\(\displaystyle \frac{L}{F}\) must have dimensions of area \([m^{2}]\), and \(\displaystyle \frac{L}{4\pi d^{2}}\) must give \([W\,m^{-2}]\).
Use SI prefixes (k, M, G, etc.) to keep numbers manageable.
\[
A = 4\pi d^{2}\;\text{m}^{2}.
\]
\[
I = \frac{L}{4\pi}\;\text{W sr}^{-1}.
\]
\[
F = \frac{I}{d^{2}} = \frac{L}{4\pi d^{2}}\;\text{W m}^{-2}.
\]
This is the inverse‑square law for radiant flux.
| Syllabus Block | Connection to the Inverse‑Square Law |
|---|---|
| 1 – Physical quantities & units | Definition of \(L\), \(F\), \(d\); unit‑check and SI prefixes. |
| 3 – Energy & power (3.1‑3.2) | Luminosity is a power; flux is power per unit area – direct application of energy conservation. |
| 5 – Work, energy and power (5.1‑5.2) | Efficiency and energy loss (e.g., interstellar extinction) modify the observed flux. |
| 7 – Oscillations & waves (7.1‑7.5) | For any wave, intensity ∝ amplitude²; therefore amplitude falls as \(1/d\) while intensity follows the \(1/d^{2}\) law. |
| 13 – Gravitational fields | Standard‑candle distances help map galaxy scales, providing data for gravitational‑field calculations. |
| 14 – Thermal physics | Black‑body law \(L = 4\pi R^{2}\sigma T^{4}\) gives intrinsic stellar luminosities; combined with the inverse‑square law to find distances. |
| 15 – Waves | Same \(1/d^{2}\) dependence appears for sound, water and electromagnetic waves. |
| 18 – Electric fields | Electric field of a point charge \(E = kQ/4\pi\varepsilon_{0}d^{2}\) mirrors the mathematical form of the flux law. |
| 22 – Quantum physics | Photon energy \(E = h\nu\) together with flux \(F\) gives photon arrival rates, relevant for detector design. |
| 23 – Nuclear physics | Supernovae release energy via nuclear reactions; their calibrated peak luminosity makes them excellent standard candles. |
| 25 – Astronomy & cosmology | Standard candles underpin the distance ladder, Hubble’s law and age‑of‑the‑Universe determinations. |
\[
d = \sqrt{\frac{L}{4\pi F}}.
\]
\[
L = 4\pi d^{2}F.
\]
\[
\frac{F{1}}{F{2}} = \left(\frac{d{2}}{d{1}}\right)^{2}.
\]
If the extinction is \(A\) magnitudes, the flux is diminished by a factor \(10^{-0.4A}\):
\[
F{\text{obs}} = F{\text{true}}\;10^{-0.4A}\quad\Longrightarrow\quad
F{\text{true}} = F{\text{obs}}\;10^{0.4A}.
\]
Neglecting this leads to an underestimate of distance because the calculated \(d\) uses a too‑small flux.
Example: \(A = 0.3\) mag → \(10^{0.4A}\approx1.30\); the true flux is 30 % larger, so the true distance is \(\sqrt{1.30}\approx1.14\) times the value obtained without extinction correction.
\[
I \propto A^{2},
\]
where \(A\) is the wave amplitude.
Consequently, amplitude falls as \(A \propto 1/d\) while intensity follows the \(1/d^{2}\) law.
Given
\[
d = \sqrt{\frac{L}{4\pi F}}.
\]
\[
d = \sqrt{\frac{1.0\times10^{43}}{4\pi\,(2.5\times10^{-13})}}
= \sqrt{\frac{1.0\times10^{43}}{3.14\times10^{-12}}}.
\]
\[
d \approx \sqrt{3.18\times10^{54}}
\approx 5.6\times10^{27}\ \text{m}.
\]
\[
d \approx \frac{5.6\times10^{27}}{9.46\times10^{15}}
\approx 5.9\times10^{11}\ \text{ly}
\approx 180\ \text{Mpc}.
\]
| Object | Typical absolute luminosity \(L\) (W) | Wavelength range | Key syllabus links |
|---|---|---|---|
| Cepheid variable | \(\sim10^{30}\)–\(10^{31}\) | Visible | Period–luminosity relation (14.2); black‑body temperature (14.1) |
| RR Lyrae star | \(\sim10^{28}\)–\(10^{29}\) | Visible | Useful for globular‑cluster distances (25.2) |
| Type Ia supernova | \(\sim10^{43}\) | Optical/UV | Standardised peak luminosity (23.3); distance ladder (25.1) |
| Tip of the Red Giant Branch (TRGB) | \(\sim10^{31}\) | Near‑infrared | Stellar evolution (14.3); colour‑magnitude diagrams (25.3) |
A Cepheid variable has \(L = 5.0\times10^{30}\ \text{W}\). Its observed flux is \(F = 1.2\times10^{-12}\ \text{W m}^{-2}\).
Calculate the distance \(d\) (show all steps, include a unit‑check).
Two identical Type Ia supernovae are observed.
Supernova A: \(F_{A}=4.0\times10^{-13}\ \text{W m}^{-2}\).
Supernova B: \(F_{B}=1.0\times10^{-13}\ \text{W m}^{-2}\).
Find the ratio \(d{B}/d{A}\).
Explain why neglecting extinction would cause the inverse‑square law to underestimate the true distance. Include a brief quantitative example using an extinction of \(A=0.3\) mag.
Sketch a point source emitting isotropically. Draw a sphere of radius \(d\) centred on the source. Indicate a small detector element \(dA\) on the sphere’s surface measuring flux \(F\). Label the quantities \(L\), \(F\), \(d\), the total area \(A=4\pi d^{2}\) and the solid angle element \(d\Omega = dA/d^{2}\). This visualises the derivation in Section 2 and highlights the relationship \(F = I/d^{2} = L/(4\pi d^{2})\).
Your generous donation helps us continue providing free Cambridge IGCSE & A-Level resources, past papers, syllabus notes, revision questions, and high-quality online tutoring to students across Kenya.