v = fλ looks simple. The marks lie in the units. Each question below trains a different skill: conversion traps, calculation chains, structured explanation, and proportional reasoning.
Convert your units before you substitute, not after. If your final number looks wildly large or wildly small, your conversion almost certainly went wrong. Write the conversion step explicitly so the examiner sees it and so you can debug your own working.
A domestic microwave oven uses electromagnetic waves of frequency 2.45 GHz to heat food. Electromagnetic waves travel at 3.0 × 10⁸ m/s.
Calculate the wavelength of these microwaves in metres.
Convert the frequency first:
1 GHz = 10⁹ Hz f = 2.45 GHz = 2.45 × 10⁹ HzRearrange:
v = fλ → λ = v ÷ fSubstitute:
λ = (3.0 × 10⁸) ÷ (2.45 × 10⁹) λ = 0.122 mλ = 0.12 m (2 s.f.)
About 12 cm. That is why microwave ovens are roughly that size inside, and why metal mesh on the door (with holes much smaller than 12 cm) blocks the waves but lets you see through.
A sound wave travelling through water has a wavelength of 75 cm. The speed of sound in water is 1500 m/s.
Calculate the frequency of the wave. Give your final answer in kHz.
Convert wavelength to metres (matching v):
λ = 75 cm = 0.75 mRearrange:
v = fλ → f = v ÷ λSubstitute:
f = 1500 ÷ 0.75 f = 2000 HzConvert to kHz (as the question asked):
1 kHz = 1000 Hzf = 2.0 kHz
Two conversions in one question. The trap is to skip the cm-to-m step and substitute 75 directly. The other trap is to leave the answer in Hz when the question asked for kHz. Both lose marks.
A student claims that the wave equation v = fλ can only be used for sound waves, because sound is the type of wave they encounter most often in everyday life.
Explain why the student is mistaken. In your answer, you must refer to at least two specific types of wave other than sound, and state a typical value for the speed of each.
Structure your answer in three parts:
1. State the claim is wrong:
The wave equation applies to all periodic waves, not just sound. It comes from the definition of speed (v = distance ÷ time) applied to one wave cycle.
2. Give two specific counter-examples with speeds:
3. Reinforce the universal nature:
v = fλ is independent of what the wave is made of. It only requires the wave to be repeating with a measurable frequency and wavelength.
Examiners want a clear claim, specific examples, and numerical values. Vague answers like "it works for light too" without a speed score partial marks at best. State the claim, name the wave, give the speed.
A laboratory laser produces red light of wavelength 633 nm. Light travels through air at approximately 3.0 × 10⁸ m/s.
(a) Express the wavelength 633 nm in metres, in standard form. [1] (b) Calculate the frequency of this light. [2] (c) Calculate the period of one oscillation of this light wave. [2](a) Convert nm to m
1 nm = 10⁻⁹ m λ = 633 × 10⁻⁹ mλ = 6.33 × 10⁻⁷ m
(b) Frequency
f = v ÷ λ f = (3.0 × 10⁸) ÷ (6.33 × 10⁻⁷)f = 4.74 × 10¹⁴ Hz
(c) Period
T = 1 ÷ f T = 1 ÷ (4.74 × 10¹⁴)T = 2.11 × 10⁻¹⁵ s
Standard form is mandatory for very small numbers. Three prefixes to memorise for waves: n = 10⁻⁹ (nm for light), μ = 10⁻⁶ (sometimes for IR), p = 10⁻¹² (rarely). For frequency it is k = 10³, M = 10⁶, G = 10⁹.
Two transverse waves, A and B, travel along the same stretched rope. They have the same speed. The wavelength of Wave A is three times the wavelength of Wave B.
(a) State and explain how the frequency of Wave A compares with the frequency of Wave B. [2] (b) Wave B has a frequency of 6.0 Hz. Calculate the frequency of Wave A. [2](a) Compare frequencies
Same speed means: v = fAλA = fBλB
If λA = 3 λB, then:
fA × 3λB = fB × λB fA = fB ÷ 3Wave A has one-third the frequency of Wave B
Because at fixed speed, f and λ are inversely proportional. Triple the wavelength means one-third the frequency.
(b) Calculate frequency of Wave A
fA = 6.0 ÷ 3fA = 2.0 Hz
When two quantities are inversely proportional at a fixed third quantity, multiplying one by a factor means dividing the other by the same factor. A reasoning shortcut worth practising.