Waves — Diagnostic Tests

Unit Tests

UT-1: Wave Properties and Calculations

Question: (a) Define the terms: amplitude, wavelength, frequency, and period. State the relationship between frequency, wavelength, and wave speed. (b) A wave has a frequency of 250 Hz and a wavelength of 1.5 m. Calculate the wave speed. (c) A sound wave in air has a speed of 340 m/s and a frequency of 440 Hz. Calculate the wavelength. (d) Explain the difference between transverse and longitudinal waves, giving an example of each.

Solution:

(a) Amplitude: The maximum displacement from the equilibrium (rest) position. Wavelength: The distance between two consecutive identical points on a wave (e.g., crest to crest). Frequency: The number of complete waves passing a point per second, measured in hertz (Hz). Period: The time taken for one complete wave cycle, measured in seconds. $T = 1/f$.

Relationship: $v = f\lambda$ (wave speed $=$ frequency $\times$ wavelength).

(b) $v = f\lambda = 250 \times 1.5 = 375$ m/s.

(c) $\lambda = v/f = 340/440 = 0.773$ m.

(d) Transverse waves: The oscillations are perpendicular to the direction of wave propagation. Examples: light, water waves, electromagnetic waves.

Longitudinal waves: The oscillations are parallel to the direction of wave propagation. They consist of compressions (high pressure) and rarefactions (low pressure). Example: sound waves.

UT-2: The Electromagnetic Spectrum

Question: (a) List the parts of the electromagnetic spectrum in order of increasing wavelength. (b) Which type of electromagnetic radiation has: (i) the highest frequency, (ii) the most energy per photon, (iii) the longest wavelength? (c) Explain why ultraviolet radiation can cause sunburn but visible light cannot. (d) State one use and one hazard for: X-rays, microwaves, and infrared radiation.

Solution:

(a) In order of increasing wavelength (decreasing frequency): gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves, radio waves.

(b) (i) Gamma rays (highest frequency). (ii) Gamma rays (energy per photon $E = hf$). (iii) Radio waves (longest wavelength).

(c) UV photons have higher frequency (shorter wavelength) than visible light, so each UV photon carries more energy ($E = hf$). UV photons have enough energy to damage DNA molecules in skin cells, causing mutations that can lead to sunburn and skin cancer. Visible light photons have insufficient energy to directly damage DNA.

(d) X-rays: Use — medical imaging (fractures, dental). Hazard — ionising radiation, can cause cancer with prolonged exposure.

Microwaves: Use — cooking food, satellite communications. Hazard — can heat body tissue (especially eyes and internal organs at high intensities).

Infrared: Use — thermal imaging, remote controls, cooking. Hazard — burns at high intensity (skin absorbs IR, causing heating).

UT-3: Reflection and Refraction

Question: (a) State the law of reflection. (b) A light ray strikes a plane mirror at an angle of incidence of $35^\circ$. Calculate the angle of reflection and the angle between the incident and reflected rays. (c) Explain why a pool of water appears shallower than it actually is, with reference to refraction. (d) Describe an experiment to demonstrate total internal reflection using a semicircular glass block.

Solution:

(a) The angle of incidence equals the angle of reflection. Both angles are measured from the normal (perpendicular) to the surface at the point of incidence.

(b) Angle of reflection $= 35^\circ$. Angle between incident and reflected rays $= 35 + 35 = 70^\circ$.

(c) Light from the bottom of the pool travels from water (denser medium, higher refractive index) to air (less dense). It refracts away from the normal as it exits. An observer above the water traces the refracted ray back in a straight line, which intersects at a point that is higher (shallower) than the actual position of the object. This is a virtual image, closer to the surface.

(d) Experiment: Direct a ray of light at the curved surface of a semicircular glass block so that it enters without refraction (along the radius). The ray hits the flat surface at an angle of incidence. Gradually increase the angle of incidence. The refracted ray bends further away from the normal until, at the critical angle, it travels along the surface. Beyond the critical angle, total internal reflection occurs — the ray is entirely reflected back into the glass, and no refracted ray exists. Measure the critical angle and compare with $\sin c = 1/n$.

---

Integration Tests

IT-1: Sound Waves and Communication (with Electricity)

Question: (a) Explain how sound waves are produced by a loudspeaker. (b) A sound wave has a frequency of 1000 Hz in air (speed 340 m/s). Calculate its wavelength. If the same sound enters water (speed 1500 m/s), calculate the new wavelength. Does the frequency change? (c) Explain why sound cannot travel through a vacuum. (d) A person stands 170 m from a cliff and claps. Calculate the time interval between the clap and hearing the echo. If the air temperature increases, explain how the time interval changes.

Solution:

(a) An alternating current from an amplifier flows through a coil in the loudspeaker’s magnetic field. The interaction between the current-carrying coil and the magnetic field creates a varying force on the cone. The cone vibrates at the same frequency as the AC signal, pushing and pulling the air molecules in front of it. These vibrations create compressions and rarefactions that propagate as a longitudinal sound wave.

(b) In air: $\lambda = 340/1000 = 0.34$ m. In water: $\lambda = 1500/1000 = 1.5$ m. The frequency does not change — it is determined by the source. The wavelength increases because the wave speed increases while frequency stays constant ($v = f\lambda$).

(c) Sound is a mechanical wave that requires a medium (particles) to propagate. The compressions and rarefactions are created by particles vibrating and colliding with neighbouring particles, transferring energy. In a vacuum, there are no particles to vibrate and transmit the wave.

(d) Distance to cliff and back $= 2 \times 170 = 340$ m. Time $= 340/340 = 1$ second.

If air temperature increases, the speed of sound increases (particles move faster, transmitting vibrations more quickly). The time interval would decrease because the sound covers the same distance at a higher speed.

IT-2: Electromagnetic Waves and Communication (with Forces)

Question: A mobile phone communicates with a cell tower 5 km away using microwaves of frequency 1800 MHz. (a) Calculate the wavelength of the microwaves. (b) Calculate the time for a signal to travel from the phone to the tower. (c) The phone transmits at a power of 1 W. If the signal spreads out equally in all directions, calculate the intensity at the tower (assume the signal is captured by a receiver area of 0.5 \text{ m^2). (d) Explain why microwave signals are used for mobile communication rather than radio waves.

Solution:

(a) $\lambda = v/f = 3 \times 10^8 / (1800 \times 10^6) = 3 \times 10^8 / 1.8 \times 10^9 = 0.167$ m.

(b) Time $= 5000 / (3 \times 10^8) = 1.67 \times 10^{-5}$ s $= 16.7$ $\mu$S.

(c) Intensity at distance 5 km: $I = P/(4\pi r^2) = 1/(4\pi \times 25,000,000) = 1/314,159,265 = 3.18 \times 10^{-9}$ W/m$^2$. Power captured by 0.5 \text{ m^2 receiver $= 3.18 \times 10^{-9} \times 0.5 = 1.59 \times 10^{-9}$ W $= 1.59$ nW.

(d) Microwaves offer a balance between range and data capacity: (1) Higher frequency than radio waves means they can carry more information (higher bandwidth). (2) Shorter wavelength allows smaller antennas (practical for handheld devices). (3) They can be directed in narrow beams between towers, reducing interference. (4) They are not absorbed by the atmosphere. Radio waves have longer wavelengths, which limits data capacity and requires larger antennas, but they can travel longer distances and penetrate obstacles better.

IT-3: Wave Interference and Applications (with Energy)

Question: (a) Explain the conditions needed for constructive and destructive interference. (b) Two coherent sources produce waves of wavelength 0.5 m. At a point P, the path difference is 1.5 m. Determine whether the interference at P is constructive or destructive. (c) Explain how noise-cancelling headphones work using the principle of superposition. (d) Ultrasound waves of frequency 2 MHz are used in medical imaging. Calculate the wavelength in tissue (speed $= 1540$ m/s) and explain why high-frequency ultrasound is preferred for detailed images.

Solution:

(a) Constructive interference: Occurs when two waves meet in phase — the path difference is a whole number of wavelengths ($n\lambda$). The amplitudes add, producing a wave with greater amplitude.

Destructive interference: Occurs when two waves meet completely out of phase (antiphase) — the path difference is a half-integer number of wavelengths ($n\lambda + \lambda/2$). The amplitudes cancel, producing zero amplitude (if waves have equal amplitude).

(b) Path difference $= 1.5$ m $= 3 \times 0.5$ m $= 3\lambda$. Since the path difference is a whole number of wavelengths, the interference is constructive.

(c) Noise-cancelling headphones have a microphone that detects ambient noise. A microchip analyses the noise waveform and generates a sound wave that is the exact inverse (antiphase) — the same amplitude but opposite phase. When this generated wave is combined with the ambient noise inside the ear cup, destructive interference occurs, significantly reducing the perceived noise. This is the principle of superposition applied in real time.

(d) $\lambda = v/f = 1540/(2 \times 10^6) = 7.7 \times 10^{-4}$ m $= 0.77$ mm.

High-frequency ultrasound has a short wavelength, which means it can detect smaller structures and produce sharper images (better resolution). This is analogous to optical microscopy — shorter wavelengths provide finer detail. However, higher frequency ultrasound is more rapidly absorbed by tissue, limiting penetration depth. Medical imaging balances resolution and penetration by selecting appropriate frequencies ( 1—15 MHz depending on the application).

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Confusing terminology or concepts that appear similar but have distinct meanings.
• Overlooking key assumptions or boundary conditions that limit applicability.