⁶ = 4.4 × 10⁻⁴ m.

A transducer sends a short pulse of ultrasound into the body, then listens for echoes reflected from boundaries between tissues.

• The time delay t between sending and receiving gives the depth
• The pulse travels there and back, so d = vt/2
• Stronger echoes come from boundaries where the media differ more
• Moving the transducer, or sending pulses in many directions, builds up a two-dimensional image

Worked example: t = 78 μs and v = 1540 m s⁻¹ gives d = 1540 × 78 × 10⁻⁶ ÷ 2 = 0.060 m.

The transducer must wait until echoes from the deepest boundary have returned before sending the next pulse. If pulses were sent more quickly, an echo from one pulse could be mistaken for an echo of the next, and the depth would be wrongly calculated.

Waves are diffracted round objects that are about the same size as, or smaller than, the wavelength. Details smaller than about one wavelength are therefore not resolved.

For 8.0 MHz in tissue, λ = 1540 ÷ 8.0 × 10⁶ = 1.9 × 10⁻⁴ m, so details smaller than about 0.19 mm are not seen.

A higher frequency gives a shorter wavelength and so a finer image, but higher-frequency ultrasound is absorbed more strongly and does not penetrate as far.

A pulse of duration t has a length vt in the tissue. Echoes from two boundaries are only separate if the extra distance travelled by the second echo, twice their separation, is greater than the pulse length.

For a 0.40 μs pulse, vt = 1540 × 0.40 × 10⁻⁶ = 6.2 × 10⁻⁴ m, so boundaries less than 0.31 mm apart give overlapping echoes. A shorter pulse duration gives better resolution along the beam.