{"id":780,"date":"2026-08-11T22:59:58","date_gmt":"2026-08-11T22:59:58","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=780"},"modified":"2026-08-24T13:03:44","modified_gmt":"2026-08-24T13:03:44","slug":"wavelength-formula-units","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/wavelength-formula-units\/","title":{"rendered":"Wavelength: Definition, Formula and Units"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\n\nWavelength is the distance between two identical points on consecutive cycles of a wave \u2014 crest to crest, or compression to compression. It is written as the Greek letter lambda (\u03bb) and measured in metres. Wavelength equals wave speed divided by frequency, so faster waves and lower frequencies both mean longer wavelengths.\n\n<\/p><\/div>\n\n<p>Stand near the speaker stack at a festival and you feel the bass in your chest long before you pick out the cymbals. That is not loudness alone. It is wavelength.<\/p>\n\n<p>A 20 Hz kick drum rolls out in waves roughly 17 metres long, so it bends around walls and bodies like water around pebbles. The hi-hat at 10 kHz arrives in ripples about 3 cm across \u2014 step behind a pillar and it disappears. Same air, same speed, wildly different lengths.<\/p>\n\n<h2>What Is Wavelength?<\/h2>\n\n<p>Wavelength is the distance over which a wave&#8217;s shape repeats itself, measured along the direction the wave travels. Drop a stone in a still pond and freeze the ripples in your mind: the gap between one ring and the next is the wavelength.<\/p>\n\n<p>More precisely, it is the distance between two consecutive points that are <strong>in phase<\/strong> \u2014 points doing exactly the same thing at the same instant. Two neighbouring crests qualify. So do two neighbouring troughs, or any two matching points one full cycle apart.<\/p>\n\n<p>That &#8220;one full cycle&#8221; part is where marks get lost. Crest to trough is only half a wavelength.<\/p>\n\n<h3>Where You Measure It on Different Waves<\/h3>\n\n<p>On a <strong>transverse<\/strong> wave \u2014 a guitar string, a water ripple, light \u2014 the particles move at right angles to the travel direction, so there are visible crests and troughs to point at.<\/p>\n\n<p>On a <strong>longitudinal<\/strong> wave such as sound in air, the particles jostle back and forth along the travel direction. There are no crests at all, only squeezed regions (compressions) and stretched ones (rarefactions). You measure compression to compression instead. If the distinction is hazy, our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/transverse-vs-longitudinal-waves\/\">transverse and longitudinal waves<\/a> sets out both cases side by side.<\/p>\n\n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/wavelength-formula-units-how-measured-crest-crest-transverse.webp\" width=\"1400\" height=\"932\" alt=\"Diagram showing how wavelength is measured: crest to crest on a transverse wave and compression to compression on a longitudinal wave\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:700px;display:block;margin:0 auto;\" \/><\/figure>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">Wavelength is measured between matching points one full cycle apart \u2014 never crest to trough, and never confused with amplitude.<\/p>\n\n<h2>The Wavelength Formula: \u03bb = v \/ f<\/h2>\n\n<p>The wavelength formula is \u03bb = v \/ f \u2014 wavelength equals wave speed divided by frequency.<\/p>\n\n<div class=\"pf-formula\">\u03bb = v \/ f<\/div>\n\n<p>Each symbol carries its own SI unit, and mixing them up is the fastest route to an answer that is wrong by a factor of a billion:<\/p>\n\n<ul>\n<li><strong>\u03bb<\/strong> (lambda) \u2014 wavelength, in <strong>metres (m)<\/strong><\/li>\n<li><strong>v<\/strong> \u2014 the speed of the wave in its medium, in <strong>metres per second (m\/s)<\/strong><\/li>\n<li><strong>f<\/strong> \u2014 frequency, in <strong>hertz (Hz)<\/strong>, meaning cycles per second<\/li>\n<\/ul>\n\n<p>The same relationship rearranges two other ways, and exams ask for all three:<\/p>\n\n<div class=\"pf-formula\">v = f \u03bb<\/div>\n\n<div class=\"pf-formula\">f = v \/ \u03bb<\/div>\n\n<p>For light and every other electromagnetic wave travelling through a vacuum, the speed is fixed at c = 299,792,458 m\/s, so the formula simplifies:<\/p>\n\n<div class=\"pf-formula\">\u03bb = c \/ f<\/div>\n\n<p>Once the numbers get awkward \u2014 gigahertz on one side, nanometres on the other \u2014 it is worth checking your arithmetic against our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/wave-speed\">Wave Speed Calculator<\/a>, which solves \u03bb = v \/ f for whichever of the three quantities you leave blank.<\/p>\n\n<h2>Wavelength Units: Metres, Nanometres and \u00c5ngstr\u00f6ms<\/h2>\n\n<p>The SI unit of wavelength is the <strong>metre (m)<\/strong>, because wavelength is simply a length. The metre itself is now defined from the speed of light, which is why wavelength and frequency measurements are so tightly linked in modern metrology \u2014 see the <a href=\"https:\/\/www.nist.gov\/si-redefinition\/definitions-si-base-units\" target=\"_blank\" rel=\"noopener\">NIST definitions of the SI base units<\/a>.<\/p>\n\n<p>In practice nobody quotes green light as 0.00000055 m. Each part of the spectrum has a working unit:<\/p>\n\n<ul>\n<li><strong>kilometres (km)<\/strong> \u2014 long-wave and AM radio<\/li>\n<li><strong>metres (m)<\/strong> \u2014 FM radio, sound, water waves<\/li>\n<li><strong>centimetres (cm)<\/strong> and <strong>millimetres (mm)<\/strong> \u2014 microwaves and radar<\/li>\n<li><strong>micrometres (\u03bcm)<\/strong> \u2014 infrared; 1 \u03bcm = 10<sup>-6<\/sup> m<\/li>\n<li><strong>nanometres (nm)<\/strong> \u2014 visible and ultraviolet light; 1 nm = 10<sup>-9<\/sup> m<\/li>\n<li><strong>\u00e5ngstr\u00f6ms (\u00c5)<\/strong> and <strong>picometres (pm)<\/strong> \u2014 X-rays and atomic spacing; 1 \u00c5 = 0.1 nm = 10<sup>-10<\/sup> m<\/li>\n<\/ul>\n\n<p>In practice, the single most common slip in exam papers is leaving wavelength in nanometres while frequency sits in hertz. Convert to metres first, every time.<\/p>\n\n<h2>How to Calculate Wavelength in 4 Steps<\/h2>\n\n<p>To calculate wavelength, find the wave speed in that medium, convert the frequency to hertz, divide speed by frequency, then check the size of your answer against the type of wave.<\/p>\n\n<ol>\n<li><strong>Find the wave speed v for that medium.<\/strong> Sound in air at 20 \u00b0C is about 343 m\/s. Light in a vacuum is 299,792,458 m\/s, usually rounded to 3.00 \u00d7 10<sup>8<\/sup> m\/s.<\/li>\n<li><strong>Convert the frequency to hertz.<\/strong> 1 kHz = 10<sup>3<\/sup> Hz, 1 MHz = 10<sup>6<\/sup> Hz, 1 GHz = 10<sup>9<\/sup> Hz.<\/li>\n<li><strong>Divide.<\/strong> \u03bb = v \/ f, working entirely in metres and seconds.<\/li>\n<li><strong>Sanity-check the magnitude.<\/strong> Radio should land in metres, microwaves in centimetres, visible light in hundreds of nanometres. An answer of 500 m for green light means a unit slipped.<\/li>\n<\/ol>\n\n<p>Try it on a 2.4 GHz Wi-Fi signal. Speed is 3.00 \u00d7 10<sup>8<\/sup> m\/s, frequency is 2.4 \u00d7 10<sup>9<\/sup> Hz, so \u03bb = 0.125 m \u2014 about 12.5 cm, roughly the length of a phone. That is why a router antenna is a few centimetres long rather than a few metres.<\/p>\n\n<h2>How Wavelength Works: Where \u03bb = v \/ f Comes From<\/h2>\n\n<p>The formula is not something to memorise blindly \u2014 it falls out of one observation. In exactly one period T, a wave advances exactly one wavelength.<\/p>\n\n<p>That is the whole derivation in a sentence. Distance equals speed multiplied by time, so:<\/p>\n\n<div class=\"pf-formula\">\u03bb = v T<\/div>\n\n<p>Period and frequency are reciprocals, T = 1 \/ f. Substitute it in and you get \u03bb = v \/ f. Nothing else is needed.<\/p>\n\n<h3>Why Frequency and Wavelength Trade Off<\/h3>\n\n<p>Keep a wave in one medium and its speed barely changes. A middle-C note and a top-C note both cross a room at about 343 m\/s.<\/p>\n\n<p>So if v is fixed and you double the frequency, the wavelength must halve \u2014 the two are inversely proportional. Think of it as the same length of ribbon being folded into twice as many pleats. Our companion piece on the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/frequency-formula\/\">frequency formula<\/a> works the same relationship from the other direction.<\/p>\n\n<p>Change the medium, though, and v itself changes \u2014 which is where most of the confusion in this topic starts.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Wave Generator Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:560px}@media(max-width:760px){.pf-sim-frame{height:840px}}<\/style><iframe src=\"\/labs\/wave.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">Drag the wavelength and frequency sliders and watch the speed readout: hold one, change the other, and \u03bb = v \/ f does the rest.<\/p>\n\n<h2>Wavelength Across the Electromagnetic Spectrum<\/h2>\n\n<p>Radio waves, visible light and gamma rays are the same phenomenon at different wavelengths \u2014 all of them travelling at c in a vacuum. NASA&#8217;s overview of <a href=\"https:\/\/science.nasa.gov\/ems\/01_intro\/\" target=\"_blank\" rel=\"noopener\">the electromagnetic spectrum<\/a> lays out how each band is used, and our own guide to the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/electromagnetic-spectrum\/\">electromagnetic spectrum<\/a> goes band by band.<\/p>\n\n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/wavelength-formula-units-across-electromagnetic-spectrum-kilometre-long.webp\" width=\"1400\" height=\"676\" alt=\"Wavelength across the electromagnetic spectrum, from kilometre-long radio waves to gamma rays below 0.01 nanometres, with the visible band expanded from 380 to 700 nanometres\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:700px;display:block;margin:0 auto;\" \/><\/figure>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">Wavelength spans fourteen orders of magnitude across the spectrum, yet every band obeys the same \u03bb = c \/ f.<\/p>\n\n<p>Put sound alongside light and the range becomes concrete. Every wavelength below is calculated from \u03bb = v \/ f:<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;\">\n<thead>\n<tr style=\"background:#0A1628;color:#FAF6EE;\">\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Wave<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Frequency<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Speed<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Wavelength<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Roughly the size of<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Deep bass note<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">20 Hz<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">343 m\/s (air)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">17 m<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">An articulated lorry<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Concert A note<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">440 Hz<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">343 m\/s (air)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">0.78 m<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">An arm&#8217;s length<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Upper hearing limit<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">20 kHz<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">343 m\/s (air)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">1.7 cm<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">A fingernail<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">FM radio<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">98.5 MHz<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">3.00 \u00d7 10<sup>8<\/sup> m\/s<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">3.04 m<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">A small car<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Wi-Fi signal<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">2.4 GHz<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">3.00 \u00d7 10<sup>8<\/sup> m\/s<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">12.5 cm<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">A smartphone<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Green light<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">5.45 \u00d7 10<sup>14<\/sup> Hz<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">3.00 \u00d7 10<sup>8<\/sup> m\/s<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">550 nm<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">A bacterium<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Medical X-ray<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">3 \u00d7 10<sup>18<\/sup> Hz<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">3.00 \u00d7 10<sup>8<\/sup> m\/s<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">0.1 nm<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">An atom<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<h2>5 Real-World Examples of Wavelength<\/h2>\n\n<h3>1. The Bass That Travels Through Walls<\/h3>\n\n<p>Your neighbour&#8217;s music always reaches you as thumping bass, never as crisp vocals. A 40 Hz bass note has a wavelength near 8.6 m, far larger than a doorway or a stud wall, so it diffracts around obstacles and passes through easily.<\/p>\n\n<p>The 5 kHz consonants ride on 7 cm waves. Those get absorbed and reflected long before they arrive.<\/p>\n\n<h3>2. Your Wi-Fi Router&#8217;s Dead Spots<\/h3>\n\n<p>At 2.4 GHz the wavelength is about 12.5 cm. Reflections from walls and appliances can arrive half a wavelength out of step \u2014 roughly 6 cm \u2014 and cancel each other out.<\/p>\n\n<p>That is why moving a laptop a hand&#8217;s width along a desk sometimes restores a signal. You have stepped out of a null.<\/p>\n\n<h3>3. The Cold Spots in a Microwave Oven<\/h3>\n\n<p>A microwave oven runs at about 2.45 GHz, giving a wavelength near 12.2 cm. The waves bounce off the metal walls and form a standing pattern with hot spots about 6.1 cm apart \u2014 half a wavelength.<\/p>\n\n<p>The turntable exists purely to drag your food through those peaks and troughs. Without it, one corner of a lasagne boils while another stays frozen.<\/p>\n\n<h3>4. Why a Leaf Looks Green<\/h3>\n\n<p>Colour is wavelength made visible. Chlorophyll absorbs strongly in the blue (around 430 nm) and red (around 660 nm) but reflects the middle of the visible band near 550 nm, which your eye reads as green.<\/p>\n\n<p>Change nothing but the wavelength and you change the colour. Nothing else about the wave needs to differ.<\/p>\n\n<h3>5. X-rays That See Bone<\/h3>\n\n<p>Medical X-rays have wavelengths near 0.1 nm \u2014 about the spacing between atoms in a solid. A wave can only resolve detail on the scale of its own wavelength, which is exactly why visible light (550 nm) cannot image an atom but X-rays can.<\/p>\n\n<p>The same principle drives electron microscopes, which push the wavelength smaller still.<\/p>\n\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/photo-1582789829727-39505f4e6f94.avif\"\n\n       alt=\"Water ripples showing the wavelength between consecutive wavefronts\"\n\n       loading=\"lazy\"\n\n       style=\"width:100%;height:auto;border-radius:4px;\" \/>\n\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">The wavelength of a water wave is the distance between two consecutive rings.<\/figcaption>\n\n<\/figure>\n\n<h2>Common Misconceptions About Wavelength<\/h2>\n\n<h3>&#8220;Wavelength Is How Tall the Wave Is&#8221;<\/h3>\n\n<p>It is not. Height is <strong>amplitude<\/strong>, measured from the rest position to a crest, and it relates to energy and loudness. Wavelength is a horizontal distance along the direction of travel.<\/p>\n\n<p>Two waves can share a wavelength exactly while one is ten times taller than the other.<\/p>\n\n<h3>&#8220;A Wave Keeps the Same Wavelength Everywhere&#8221;<\/h3>\n\n<p>This is the big one. When a wave crosses into a new medium, its <strong>frequency stays the same<\/strong> but its speed changes \u2014 and since \u03bb = v \/ f, the wavelength must change with it.<\/p>\n\n<p>Light entering water slows from 3.00 \u00d7 10<sup>8<\/sup> m\/s to about 2.25 \u00d7 10<sup>8<\/sup> m\/s, so its wavelength shrinks by the same factor. The frequency cannot change, because the source is still shaking the field at the same rate. Our article on <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/wave-speed-different-media\/\">wave speed in different media<\/a> follows this through in detail.<\/p>\n\n<h3>&#8220;Longer Wavelength Means a Faster Wave&#8221;<\/h3>\n\n<p>Within a single medium, no. The speed is set by the medium&#8217;s properties, not by the wave.<\/p>\n\n<p>A 20 Hz note and a 20 kHz note both travel through air at 343 m\/s. Long wavelength simply means the crests are spread out, not that they arrive sooner.<\/p>\n\n<h3>&#8220;Colour Changes When Light Enters Glass&#8221;<\/h3>\n\n<p>Look through a window at a red bus. It is still red. Since colour is set by frequency \u2014 which is unchanged \u2014 the light keeps its colour even though its wavelength inside the glass is genuinely shorter.<\/p>\n\n<p>This is the practical reason physicists usually quote a &#8220;wavelength of light&#8221; as its vacuum value unless a medium is stated.<\/p>\n\n<h2>How Wavelength Relates to Energy, Refraction and Matter Waves<\/h2>\n\n<p>Wavelength is the hinge between wave behaviour and quantum behaviour, which is why it turns up in so many other formulae.<\/p>\n\n<h3>Photon Energy<\/h3>\n\n<p>Shorter wavelength means a more energetic photon, with h the Planck constant:<\/p>\n\n<div class=\"pf-formula\">E = h c \/ \u03bb<\/div>\n\n<p>This is why ultraviolet burns skin and radio waves do not, despite both being light. The <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/modern-physics\/photon-energy-formula\/\">photon energy formula<\/a> covers the calculation properly.<\/p>\n\n<h3>Refraction<\/h3>\n\n<p>In a medium of refractive index n, the wavelength contracts while frequency holds:<\/p>\n\n<div class=\"pf-formula\">\u03bb (in medium) = \u03bb (in vacuum) \/ n<\/div>\n\n<p>Because different wavelengths bend by slightly different amounts, white light fans out into a spectrum \u2014 the mechanism behind a prism and a rainbow alike.<\/p>\n\n<h3>Matter Waves<\/h3>\n\n<p>Every moving particle has a wavelength too, given by the de Broglie relation with m its mass and v its speed:<\/p>\n\n<div class=\"pf-formula\">\u03bb = h \/ (m v)<\/div>\n\n<p>For everyday objects the result is absurdly small and unmeasurable. For an electron it is comparable to atomic spacing, which is why electrons diffract \u2014 see <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/modern-physics\/de-broglie-wavelength\/\">the de Broglie wavelength<\/a> for the full treatment.<\/p>\n\n<h3>Wave Number<\/h3>\n\n<p>In more advanced work you will meet the wave number k, which counts radians of phase per metre:<\/p>\n\n<div class=\"pf-formula\">k = 2\u03c0 \/ \u03bb<\/div>\n\n<h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A tuning fork sounds concert A at 440 Hz. Sound travels through air at 343 m\/s. What is the wavelength?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Use the wavelength formula, \u03bb = v \/ f.<\/p>\n<p>Step 2: Substitute with units, \u03bb = 343 m\/s \u00f7 440 Hz.<\/p>\n<p>Step 3: Solve, \u03bb = 0.7795 m.<\/p>\n<p><strong>Answer: \u03bb = 0.78 m (2 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">An FM station broadcasts at 98.5 MHz. Taking the speed of radio waves as 3.00 \u00d7 10^8 m\/s, find the wavelength.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Convert the frequency to hertz, f = 98.5 MHz = 9.85 \u00d7 10<sup>7<\/sup> Hz.<\/p>\n<p>Step 2: Apply \u03bb = c \/ f = (3.00 \u00d7 10<sup>8<\/sup> m\/s) \u00f7 (9.85 \u00d7 10<sup>7<\/sup> Hz).<\/p>\n<p>Step 3: Solve, \u03bb = 3.045 m.<\/p>\n<p><strong>Answer: \u03bb = 3.04 m (3 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">Green light has a wavelength of 550 nm in a vacuum. What is its frequency?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Rearrange to f = c \/ \u03bb.<\/p>\n<p>Step 2: Convert the wavelength to metres, 550 nm = 550 \u00d7 10<sup>\u22129<\/sup> m = 5.50 \u00d7 10<sup>\u22127<\/sup> m.<\/p>\n<p>Step 3: Solve, f = (3.00 \u00d7 10<sup>8<\/sup> m\/s) \u00f7 (5.50 \u00d7 10<sup>\u22127<\/sup> m) = 5.45 \u00d7 10<sup>14<\/sup> Hz.<\/p>\n<p><strong>Answer: f = 5.45 \u00d7 10<sup>14<\/sup> Hz<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">You watch a harbour wall and count 12 complete waves passing in 20 s. The crests are 1.5 m apart. How fast are the waves moving?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Frequency is cycles per second, f = 12 \u00f7 20 s = 0.60 Hz.<\/p>\n<p>Step 2: Use the rearranged formula, v = f \u03bb = 0.60 Hz \u00d7 1.5 m.<\/p>\n<p>Step 3: Solve, v = 0.90 m\/s.<\/p>\n<p><strong>Answer: v = 0.90 m\/s<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">Light of vacuum wavelength 600 nm enters glass of refractive index n = 1.50. Find its speed, frequency and wavelength inside the glass.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Speed in the medium is v = c \/ n = (3.00 \u00d7 10<sup>8<\/sup> m\/s) \u00f7 1.50 = 2.00 \u00d7 10<sup>8<\/sup> m\/s.<\/p>\n<p>Step 2: Frequency is set by the source and does not change, f = c \/ \u03bb = (3.00 \u00d7 10<sup>8<\/sup>) \u00f7 (6.00 \u00d7 10<sup>\u22127<\/sup>) = 5.00 \u00d7 10<sup>14<\/sup> Hz.<\/p>\n<p>Step 3: Wavelength in glass, \u03bb = v \/ f = (2.00 \u00d7 10<sup>8<\/sup>) \u00f7 (5.00 \u00d7 10<sup>14<\/sup>) = 4.00 \u00d7 10<sup>\u22127<\/sup> m. Equivalently \u03bb = 600 nm \u00f7 1.50.<\/p>\n<p><strong>Answer: v = 2.00 \u00d7 10<sup>8<\/sup> m\/s, f = 5.00 \u00d7 10<sup>14<\/sup> Hz (unchanged), \u03bb = 400 nm<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A guitar string 0.650 m long vibrates in its fundamental mode at 196 Hz. Find the wavelength of the standing wave and the wave speed along the string.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: In the fundamental there is a node at each fixed end and one antinode between, so the string holds half a wavelength: \u03bb = 2L.<\/p>\n<p>Step 2: Substitute, \u03bb = 2 \u00d7 0.650 m = 1.30 m.<\/p>\n<p>Step 3: Apply v = f \u03bb = 196 Hz \u00d7 1.30 m = 254.8 m\/s.<\/p>\n<p><strong>Answer: \u03bb = 1.30 m and v = 255 m\/s (3 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">Blue light has a wavelength of 450 nm. Calculate the energy of one photon, in joules and in electronvolts. Take h = 6.626 \u00d7 10^-34 J s.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Use E = h c \/ \u03bb.<\/p>\n<p>Step 2: Substitute with the wavelength in metres, E = (6.626 \u00d7 10<sup>\u221234<\/sup> J s \u00d7 3.00 \u00d7 10<sup>8<\/sup> m\/s) \u00f7 (4.50 \u00d7 10<sup>\u22127<\/sup> m).<\/p>\n<p>Step 3: Solve, E = 4.414 \u00d7 10<sup>\u221219<\/sup> J. Convert by dividing by 1.602 \u00d7 10<sup>\u221219<\/sup> J\/eV, giving 2.755 eV.<\/p>\n<p><strong>Answer: E = 4.41 \u00d7 10<sup>\u221219<\/sup> J, or 2.76 eV (3 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 8<\/div><div class=\"pf-problem-question\">An electron travels at 2.50 \u00d7 10^6 m\/s. Find its de Broglie wavelength. Take the electron mass as 9.11 \u00d7 10^-31 kg.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Use the de Broglie relation, \u03bb = h \/ (m v).<\/p>\n<p>Step 2: Find the momentum first, m v = 9.11 \u00d7 10<sup>\u221231<\/sup> kg \u00d7 2.50 \u00d7 10<sup>6<\/sup> m\/s = 2.278 \u00d7 10<sup>\u221224<\/sup> kg m\/s.<\/p>\n<p>Step 3: Divide, \u03bb = (6.626 \u00d7 10<sup>\u221234<\/sup> J s) \u00f7 (2.278 \u00d7 10<sup>\u221224<\/sup> kg m\/s) = 2.909 \u00d7 10<sup>\u221210<\/sup> m.<\/p>\n<p><strong>Answer: \u03bb = 2.91 \u00d7 10<sup>\u221210<\/sup> m, or 0.291 nm \u2014 comparable to atomic spacing, which is why electrons diffract through crystals.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is wavelength in simple terms?<\/summary><div class=\"pf-faq-item-answer\">\n\nWavelength is the distance between one wave and the next identical point on the following wave, such as crest to crest. Think of ripples on a pond: the gap between two neighbouring rings is the wavelength. It is written as the Greek letter lambda (\u03bb) and measured in metres.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the SI unit of wavelength?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe SI unit of wavelength is the metre (m), because wavelength is a distance. Convenient sub-units are used in practice: nanometres (nm) for visible light, micrometres (\u03bcm) for infrared, centimetres for microwaves, and metres or kilometres for radio waves. Always convert to metres before using \u03bb = v \/ f.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>How do you calculate wavelength from frequency?<\/summary><div class=\"pf-faq-item-answer\">\n\nDivide the wave speed by the frequency, using \u03bb = v \/ f. For sound in air use v = 343 m\/s at 20 \u00b0C; for light in a vacuum use c = 3.00 \u00d7 10<sup>8<\/sup> m\/s. Convert the frequency to hertz first, so a 2.4 GHz signal becomes 2.4 \u00d7 10<sup>9<\/sup> Hz, giving \u03bb = 0.125 m.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does wavelength change when a wave enters a new medium?<\/summary><div class=\"pf-faq-item-answer\">\n\nYes. Frequency is fixed by the source and does not change, but the speed depends on the medium, so wavelength changes in proportion. Light entering glass of refractive index 1.5 has its wavelength cut to two-thirds of the vacuum value, while its frequency \u2014 and therefore its colour \u2014 stays exactly the same.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the difference between wavelength and amplitude?<\/summary><div class=\"pf-faq-item-answer\">\n\nWavelength is a horizontal distance along the wave&#8217;s direction of travel, measured crest to crest. Amplitude is the vertical displacement from the rest position to a crest. Wavelength determines pitch or colour; amplitude determines loudness or brightness. Two waves can share the same wavelength while having completely different amplitudes.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the wavelength range of visible light?<\/summary><div class=\"pf-faq-item-answer\">\n\nVisible light runs from roughly 380 nm at the violet end to about 700 nm at the red end. Wavelengths just below 380 nm are ultraviolet and just above 700 nm are infrared, and neither is detectable by the human eye. Green sits near the middle of the band at about 550 nm.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Are wavelength and frequency inversely proportional?<\/summary><div class=\"pf-faq-item-answer\">\n\nYes, provided the wave speed stays constant. Since \u03bb = v \/ f, doubling the frequency halves the wavelength when v is fixed. This holds for all electromagnetic waves in a vacuum, where v is always c. If the wave changes medium, v changes too and the simple inverse relationship no longer applies across the boundary.\n\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>Wavelength is the distance between two matching points on a wave, such as crest to crest. This guide covers the \u03bb = v\/f formula, SI units, five real-world examples and eight worked problems.<\/p>\n","protected":false},"author":1,"featured_media":781,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"class_list":["post-780","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-waves"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/780","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=780"}],"version-history":[{"count":7,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/780\/revisions"}],"predecessor-version":[{"id":1469,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/780\/revisions\/1469"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/781"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=780"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=780"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=780"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}