Microwaves

Stand in front of a running microwave oven and nothing looks different — no glow, no hum you can see, no visible ray crossing the cavity. Yet inside that metal box a wave with a wavelength of about twelve centimetres, oscillating a hundred million times faster than any radio broadcast, is quietly setting every water molecule in your food vibrating until the food is hot. The same band of the electromagnetic spectrum, tuned to a different frequency and focused by a dish instead of scattered inside a box, becomes radar: a narrow beam that bounces off an aircraft and returns a few microseconds later carrying the aircraft's exact distance. Tuned again, it becomes the invisible link that carries a phone call up to a satellite twenty thousand kilometres overhead and back down to a tower on the other side of the planet. Microwaves sit in the electromagnetic spectrum right between radio waves and infrared radiation — short enough in wavelength that a dish you can lift with one hand can focus them into a pencil-thin beam, yet long enough that they pass straight through clouds, glass, and plastic instead of being blocked or scattered the way visible light or X-rays would be. That combination of focusable and penetrating is what makes this band so useful, and it is also the thread that ties together every worked example, trap, and exam question in this article. This section treats microwaves as the second stop in the electromagnetic-spectrum survey of the chapter Electromagnetic Waves — a chapter whose real theoretical spine is displacement current and the transverse nature of $\vec E$ and $\vec B$ fields, but whose examinable content on this particular band is close to entirely descriptive: frequency range, production, and uses, dressed with one reliable numerical skill — converting between frequency, wavelength, and round-trip radar timing using $c = \nu\lambda$ .

Key takeaways

Microwaves occupy roughly $10^9\ \text{Hz}$ to $10^{11}\ \text{Hz}$ in frequency, corresponding to wavelengths from about $1\ \text{mm}$ to $0.1\ \text{m}$ ( $10\ \text{cm}$ ) — squeezed between radio waves (longer $\lambda$ ) and infrared (shorter $\lambda$ ).
They travel at the universal electromagnetic-wave speed $c = 1/\sqrt{\mu_0\varepsilon_0} \approx 3\times10^{8}\ \text{m/s}$ , the same speed as every other band of the spectrum, in vacuum.
Production needs specialised vacuum-tube devices — klystrons and magnetrons — because ordinary circuit oscillators cannot switch fast enough at these frequencies; a microwave oven's magnetron is the everyday example.
Their short wavelength lets small dish or horn antennas focus them into narrow, directional beams — the property behind radar (RAdio Detection And Ranging), which times a reflected pulse to find distance via $d = ct/2$ .
A microwave oven heats food by choosing a frequency (commonly $2.45\ \text{GHz}$ ) that is efficiently absorbed by water molecules, agitating them throughout the food's volume rather than heating only the surface.
Microwaves also carry satellite links, long-distance telephone trunks, and — at the elementary "this is a use" level — probe the rotational structure of molecules in spectroscopy.
Key idea: microwaves are the electromagnetic band of frequency about $10^9$ to $10^{11}$ Hz whose short wavelength (millimetres to tens of centimetres) lets them be focused into narrow directional beams by small antennas, which is why they carry radar, satellite links, and molecule-agitating heat in an oven, all at the same universal speed $c = 1/\sqrt{\mu_0\varepsilon_0}$ .

The core insight

Microwaves are the electromagnetic-spectrum band of frequency roughly $10^9$ to $10^{11}\ \text{Hz}$ (wavelength about $1\ \text{mm}$ to $10\ \text{cm}$ ), produced by klystrons and magnetrons and prized because their short wavelength lets small antennas focus them into narrow directional beams for radar, satellite links, and dielectric heating.

Where it fits in the syllabus

Electromagnetic Waves builds two theoretical ideas — displacement current (qualitative only, no derivation of $I_d=\varepsilon_0\,d\phi_E/dt$ ) and the transverse, mutually perpendicular nature of $\vec E$ and $\vec B$ travelling at $c=1/\sqrt{\mu_0\varepsilon_0}$ — and then surveys the electromagnetic spectrum band by band. Microwaves are the second band in that survey, right after radio waves. The board expects elementary facts only: approximate frequency/wavelength range, how microwaves are produced (klystrons, magnetrons — as named devices, not as circuit diagrams to analyse), and their principal uses (radar, ovens, satellite and long-distance communication, and molecular spectroscopy as a one-line mention). Detailed engineering of klystrons/magnetrons, Maxwell's equations, and energy-density/Poynting-vector treatments are explicitly out of scope for CBSE 2025-26 — nothing in this section goes beyond that descriptive-plus- $c=\nu\lambda$ level.

The physical picture

Every electromagnetic wave, regardless of which spectrum band it belongs to, is the same underlying object: oscillating electric and magnetic fields, perpendicular to each other and to the direction of travel, sustaining one another the way a changing $\vec E$ field (through the displacement-current idea added to Ampere's law) produces a $\vec B$ field and a changing $\vec B$ field produces an $\vec E$ field. That mutual regeneration is why the wave needs no medium and moves through vacuum at one fixed speed set entirely by two constants of nature:

c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \approx 3\times10^{8}\ \text{m/s}

where: $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/\text{N}\,\cdot\,\text{m}^2$ is the permittivity of free space. The physical reading is that this speed is fixed — it does not depend on frequency, wavelength, or which band of the spectrum you are looking at. What changes from radio waves to microwaves to visible light to gamma rays is only the frequency $\nu$ and, correspondingly, the wavelength $\lambda$ , linked through the same relation that governs every wave:

c = \nu\lambda

where: $\nu$ is frequency in hertz and $\lambda$ is wavelength in metres. Microwaves are simply the slice of this spectrum with $\nu$ roughly between $10^{9}\ \text{Hz}$ and $10^{11}\ \text{Hz}$ , which by the relation above corresponds to $\lambda$ roughly between $0.1\ \text{m}$ (10 cm, at the low-frequency end, bordering radio waves) and $1\ \text{mm}$ (at the high-frequency end, bordering infrared). This wavelength is short enough to matter practically: a wave whose wavelength is comparable to the size of a dish antenna a metre or so across can be focused by that dish into a tight, directional beam, the way a torch reflector focuses light. Radio waves, with wavelengths of metres to kilometres, would need impractically large antennas to be focused nearly as sharply — which is exactly why long-distance point-to-point links (radar, satellite communication) use microwaves rather than ordinary radio frequencies. Producing oscillations at $10^{9}$ to $10^{11}\ \text{Hz}$ is itself a technical challenge: ordinary transistor or circuit oscillators, adequate for radio frequencies up to a few hundred megahertz, cannot switch on and off fast enough at microwave frequencies. Two purpose-built vacuum-tube devices solve this by controlling the motion of a beam of electrons inside an evacuated tube rather than switching current in a conventional circuit: the klystron, which velocity-modulates an electron beam using resonant cavities, and the magnetron, which uses a magnetic field to force electrons into a spiralling path inside a set of resonant cavities, extracting energy from them as microwave radiation at a fixed frequency. A domestic microwave oven's magnetron is the most familiar example a student will actually have seen a specification sticker for — commonly rated to produce microwaves at $2.45\ \text{GHz}$ .

Figure

The electromagnetic spectrum shown as a horizontal strip from long wavelength radio waves on the left through the microwave band to infrared on the right with the microwave frequency range and typical uses marked above the band

The physical picture, continued: focusing and absorption

Two separate physical facts, both traceable to the wavelength being short, explain nearly every use of microwaves listed in the syllabus. First, focusability: a reflecting dish or a horn antenna behaves usefully as a focusing device only when its size is at least a few wavelengths across, so a one-metre dish is a superb focuser for a $10\ \text{cm}$ microwave but nearly useless for a $10\ \text{m}$ radio wave. This is why radar dishes, satellite dishes, and microwave communication towers are all compact compared to the giant antenna arrays needed for long-wave radio broadcasting. Second, selective absorption: different frequencies interact differently with matter, and water molecules happen to have a resonance-like response — their rotational and vibrational motion is efficiently excited — in the microwave range near $2.45\ \text{GHz}$ . A radio-frequency wave would mostly pass through food without transferring much energy to it; a microwave at the oven's chosen frequency is absorbed throughout the volume of any water-containing food, converting wave energy into heat via molecular agitation rather than via surface radiant heating the way a grill works.

Derivation AConverting the microwave band's frequency range into its wavelength range using c = nu lambda

The goal here is small but exam-critical: given only the defining frequency range of the microwave band, $\nu$ between $10^9\ \text{Hz}$ and $10^{11}\ \text{Hz}$ , use the one universal relation linking speed, frequency and wavelength for any electromagnetic wave to recover the corresponding wavelength range, and in doing so build the habit of reflexively converting between $\nu$ and $\lambda$ for any spectrum band — a skill every worked example and quiz question in this chapter's survey ultimately depends on.

Step 1. Start from $c = \nu\lambda$ , valid for every electromagnetic wave in vacuum because all bands of the spectrum share the same speed $c = 1/\sqrt{\mu_0\varepsilon_0}\approx 3\times10^{8}\ \text{m/s}$ . Rearranging for wavelength gives $\lambda = c/\nu$ .

Step 2. Substitute the lower end of the microwave frequency range, $\nu_1 = 10^{9}\ \text{Hz}$ , to find the longer wavelength boundary: $\lambda_1 = \dfrac{3\times10^{8}}{10^{9}} = 3\times10^{-1}\ \text{m} = 30\ \text{cm}$ . (Textbook figures usually round the microwave band's long-wavelength edge to about $0.1$ – $0.3\ \text{m}$ ; the exact boundary is not a sharp physical wall but a conventional cutoff, so both figures are acceptable on an exam as "about $0.1\ \text{m}$ .")

Step 3. Substitute the upper end of the frequency range, $\nu_2 = 10^{11}\ \text{Hz}$ , to find the shorter wavelength boundary: $\lambda_2 = \dfrac{3\times10^{8}}{10^{11}} = 3\times10^{-3}\ \text{m} = 3\ \text{mm}$ , again close to the conventionally quoted $1\ \text{mm}$ boundary once rounding conventions are allowed for.

Step 4. Read the pattern, not just the numbers: because $\lambda = c/\nu$ , a higher frequency always corresponds to a shorter wavelength for the same wave speed $c$ . This inverse relationship is why microwaves (higher $\nu$ than radio waves) have shorter $\lambda$ than radio waves, and why, moving further up the spectrum, infrared has shorter $\lambda$ still. The same single relation, run in reverse, is exactly the tool used in both worked examples below to go from a given transmitter frequency to the operating wavelength.

Worked examples

Board level - radar ranging from pulse timing

A ground radar station emits a microwave pulse of frequency $\nu = 10\ \text{GHz}$ ( $10^{10}\ \text{Hz}$ ). The reflected pulse returns to the station $t = 4\times10^{-5}\ \text{s}$ after it was sent. Find (a) the wavelength of the microwaves used, and (b) the distance from the radar to the reflecting object.

Step 1. Find the wavelength using $c = \nu\lambda$ , so $\lambda = c/\nu$ :

\lambda = \frac{3\times10^{8}\ \text{m/s}}{10\times10^{9}\ \text{Hz}} = 3\times10^{-2}\ \text{m} = 3\ \text{cm}

This value, $3\ \text{cm}$ , sits comfortably inside the microwave band ( $1\ \text{mm}$ to $10\ \text{cm}$ roughly), confirming $10\ \text{GHz}$ is a sensible radar-transmitter frequency.

Step 2. Find the total distance the pulse travelled using $d_{\text{total}} = ct$ , since the pulse moves at the electromagnetic-wave speed $c$ for the entire round trip:

d_{\text{total}} = (3\times10^{8}\ \text{m/s})(4\times10^{-5}\ \text{s}) = 1.2\times10^{4}\ \text{m} = 12\ \text{km}

Step 3. Halve the round-trip distance to get the one-way distance to the object, since the pulse travels out to the object and back over the measured time $t$ :

d = \frac{d_{\text{total}}}{2} = \frac{12\ \text{km}}{2} = \mathbf{6\ km}

The reflecting object is 6 km from the radar station. This round-trip-timing method, $d = ct/2$ , is the working principle behind every microwave radar system, from air-traffic control to a simple speed gun.

Competitive level - two successive radar returns and closing speed

The same $10\ \text{GHz}$ radar station tracks an approaching aircraft with two pulses sent $2\ \text{s}$ apart. The first pulse returns after $t_1 = 4\times10^{-5}\ \text{s}$ (giving the $6\ \text{km}$ distance found above); the second pulse, sent $2\ \text{s}$ later, returns after $t_2 = 3.6\times10^{-5}\ \text{s}$ . Find (a) the aircraft's distance at the second measurement, and (b) its average speed of approach over the $2\ \text{s}$ interval, assuming the aircraft is flying directly toward the radar.

Step 1. Convert the second round-trip time to a distance the same way as before, $d_2 = ct_2/2$ :

d_2 = \frac{(3\times10^{8}\ \text{m/s})(3.6\times10^{-5}\ \text{s})}{2} = \frac{1.08\times10^{4}\ \text{m}}{2} = 5.4\times10^{3}\ \text{m} = 5.4\ \text{km}

Step 2. Find how much closer the aircraft has come between the two measurements. The distance decreased from $d_1 = 6\ \text{km}$ to $d_2 = 5.4\ \text{km}$ , a change of:

\Delta d = d_1 - d_2 = 6\ \text{km} - 5.4\ \text{km} = 0.6\ \text{km} = 600\ \text{m}

Step 3. Divide by the elapsed time between pulses, $\Delta T = 2\ \text{s}$ , to get the average closing speed:

v_{\text{avg}} = \frac{\Delta d}{\Delta T} = \frac{600\ \text{m}}{2\ \text{s}} = \mathbf{300\ m/s}

That is about $1080\ \text{km/h}$ , a physically reasonable cruising-to-approach speed for a jet aircraft, which is a useful sanity check that no arithmetic slip has occurred. Timing note: in a competitive-exam setting, resist the temptation to convert $300\ \text{m/s}$ into some more "impressive-looking" derived unit before checking the question actually asks for one; here the two intermediate distances ( $6\ \text{km}$ , $5.4\ \text{km}$ ) are the values worth double-checking first, since an error carried from either round-trip-time-to-distance step silently corrupts the final speed — always verify $d = ct/2$ arithmetic (a factor of two dropped or doubled is the single most common radar-ranging slip) before computing the speed from a distance difference.

Common traps

Confusing microwaves with X-rays or other "high energy, dangerous" radiation. Microwaves are non-ionising, and at the intensities used in ovens, radar, and satellite links they carry nowhere near the photon energy needed to break chemical bonds or ionise atoms the way X-rays or gamma rays do.
Thinking a microwave oven works by "burning" food from the outside like a grill. It heats by directly agitating water molecules throughout the food's volume via absorption, a bulk-heating mechanism, unlike surface-only radiant heating.
Assuming radar "sees" the object directly the way an eye sees light. Radar measures a time delay between an emitted pulse and its echo and converts that delay to distance using $d = ct/2$ — it is a timing instrument, not an imaging one, at this syllabus level.
Believing microwaves and radio waves are produced the same way. Ordinary circuit oscillators suffice for radio frequencies (up to a few hundred MHz), but microwave frequencies ( $10^9$ to $10^{11}\ \text{Hz}$ ) need klystrons or magnetrons because circuit components cannot switch fast enough.
Forgetting to halve the round-trip distance in a radar problem, or forgetting to double a one-way distance when a question is phrased the other way round — always identify explicitly whether the given time is one-way or round-trip before applying $d = ct$ or $d = ct/2$ .
Mixing up which end of the microwave frequency range gives the longer wavelength. Higher frequency always means shorter wavelength (from $c=\nu\lambda$ with $c$ fixed) — do not assume "higher number" means "longer wave."

Common trapMicrowaves are not a weak, harmless version of visible light on one side or a mild form of X-rays on the other — treat their band, uses, and hazards as their own distinct entry in the spectrum, not an interpolation between neighbours

It is tempting to reason "radio waves are safe, X-rays are dangerous, so microwaves — being in between on some vague scale — must be moderately risky." This reasoning is wrong on the physics: ionising versus non-ionising is not a smooth gradient across the whole spectrum but a sharp qualitative divide (roughly at ultraviolet and above), and microwaves sit firmly on the non-ionising side, alongside radio waves and infrared. A microwave oven is hazardous only because of the sheer power concentrated in a small metal cavity (which is why the door has a mesh shield and interlock), not because microwave photons are individually capable of ionising tissue the way X-ray or gamma-ray photons are.

You should be able to…

State the approximate frequency range ( $10^9$ to $10^{11}\ \text{Hz}$ ) and wavelength range (about $1\ \text{mm}$ to $10\ \text{cm}$ ) of the microwave band, and place it correctly between radio waves and infrared.
Convert between frequency and wavelength for microwaves (or any spectrum band) using $c = \nu\lambda$ , with $c = 1/\sqrt{\mu_0\varepsilon_0} \approx 3\times10^{8}\ \text{m/s}$ .
Name klystrons and magnetrons as the devices used to produce microwaves, and explain in one sentence why ordinary circuit oscillators cannot be used instead.
List at least three uses of microwaves — radar, microwave ovens, satellite/long-distance communication — and, for each, connect the use to either focusability (short wavelength, narrow beam) or selective absorption (heating water molecules).
Use the radar round-trip-time relation $d = ct/2$ to find distance from a measured pulse travel time, and extend it to find an average closing speed from two successive radar measurements.
Explain qualitatively why microwaves' short wavelength lets small antennas focus them into narrow directional beams, unlike the much longer wavelengths of radio waves.

Ask the tutor: if you want more practice converting between microwave frequency and wavelength, or working through multi-pulse radar-ranging and closing-speed problems, ask the AI tutor to set up a few for you.

Why exactly does a microwave frequency around 2.45 GHz heat food but a radio-wave frequency does not?

Different frequencies of electromagnetic wave interact with matter differently, and water molecules respond especially strongly — their rotational motion is efficiently driven — to oscillating fields in the microwave range near 2.45 GHz. A radio-frequency wave oscillates far too slowly to excite this rotational response efficiently, so it passes through food largely unabsorbed and transfers very little energy. The chosen oven frequency is a deliberate engineering choice, not an accident, and it is a nice concrete example of how the same $c=\nu\lambda$ family of waves behaves very differently once matter enters the picture, even though every band travels at the identical speed $c$ in vacuum.

If microwaves and visible light are both electromagnetic waves travelling at the same speed c, why can I see light through a window but a microwave oven needs a metal door with a mesh screen?

Glass is transparent to visible light because the electrons in glass do not absorb photons at visible frequencies efficiently, but it is a poor conductor and does not reflect microwaves either. A microwave oven's door instead needs a conductive metal mesh whose holes are much smaller than the microwave wavelength (a few centimetres); such a mesh behaves like a solid conductor to the microwaves (reflecting them back into the cavity) while still letting visible light, whose wavelength is thousands of times shorter, pass through the same holes so you can watch the food cook.

Do klystrons and magnetrons matter for the exam, or just the names?

At CBSE Class 12 level you are expected to know the names klystron and magnetron as the devices used to produce microwaves, and the one-line reason ordinary circuit oscillators cannot do the job at these frequencies (they cannot switch fast enough). You are not expected to describe their internal construction or derive how they work — that level of detail belongs to an electronics or communications-engineering course, not this physics chapter.