ARCHETYPE: EM-Wave Numericals — E₀/B₀, λf = c & Intensity

The EM-waves chapter carries a compact, predictable family of numericals, and this gym drills all six of its branches: the $\lambda f = c$ conversion (with band placement as the follow-up), the read-the-wave-equation problem (handed an expression like $E = E_0\sin(kx - \omega t)$ , extract everything), the $E_0/B_0 = c$ amplitude conversion, the intensity chain (power → intensity → field amplitudes), the radiation-pressure force, and the photon-count-per-second staple. Six branches, one discipline: every problem is two or three substitutions if you route the givens through the right chain in the right order. Nothing is derived here — the physics belongs to The Nature of Electromagnetic Waves (structure, speed), The Electromagnetic Spectrum (band placement), and Energy, Intensity & Radiation Pressure (all the energy formulas). What this gym adds is the exam mechanics: which form of which formula eats which given, the unit conversions that break most attempts ( $\mu$ m to m, GHz to Hz, mW to W), the audits that catch a dropped $2\pi$ or a peak/rms slip, and the timing — ninety seconds is the budget for most of these, and the worked ladder is calibrated to it. One meta-lesson runs through the module: an EM-wave numerical is usually a chain, not a formula lookup. Power becomes intensity, intensity becomes a field, the field becomes its partner field. Learn the chains as chains and the questions assemble themselves.

Key takeaways

Branch 1: $\lambda f = c$ — convert, then place the answer on the spectrum ladder (the follow-up mark).
Branch 2: given $E = E_0\sin(kx - \omega t)$ : $\lambda = 2\pi/k$ , $f = \omega/2\pi$ , speed check $\omega/k = c$ , travel direction from the sign, $B_0 = E_0/c$ , and $\vec B$ perpendicular to both $\vec E$ and travel.
Branch 3: amplitudes always exchange at $E_0 = cB_0$ — both directions.
Branch 4: the intensity chain — point source: $I = P/4\pi r^2$ ; beam: $I = P/A$ ; then $E_0 = \sqrt{2I/\varepsilon_0 c}$ .
Branch 5: pressure/force — $F = IA/c$ absorbed, $2IA/c$ mirrored (underline the surface word).
Branch 6: photon count — $N$ per second $= P/(hf) = P\lambda/hc$ (survey-level quantum, the JEE-Main staple).
Key idea: every EM-wave numerical is a short chain of substitutions — identify the branch, run its chain in order, audit units and the peak/rms convention.

The core insight

EM-wave numericals reduce to six chains — wavelength-frequency conversion, wave-equation reading ( $k$ and $\omega$ to $\lambda$ , $f$ , direction), the $E_0 = cB_0$ exchange, power-to-intensity-to-amplitude, pressure-force, and photon counting — each two or three substitutions long, with the marks living in unit discipline, the $2\pi$ s of $k$ and $\omega$ , and the absorbed-versus-reflected factor.

Where this lives today

These six chains are working formulas across a surprising span of jobs. Telecom engineers run branch 1 daily (channel wavelengths, antenna sizing); laser-safety officers run branch 4 (beam intensity → exposure limits — the reason pointer classes exist); photonics labs run branch 6 in reverse (photon-counting detectors literally count $P/hf$ ); and spacecraft dynamicists budget branch 5's forces — solar pressure measurably perturbs GPS satellite orbits and had to be modelled before GPS could be accurate. At the exotic end, laser cooling and optical tweezers are branch 5 run at the atomic scale, and LightSail 2 flew the mirror factor as propulsion in 2019. The wave equation you learn to read in branch 2 is the exact format in which electromagnetic simulation software reports its answers.

The physical picture — one wave, six handles

Every question in this gym grabs the same object — the traveling crossed-comb wave — by one of six handles. The exam tells you which handle by what it gives you:

Figure

The six branches and their tells — route the givens, then substitute

The chains, with their traps

Branch 1 — $\lambda f = c$ . One multiplication, three ways to lose it: unit conversion (nm $= 10^{-9}$ m, GHz $= 10^9$ Hz — convert before substituting), the band-placement follow-up (answer in hand, place it with the spectrum's anchors), and the vacuum caveat ( $c$ is the vacuum speed; in glass the wave slows and $\lambda$ shortens while $f$ stays fixed — one honesty line here, the full story in optics).

Branch 2 — reading the wave equation. The given $E_y = E_0\sin(kx - \omega t)$ is a loaded sentence; unpack it in fixed order. The number multiplying $x$ is $k$ (rad/m), so $\lambda = 2\pi/k$ ; the number multiplying $t$ is $\omega$ (rad/s), so $f = \omega/2\pi$ . Audit immediately: $\omega/k$ must equal $3\times10^8$ m/s, or you misread a coefficient — this ten-second check is the branch's seatbelt. Direction: $(kx - \omega t)$ travels toward $+x$ ; $(kx + \omega t)$ toward $-x$ . Then the partner field: magnitude $B_0 = E_0/c$ , direction perpendicular to both $\vec E$ and travel, oriented so the wave's fields and velocity form the right-handed set drilled in Direction Triage — with $\vec E$ along $y$ and travel along $+x$ , $\vec B$ lies along $z$ , in phase: $B_z = B_0\sin(kx-\omega t)$ , same argument, same sign.

Figure

Dissecting the wave expression — every symbol is a question the exam can ask

Branch 3 — the exchange rate. $E_0 = cB_0$ in both directions; the traps are pure decimal handling ( $10^{-7}$ T territory) and the false instinct that the "small" $B$ can be neglected — its energy share is equal (the energy module's fifty-fifty), only its number is small.

Branch 4 — the intensity chain. Two openings: point source → $I = P/4\pi r^2$ ; beam of cross-section A → $I = P/A$ (no $4\pi$ , no $r^2$ — a laser does not dilute). Then $I = \tfrac12\varepsilon_0E_0^2c$ inverted: $E_0 = \sqrt{2I/\varepsilon_0c}$ . Keep the peak/rms convention straight: the $\tfrac12$ belongs with $E_0$ ; using $E_{\text{rms}}$ drops it (the AC chapter's discipline, transplanted).

Branch 5 — force. $F = IA/c$ (absorbed) or $2IA/c$ (mirrored), with $A$ the illuminated area and a cosine if the surface tilts. For a beam hitting a surface bigger than itself, $IA$ is just the beam's power $P$ : $F = P/c$ or $2P/c$ — the cleanest form, worth recognizing.

Branch 6 — photon counting. Survey-level quantum, standard on JEE-Main: energy per photon $E_{\text{ph}} = hf = hc/\lambda$ ; photons per second $N = P/E_{\text{ph}}$ . The numbers teach the lesson: even milliwatts deliver $10^{16}$ photons a second, which is why light looks continuous.

Derivation AThe Power to Force Pipeline Assembled Once

The goal is to chain branches 4, 3 and 5 into the single pipeline that JEE composites actually set — from a source's power all the way to the force on a mirror — so that the composite question becomes substitution rather than navigation. Every link is a cited result; the assembly is the archetype's master pattern.

Step 1. From power to local intensity. Point source at distance $r$ : $I = P/4\pi r^2$ (the swelling sphere); collimated beam of area $A_b$ : $I = P/A_b$ (no dilution). Why the fork matters: choosing the wrong opening is the most common composite error — the word "laser" or "beam" always selects the second road.

Step 2. From intensity to fields, when asked: $E_0 = \sqrt{2I/\varepsilon_0c}$ , then $B_0 = E_0/c$ . Why this order: the intensity formula is written in $E_0$ ; extracting $B_0$ first would mean carrying $c^2$ around for no benefit.

Step 3. From intensity to force on a surface of illuminated area $A$ : $F = IA/c$ absorbed, $2IA/c$ mirrored. If the surface swallows the whole beam, $IA = P$ and the force is $P/c$ or $2P/c$ directly — skip the intensity entirely. Why the shortcut is safe: force cares about total momentum flux, and the beam's total is set by its power, however its area is distributed.

Step 4. Audit the pipeline with magnitudes: fields come out in volts-per-metre (tens for bulbs at metres, ~kV/m for sunlight and laser pointers), forces come out microscopic ( $P/c$ puts 60 W at $2\times10^{-7}$ N). A force of newtons from a lamp, or a field of $10^6$ V/m from a bulb, means a dropped power of ten — the anchor bands are the audit. Verified: 60 W point source at 2.0 m — $I = 60/(4\pi\times4) = 1.19$ W/m², $E_0 = \sqrt{2\times1.19/(8.854\times10^{-12}\times3\times10^8)} = 30$ V/m, $B_0 = 1.0\times10^{-7}$ T; 60 W beam on a mirror — $F = 2P/c = 4.0\times10^{-7}$ N (all recomputed ✓).

Worked examples

Board level — convert and place (CBSE pattern, 2 marks)

An FM station broadcasts at 90.4 MHz. Find the wavelength, and name the spectrum band this belongs to.

Step 1. Convert and substitute: $\lambda = \dfrac{c}{f} = \dfrac{3\times10^8}{90.4\times10^6} = \boxed{3.32\ \mathrm{m}}$ .

Step 2. Placement: metre-scale wavelength → radio band (the spectrum's m-to-km anchor). Both the number and the band name carry marks; the placement sentence is not decoration.

Board level — read the wave equation (CBSE/JEE-Main pattern, 3 marks)

The electric field of a wave in vacuum is $E_y = 60\sin\big(1.0\times10^7\,x - 3.0\times10^{15}\,t\big)$ V/m (SI units). Find (a) the wavelength and frequency, (b) the direction of travel, (c) the amplitude and direction of the magnetic field, and write its expression.

Step 1. Extract: $k = 1.0\times10^7$ rad/m → $\lambda = \dfrac{2\pi}{k} = \boxed{6.28\times10^{-7}\ \mathrm{m}}$ (628 nm — visible, red-orange). $\omega = 3.0\times10^{15}$ rad/s → $f = \dfrac{\omega}{2\pi} = \boxed{4.77\times10^{14}\ \mathrm{Hz}}$ . Audit: $\omega/k = 3.0\times10^{15}/1.0\times10^7 = 3\times10^8$ m/s ✓.

Step 2. The argument $(kx - \omega t)$ → travel along $\boxed{+x}$ .

Step 3. $B_0 = \dfrac{E_0}{c} = \dfrac{60}{3\times10^8} = \boxed{2.0\times10^{-7}\ \mathrm{T}}$ , along $z$ (perpendicular to $\vec E \parallel y$ and to travel $\parallel x$ , right-handed set), in phase: $B_z = 2.0\times10^{-7}\sin\big(1.0\times10^7x - 3.0\times10^{15}t\big)$ T.

Competitive level — the full pipeline (JEE-Main pattern, ~2.5 minutes)

A 60 W lamp radiates uniformly in all directions. At 2.0 m from the lamp, find (a) the intensity, (b) the peak electric and magnetic fields. (c) If instead a 60 W collimated beam falls entirely on a perfect mirror, find the force on the mirror.

Step 1. Point source: $I = \dfrac{60}{4\pi(2.0)^2} = \boxed{1.19\ \mathrm{W/m^2}}$ .

Step 2. Fields: $E_0 = \sqrt{\dfrac{2I}{\varepsilon_0c}} = \sqrt{\dfrac{2\times1.19}{8.854\times10^{-12}\times3\times10^8}} = \boxed{30\ \mathrm{V/m}}$ ; $B_0 = \dfrac{30}{3\times10^8} = \boxed{1.0\times10^{-7}\ \mathrm{T}}$ .

Step 3. Beam on mirror — the shortcut form: $F = \dfrac{2P}{c} = \dfrac{2\times60}{3\times10^8} = \boxed{4.0\times10^{-7}\ \mathrm{N}}$ . Note the two openings in one question: dilution for the lamp (part a), no dilution for the beam (part c) — the fork the derivation flagged. Timing note: ~2.5 minutes. The mark distribution follows the pipeline stations; the classic slips are $4\pi r^2$ with $r$ unsquared, and forgetting the mirror's factor 2. The anchor audit ("tens of V/m from a bulb at metres" ✓, "sub-μN force" ✓) certifies both parts in five seconds.

Competitive level — photons per second (JEE-Main staple, ~90 seconds)

A 10 mW laser emits at 663 nm. How many photons leave it per second? ( $h = 6.63\times10^{-34}$ J·s)

Step 1. Energy per photon: $E_{\text{ph}} = \dfrac{hc}{\lambda} = \dfrac{6.63\times10^{-34}\times3\times10^8}{663\times10^{-9}} = \boxed{3.0\times10^{-19}\ \mathrm{J}}$ (≈ 1.9 eV — a red photon, consistent with the spectrum ladder).

Step 2. Count: $N = \dfrac{P}{E_{\text{ph}}} = \dfrac{10\times10^{-3}}{3.0\times10^{-19}} = \boxed{3.3\times10^{16}\ \text{photons/s}}$ — thirty-three million billion per second from a pocket pointer, which is why light looks like a continuous stream and why the photon graininess took until 1900 to notice. Timing note: ~90 seconds, half of it decimal hygiene (mW → W, nm → m). Setters love $\lambda = 663$ nm precisely because $hc/\lambda$ comes out clean with $h = 6.63\times10^{-34}$ — when the numbers cancel neatly, you are on the intended path.

Exam traps

Convert first, substitute second. nm, μm, GHz, mW — all conversions happen before the formula, in the margin, visibly. Most lost marks in this gym are decimal, not conceptual.

The $2\pi$ pair. $\lambda = 2\pi/k$ , $f = \omega/2\pi$ ; audit with $\omega/k = c$ before answering anything else about a wave equation.

Point source or beam? $4\pi r^2$ for isotropic sources; $P/A_b$ for lasers and beams. The word "beam" bans the inverse square.

Peak vs rms carries a $\sqrt2$ . $I = \tfrac12\varepsilon_0E_0^2c = \varepsilon_0E_{\text{rms}}^2c$ ; the options will contain both answers, $\sqrt2$ apart.

The mirror's factor 2. Underline the surface word; $(1+\rho)$ for partial reflectors.

Anchor audits close every part. Bulb fields: tens of V/m. Sunlight/laser-pointer fields: ~kV/m. Radiation forces: micronewtons and below for household powers. Outside the band → recompute.

Boards and depth. Tagged CBSE, JEE-Main, JEE-Advanced and NEET. CBSE's fair game is branches 1–3 (conversion, wave-equation reading, the exchange rate) — its papers have asked all three as 2–3-markers despite the chapter's qualitative framing. NEET adds branch 6's photon counting (dual-nature overlap) and quick intensity reads. JEE-Main runs every branch and loves the composites; JEE-Advanced chains them into multi-part structures where the energy module's pressure bookkeeping joins in. ISC stops at the qualitative spectrum (no numericals — the reason this gym carries no ICSE tag), and the international boards' spectrum demands are matching/ordering, served by the previous gym. Constants: $c = 3\times10^8$ m/s, $\varepsilon_0 = 8.854\times10^{-12}$ , $h = 6.63\times10^{-34}$ J·s.

Origins · 1 minPyotr Lebedev and how this was discovered

Pyotr Lebedev

1866–1912 · Moscow University

Lebedev measured the gentlest push in classical physics — and the numbers this gym computes are the ones his torsion balance had to feel. Radiation pressure had been on theory's books since Maxwell (sunlight at micropascals; a laboratory arc lamp far less), but every attempt to detect it drowned in artifacts: residual gas heated by the light convects, and unevenly warmed vanes twist by the radiometric effect — both forces orders of magnitude larger than the light's own momentum. Lebedev's 1900 masterpiece was artifact engineering: bulbs pumped to the best vacuum of the day, vane pairs of blackened and mirrored faces (absorbed versus reflected — the factor of two, built into the apparatus), light flipped between faces to cancel slow drifts, and thermal effects measured separately and subtracted. The result confirmed Maxwell's $I/c$ — and the mirrored faces' doubling — to within experimental error, announced at the 1900 Paris congress to real sensation; Nichols and Hull's Dartmouth measurement confirmed it within a few percent three years later. Kelvin reportedly said Lebedev had made him surrender on light pressure. Lebedev built Russia's first true physics research school, resigned from Moscow University in the 1911 Kasso-affair protest along with a third of the faculty, and died months later at 46. Comet tails, solar sails and this gym's micronewtons all cash the cheque he cleared.

History

1873

— Maxwell's treatise prices radiation pressure at I/c; too small for any instrument of the day

1876

— Bartoli re-derives the pressure from thermodynamics alone — theory doubles down

1900

— Lebedev's torsion balance in Moscow detects light pressure on solid bodies, absorbed and mirrored faces distinguished

1903

— Nichols and Hull at Dartmouth confirm to a few percent with an independent design

1910

— Lebedev extends the measurement to gases — harder still

2019

— LightSail 2 raises its orbit on sunlight alone: the gym's F = 2IA/c, flown as propulsion

Going further

The formulas drilled here are earned in three parents: The Nature of Electromagnetic Waves (structure, $c$ , the wave form), The Electromagnetic Spectrum (band placement and anchors) and Energy, Intensity & Radiation Pressure (the whole energy suite). The right-hand direction discipline is Direction Triage's. This gym closes the physics content of the E&M track — what remains are the two capstone techniques, Circuit Solving at Speed and Option Elimination in E&M, which weld the entire track into exam-day reflexes. The medium-speed twist and polarisation numericals await the optics track; the photon's full story opens modern physics (UG deep dive: monochromatic plane waves; momentum and radiation pressure). Provenance: fresh archetype per the HS map; all six chains' worked numbers composed for this article and recomputed (628 nm wave-equation set chosen for the clean ω/k audit; 663 nm photon count for the clean hc/λ).

How do I decide between I = P/4πr² and I = P/A when both a distance and an area appear?

Ask what the source does with its power. An isotropic emitter (bulb, star, antenna treated as a point) spreads power over ever-larger spheres — geometry forces $4\pi r^2$ , and the given distance is your $r$ . A collimated beam (laser, searchlight treated ideally) keeps its power inside a fixed cross-section — the given area is your $A_b$ , and distance is irrelevant (ideally). When a question supplies both — say, a lamp at 2 m illuminating a small mirror of area $A$ — they serve different stations: the distance builds the local intensity, the mirror's area then collects force $2IA/c$ . One quantity per station; the chain never needs both in the same formula.

Why does ω/k have to equal c — and what if the numbers give something else?

Because $\omega/k = (2\pi f)/(2\pi/\lambda) = f\lambda$ , which for any EM wave in vacuum is the wave speed. The audit works as a misread detector: coefficients in exam expressions are chosen so the ratio lands exactly on $3\times10^8$ ; getting $3\times10^7$ or $3\times10^9$ means you swapped $k$ and $\omega$ or dropped a power. If a problem deliberately gives $\omega/k < c$ , it is telling you the wave travels in a medium (glass, water) — then $v = \omega/k$ is the medium speed, $f$ is unchanged from vacuum, and $\lambda$ has shrunk. That twist is optics-track territory, but JEE has used it; the audit is how you notice.

In writing the partner B field, how do I get its direction right rather than guessing z?

Use the one right-hand system from Direction Triage — no new rule needed. The wave's trio must satisfy: $\vec E \times \vec B$ points along the travel direction. Given $\vec E \parallel +y$ and travel $\parallel +x$ : test $\vec B \parallel +z$ : $\hat y \times \hat z = \hat x$ ✓. Had travel been $-x$ with the same $\vec E$ , then $\vec B$ must flip to $-z$ . Two habits: write the trio test explicitly (one line, one mark), and remember the in-phase rule — same sine, same argument, no quarter-cycle shifts (the nature module's warning). Direction plus phase is the full answer; the magnitude was one division.

Is the photon-count branch really fair game if photons belong to the modern-physics chapter?

Yes, at survey level, and it appears on JEE-Main and NEET attached to this chapter's numbers. What you need: energy comes in packets $E_{\text{ph}} = hf = hc/\lambda$ (quoted), so a source of power $P$ emits $N = P/E_{\text{ph}}$ packets per second. What you do not need yet: why packets exist, the photoelectric effect, momentum $h/\lambda$ — the modern-physics track earns those. The crossover fact worth absorbing now: the wave description (this chapter) and the photon count describe the same beam, and for $10^{16}$ photons a second the graininess is invisible — which is precisely why the wave picture works so well.

My field answer was off by exactly √2 from the key — which convention did I break?

You used the rms formula with a peak amplitude or vice versa — the classic. $I = \tfrac12\varepsilon_0E_0^2c$ (peak, with the half) and $I = \varepsilon_0E_{\text{rms}}^2c$ (rms, no half) are the same fact in two currencies, related by $E_0 = \sqrt2E_{\text{rms}}$ . Exam protocol: JEE questions asking for "amplitude" or "peak value" want $E_0$ — run $E_0 = \sqrt{2I/\varepsilon_0c}$ by default and label your answer. If both candidate numbers appear among the options (they will, $\sqrt2$ apart), the question's noun — amplitude versus rms — is the tiebreak. Same discipline, same trap, same fix as the AC chapter's 230-versus-325 V.

If CBSE's EM-waves chapter is officially qualitative, why is this numericals gym tagged CBSE at all?

Because CBSE's "qualitative" applies to the wave theory — transverse nature, characteristics — not to the arithmetic riding on $c = f\lambda$ and the wave expression, which CBSE papers ask routinely (a 2-marker converting a frequency, or extracting $\lambda$ , $f$ and $B_0$ from a given $E(x,t)$ , appears across recent boards). Branches 1–3 are therefore firmly CBSE-fair; that is the tag's meaning. Branches 4–6 (intensity chains, pressure forces, photon counts) escalate toward JEE-Main/NEET territory — a CBSE student short on time should master the first three branches cold and read the rest once. The ISC case is different: its syllabus keeps even the numericals out (ordering only), which is why this gym carries no ICSE tag while its MCQ sibling does.