Intensity & Spreading Lab
H2 Physics · Waves — one source power P, four geometries, four spreading laws
I vs r — all four geometries, relative to the value at 1 m
spherehemisphereripplebeam
ⓘ more
Normalised to I(1 m) so all four overlay. Hemisphere is dashed — same 1/r² shape as the sphere, but its absolute I is ×2 at every r. Flip to log–log to see the slopes.
A vs r — amplitude, relative to 1 m (A ∝ √I)
ⓘ more
Same overlay for amplitude: A ∝ 1/r (sphere, hemisphere), 1/√r (ripple), constant (beam) — log–log slopes −1, −½, 0.
Try first: press r ×2 — watch I divide by 4.
Geometry — the core control
I = P / 4πr²
A ∝ 1/r
energy spread over a sphere of area 4πr²
Valid for: point source · uniform emission · no absorption
Controls
At the detector
- I (W/m²)
- 6.37e-3
- A(r) / A(1 m)
- 0.200
- √( I(r) / I(1 m) )
- 0.200
- P_rx = I × A_det
- 1.59e-3 W
Middle rows always agree: A ∝ √I.
Power through the current wavefront
- wavefront radius
- 0.2 m
- area 4πr²
- 0.5 m²
- I × area
- 2.000 W
- source P
- 2.000 W
Equal at every radius — spread thinner, never lost.
What this confirms
- ›“Intensity, I” — I is rate of energy per unit area: every I readout here is literally P divided by the wavefront it crosses.
- ›“Geometry decides the law” — the toggle proves each case: sphere 1/r², hemisphere 1/r² at double strength, ripple 1/r, beam constant — log–log slopes −2, −1, 0.
- ›“The intensity chain from SHM” — A ∝ √I: the A readout and √(I/I₁) stay identical no matter what you drag.
- ›“Receiver questions” — the area ×2 button doubles power collected while I sits still: detector size changes P_rx, never intensity.
- ›“Boundary/dilution: energy NOT lost” — I × wavefront area equals P exactly as the shell expands: dimmer only because the same power is spread thinner.
- ›“Proportionality → ratio, always” — the doubling buttons answer with pure ratios (÷4, ÷2, ×2) — no absolute numbers needed, exactly how exam questions want it.