Ultraviolet rays

Hold a hundred-rupee note under the blue-purple light of a shopkeeper's little lamp and hidden numbers and threads suddenly glow — not because the shopkeeper's lamp is magic, but because it is quietly firing radiation your eyes cannot see at all. That lamp emits ultraviolet (UV) rays, the band of the electromagnetic spectrum sitting just beyond the violet edge of visible light, invisible to the human eye yet energetic enough to make certain inks and minerals fluoresce, to kill bacteria in a water-purifier, and — in far larger doses from the Sun — to tan and burn your skin. UV radiation is one stop on the electromagnetic spectrum's journey from radio waves to gamma rays, a journey this chapter surveys after establishing that light, radio, X-rays and everything between are the same phenomenon: mutually perpendicular electric and magnetic fields travelling together at the same universal speed $c$ . What changes from one band to the next is only frequency and wavelength, yet that single change in scale is enough to turn a harmless radio signal into a sunscreen-worthy hazard. UV sits at the point in the spectrum where "just light, but a bit shorter" starts to mean "chemically and biologically active." This article stays inside the same modest toolkit as the rest of the electromagnetic-waves survey — no calculus, no Maxwell's equations, just $c = \nu\lambda$ and a healthy respect for orders of magnitude. We will pin down where UV sits between violet light and X-rays, see how it is produced (very hot bodies, and purpose-built discharge lamps), work two board-and-competitive-level numerical examples building on wavelengths you already trust, and connect the physics to the everyday facts every student is expected to know: the ozone layer's protective role, sterilisation, fluorescence detection, and why sitting behind a glass window is not the same as standing in direct sunlight.

Key takeaways

UV is the band just beyond violet light. Wavelength roughly $400\ \text{nm}$ down to about $1\ \text{nm}$ ; frequency roughly $7\times10^{14}$ to $3\times10^{17}\ \text{Hz}$ — shorter wavelength and higher frequency than any colour the eye can see.
It travels at the same universal speed as every electromagnetic wave. $c = 1/\sqrt{\mu_0\varepsilon_0} \approx 3\times10^8\ \text{m/s}$ in vacuum, exactly as for radio waves, visible light, or X-rays — only $\nu$ and $\lambda$ differ between bands.
Production needs a hot or energetic source. Very hot bodies (the Sun's surface, roughly $6000\ \text{K}$ ) and specially built electric discharge lamps (mercury-vapour lamps) both radiate strongly in the UV.
The ozone layer is Earth's UV shield. Stratospheric ozone absorbs most — not all — of the Sun's UV before it reaches the ground; ozone depletion is a concern precisely because it raises UV exposure at the surface.
UV does real chemistry that visible light cannot. It is energetic enough to disinfect water and sterilise instruments (germicidal UV-C), make certain substances fluoresce (currency verification, mineral ID), and damage skin and DNA on overexposure.
Ordinary glass blocks most UV. This is why sunlight through a window feels warm but tans you far less than standing directly outdoors.
Key idea: UV is the electromagnetic band from about 400 nm down to 1 nm, travelling at the same speed c = 1/√(μ₀ε₀) as every other band, produced by very hot sources and discharge lamps, and energetic enough — through c = νλ giving frequencies up to 3×10^17 Hz — to sterilise, cause fluorescence, and damage tissue, which is exactly why the ozone layer's partial absorption of it matters.

The core insight

Ultraviolet rays are the electromagnetic band lying between violet visible light and X-rays, roughly $400\ \text{nm}$ down to $1\ \text{nm}$ in wavelength, travelling at the same speed $c=1/\sqrt{\mu_0\varepsilon_0}\approx3\times10^8\ \text{m/s}$ as every other band but at a frequency high enough (up to about $3\times10^{17}\ \text{Hz}$ , via $c=\nu\lambda$ ) to sterilise, fluoresce and damage living tissue — effects the ozone layer partly shields us from.

Where it fits in the syllabus

Ultraviolet rays are the fifth band in the electromagnetic-spectrum survey of the chapter Electromagnetic Waves, sitting between visible light and X-rays. The prescribed CBSE 2025-26 scope asks only for "elementary facts about production and uses" — the approximate frequency/wavelength range, how UV is produced (hot bodies, discharge lamps), and its principal effects and applications (ozone-layer absorption, sterilisation, fluorescence, biological hazard). Nothing about Maxwell's equations, displacement current, or the energy carried by the wave (Poynting vector, radiation pressure) is examinable — those sit above the board ceiling and are quoted, never derived, anywhere in this chapter. Photon-level reasoning ( $E=h\nu$ ) belongs formally to the later Dual Nature of Radiation and Matter chapter, but a qualitative mention here — "higher frequency means more energetic photons" — is standard board practice and is used sparingly below.

The physical picture

Every electromagnetic wave, whatever its band, is built from the same two ingredients: an oscillating electric field $\vec E$ and an oscillating magnetic field $\vec B$ , perpendicular to each other and to the direction the wave travels, both oscillating in step (qualitative picture only — no field equations are needed at this level). What distinguishes ultraviolet light from a radio wave or an X-ray is nothing structural — only how fast the fields oscillate, and correspondingly how short the wave is. The one number that ties frequency and wavelength together, for any electromagnetic wave in vacuum, is the wave's speed:

c = \nu\lambda

where: $c$ is the speed of light in vacuum, $\nu$ is the frequency in hertz, and $\lambda$ is the wavelength in metres. This relation is the everyday working tool of the whole chapter — given either $\nu$ or $\lambda$ for a wave, the other follows immediately, and it is exactly this relation that lets a physicist look at a measured wavelength and say "that's ultraviolet" rather than "that's violet light" or "that's an X-ray." The speed itself is not something this chapter derives — it is a quoted result of Maxwell's full electromagnetic theory (out of scope here), fixed entirely by two constants of the vacuum:

c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}

where: $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space (from magnetism) and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/\text{N}\,\cdot\,\text{m}^2$ is the permittivity of free space (from electrostatics) — the same two constants a student has already met in Coulomb's law and Ampere's law. The physical reading is striking: the speed of light is not an independent fact about optics at all, it is built entirely out of electric and magnetic constants, which is historically exactly how Maxwell first realised that light itself is an electromagnetic wave. Substituting the values gives $c \approx 3\times10^8\ \text{m/s}$ , and — crucially for this article — that same value of $c$ applies to ultraviolet rays, visible light, radio waves and X-rays alike. Nothing in the electromagnetic spectrum travels faster or slower than another band in vacuum; only $\nu$ and $\lambda$ change, trading off against each other so that their product stays fixed at $c$ . Ultraviolet rays occupy the stretch of the spectrum with wavelength roughly $400\ \text{nm}$ (where visible violet light ends) down to about $1\ \text{nm}$ (where soft X-rays begin), corresponding through $c=\nu\lambda$ to frequencies roughly $7\times10^{14}\ \text{Hz}$ to $3\times10^{17}\ \text{Hz}$ . That is a wide band — nearly three orders of magnitude in frequency — which is why UV is loosely divided in broader usage into UV-A (nearest to visible, least energetic), UV-B, and UV-C (nearest to X-rays, most energetic and most biologically damaging); the CBSE syllabus does not require this sub-division in detail, only the overall picture that "shorter wavelength UV is more energetic and more hazardous."

Figure

The electromagnetic spectrum drawn as a horizontal bar from radio waves on the left through microwaves infrared visible light ultraviolet X-rays to gamma rays on the right, with wavelength decreasing and frequency increasing left to right, and the ultraviolet segment highlighted between the violet edge of visible light and the start of X-rays

Derivation AFrom the wavelength range to the frequency range of UV rays

The goal of this derivation is to take the wavelength boundaries of the ultraviolet band — roughly $400\ \text{nm}$ at the violet end and about $1\ \text{nm}$ at the X-ray end — and, using only the quoted speed of light and the relation $c=\nu\lambda$ , produce the corresponding frequency boundaries. This is exactly the reasoning a student must reproduce whenever asked "what is the approximate frequency range of ultraviolet radiation," and it is the same tool used later to classify any given wavelength or frequency into its correct spectral band.

Step 1. Start from the two known constants for this chapter: the speed of light $c = 1/\sqrt{\mu_0\varepsilon_0} \approx 3\times10^8\ \text{m/s}$ , quoted rather than derived, and the defining relation $c=\nu\lambda$ , which rearranges to $\nu = c/\lambda$ .

Step 2. Convert the longer wavelength boundary, $\lambda_1 = 400\ \text{nm} = 4\times10^{-7}\ \text{m}$ , into a frequency. $\nu_1 = \dfrac{c}{\lambda_1} = \dfrac{3\times10^8}{4\times10^{-7}} = 7.5\times10^{14}\ \text{Hz}$ . This is the frequency at which ultraviolet begins, immediately above violet visible light.

Step 3. Convert the shorter wavelength boundary, $\lambda_2 = 1\ \text{nm} = 1\times10^{-9}\ \text{m}$ , into a frequency. $\nu_2 = \dfrac{c}{\lambda_2} = \dfrac{3\times10^8}{1\times10^{-9}} = 3\times10^{17}\ \text{Hz}$ . This is the frequency at which ultraviolet ends and soft X-rays begin.

Step 4. Read off the result. As wavelength shrinks from $400\ \text{nm}$ to $1\ \text{nm}$ — a factor of $400$ — frequency grows from $7.5\times10^{14}\ \text{Hz}$ to $3\times10^{17}\ \text{Hz}$ , also a factor of $400$ , exactly as $c=\nu\lambda$ demands: whatever factor $\lambda$ shrinks by, $\nu$ must grow by the same factor to keep the product fixed at $c$ . This inverse proportionality is the single algebraic fact behind every "identify the band" question in this chapter, and it confirms the frequency range quoted throughout this article, roughly $7\times10^{14}$ to $3\times10^{17}\ \text{Hz}$ .

Worked examples

Board level - identifying the germicidal UV wavelength used in water purifiers

A UV water-purification lamp is quoted as emitting mainly at $\lambda = 254\ \text{nm}$ . Find its frequency and confirm this lies within the ultraviolet band, closer to the X-ray end or the visible end.

Step 1. Convert the wavelength to metres. $\lambda = 254\ \text{nm} = 2.54\times10^{-7}\ \text{m}$ .

Step 2. Apply $\nu = c/\lambda$ . $\nu = \dfrac{3\times10^8}{2.54\times10^{-7}} \approx 1.18\times10^{15}\ \text{Hz}$ .

Step 3. Compare with the UV boundaries found above, $7.5\times10^{14}\ \text{Hz}$ to $3\times10^{17}\ \text{Hz}$ . Since $7.5\times10^{14} < 1.18\times10^{15} < 3\times10^{17}$ , the lamp's output is comfortably inside the ultraviolet band. Because $1.18\times10^{15}\ \text{Hz}$ is much closer to the lower boundary ( $7.5\times10^{14}\ \text{Hz}$ , the visible end) than to the upper boundary ( $3\times10^{17}\ \text{Hz}$ , the X-ray end), this wavelength sits in the lower-frequency part of the UV band, in the region loosely called UV-C — consistent with $254\ \text{nm}$ being the standard germicidal wavelength used in real water-purification and surgical-instrument sterilisers, matching the sterilisation use-case named in the syllabus.

The answer is $\nu \approx 1.18\times10^{15}\ \text{Hz}$ , well inside the ultraviolet band and close to the visible-light boundary.

Competitive level - comparing photon energy of sunburn-causing UV against visible violet light

Sunlight reaching the ground carries both violet visible light ( $\lambda_{violet} = 400\ \text{nm}$ ) and a smaller amount of UV-B radiation that has slipped through the ozone layer ( $\lambda_{UV} = 310\ \text{nm}$ , a wavelength strongly linked to sunburn). Find the frequency of each, and use $E=h\nu$ (quoted only, not derived at this level) to compare how many times more energetic a UV-B photon is than a violet photon.

Step 1. Frequency of the violet light. $\nu_{violet} = \dfrac{c}{\lambda_{violet}} = \dfrac{3\times10^8}{4\times10^{-7}} = 7.5\times10^{14}\ \text{Hz}$ .

Step 2. Frequency of the UV-B radiation. $\nu_{UV} = \dfrac{c}{\lambda_{UV}} = \dfrac{3\times10^8}{3.10\times10^{-7}} \approx 9.68\times10^{14}\ \text{Hz}$ .

Step 3. Take the ratio of frequencies (equivalently, of photon energies, since $E=h\nu$ makes energy directly proportional to frequency for a fixed $h = 6.63\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ ). $\dfrac{\nu_{UV}}{\nu_{violet}} = \dfrac{9.68\times10^{14}}{7.5\times10^{14}} \approx 1.29$ .

Step 4. Interpret the result physically. The UV-B photon is about $1.29$ times, or roughly $29\%$ , more energetic than the violet photon. That may sound modest, but it is precisely the extra energy per photon that pushes UV-B past the threshold needed to damage the chemical bonds in skin cells and DNA — a threshold ordinary visible light, however bright, does not cross. The UV-B photon carries about 1.29 times the energy of a violet visible photon, the extra margin responsible for its capacity to cause sunburn. Timing note: in a competitive exam, do not compute $h\nu$ in joules for a ratio question — the ratio of two frequencies already equals the ratio of two photon energies, since $h$ cancels; converting to joules and back wastes time you do not have.

Common traps

Thinking the eye can detect UV at very low intensity, "just faintly." UV lies entirely beyond the violet edge of the visible range at every intensity; the human eye has no photoreceptor response there at all, regardless of brightness.
Assuming the ozone layer blocks all incoming UV. It absorbs most, not all, of the Sun's UV, and the amount that reaches the ground still causes tanning, sunburn, and long-term skin damage — the layer reduces the hazard, it does not eliminate it.
Confusing UV's biological effects with the deep, ionising damage of X-rays or gamma rays. UV photons are considerably less energetic per photon (the worked example above shows a UV photon barely $29\%$ more energetic than violet light, nowhere near the huge jump to X-ray energies), so UV's hazard is chiefly a skin/surface-tissue effect, not the deep-body ionising damage of X-rays.
Forgetting that ordinary window glass blocks most UV. A common exam and real-life misconception is that sitting near a sunny window gives the same UV exposure as standing directly outdoors — glass absorbs a large part of incident UV, which is why sunburn from indoor window light is rare even on a bright day.
Mixing up which end of the UV band is more energetic. Shorter wavelength always means higher frequency and higher photon energy — so the $1\ \text{nm}$ end (nearer X-rays) is the more hazardous, more penetrating end, not the $400\ \text{nm}$ end nearer visible light.

Common trapUV is not just "slightly stronger violet light" — it is a qualitatively different hazard because of photon energy, and the ozone layer only partly blocks it

The single biggest mark-losing habit is treating ultraviolet as though it were merely a more intense version of violet visible light — as if turning up the brightness of a violet lamp would eventually make it act like UV. It would not. The defining property of UV is its higher frequency, not its intensity: a dazzlingly bright violet lamp still emits photons of the same, comparatively low, violet-light energy no matter how many of them there are, while even a dim UV source emits individual photons energetic enough to trigger chemical and biological effects that visible light — at any brightness — simply cannot. Layered onto this is the ozone-layer misconception: many students write as though the ozone layer is a perfect UV filter, which is not true and not what the syllabus states — it absorbs "most," and the UV that does get through is exactly what causes sunburn, skin ageing, and the increased skin-cancer risk associated with ozone depletion. Get the mechanism right: it is frequency (via $\nu=c/\lambda$ and, qualitatively, $E=h\nu$ ), not intensity, that makes UV special, and the ozone shield is partial, not absolute.

You should be able to…

State the approximate wavelength range (about $400\ \text{nm}$ to $1\ \text{nm}$ ) and frequency range (about $7\times10^{14}$ to $3\times10^{17}\ \text{Hz}$ ) of ultraviolet radiation, and place it correctly between visible light and X-rays on the spectrum.
Use $c=\nu\lambda$ , with $c=1/\sqrt{\mu_0\varepsilon_0}\approx3\times10^8\ \text{m/s}$ , to convert confidently between the wavelength and frequency of any given UV wave, and to check whether a computed value actually falls inside the UV band.
Explain how UV radiation is produced — very hot bodies such as the Sun's surface, and electric discharge lamps such as mercury-vapour lamps.
Explain the ozone layer's partial protective role, and state at least two practical uses of UV (sterilisation/disinfection, fluorescence-based detection such as currency verification).
Explain, qualitatively via the idea that higher frequency means higher photon energy, why UV radiation can damage biological tissue and DNA in a way visible light generally cannot.
Recognise the common traps: UV is not "brighter violet," the ozone layer is not a perfect shield, and ordinary window glass blocks a large part of incident UV.

Ask the tutor: if you would like to see how the UV-A, UV-B and UV-C sub-bands differ in penetration and hazard, or how a mercury-vapour discharge lamp actually generates UV light, ask the AI tutor to walk you through it.

If UV is invisible, how do those currency-checking lamps make ink glow?

The glow is called fluorescence, and it is not the UV light itself becoming visible — your eyes still cannot see UV directly. Certain inks and fibres used in currency notes contain substances that absorb the high-energy UV photons hitting them and then re-emit that absorbed energy as lower-energy visible-light photons, which you do see, typically as a bright blue or green glow. So the lamp is invisible UV going in, and the note's special ink is doing the actual light conversion — a chemical trick, not a change in what your eye can detect.

Why does the ozone layer specifically block UV and not other bands like radio waves or visible light?

Ozone molecules ( $O_3$ ) happen to have internal energy levels that match almost exactly the photon energies carried by UV radiation, so ozone absorbs UV very efficiently by using that energy to break apart or excite the molecule — a resonance-like matching that is a chemistry topic beyond this syllabus, but the physics takeaway is simple: absorption depends on whether a photon's energy matches something the absorbing material can "use." Visible light and radio waves carry energies that ozone (and most of the rest of the atmosphere) largely does not absorb, so they pass through far more freely, which is exactly why we can see the Sun and use radio communication at all.

Is there a formula I need to memorise to convert between UV wavelength and frequency for the exam?

Only one, and you already know it from earlier in this chapter: $c=\nu\lambda$ , rearranged as $\nu=c/\lambda$ or $\lambda=c/\nu$ , with $c\approx3\times10^8\ \text{m/s}$ . There is no separate "UV formula" — the same relation that works for radio waves or visible light works identically for ultraviolet; only the numbers you plug in (roughly nanometre-scale wavelengths, roughly $10^{14}$ to $10^{17}$ Hz frequencies) tell you that you are in the UV band. Memorise the approximate boundary numbers ( $400\ \text{nm}$ and $1\ \text{nm}$ , or $7.5\times10^{14}\ \text{Hz}$ and $3\times10^{17}\ \text{Hz}$ ) so you can quickly check whether a computed answer "looks like" UV.

Does more UV exposure always mean a worse sunburn, or does it depend on which part of the UV band it is?

Both intensity and which part of the band matter, but the wavelength dependence is the more important physics point for this chapter. The lower-frequency end of the UV band nearer visible light (often called UV-A in broader usage) is the least energetic per photon and the least damaging in short exposures, while the higher-frequency end nearer X-rays (UV-C) carries the most energetic photons and would be the most dangerous — except that UV-C is almost entirely absorbed by the atmosphere before reaching the ground, so it rarely reaches you outdoors. Most sunburn is caused by the intermediate band, UV-B, which is energetic enough to damage skin and is only partially absorbed by ozone. The CBSE syllabus does not require the UV-A/B/C sub-division in detail, but the underlying idea — shorter wavelength UV is more energetic and, where it reaches you, more hazardous — is exactly the physics behind $\nu=c/\lambda$ and the qualitative link to $E=h\nu$ .