recall and use the inverse square law for radiant flux intensity F in terms of the luminosity L of the source F = L / (4πd 2)

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Standard Candles

Standard Candles

In astrophysics a standard candle is an astronomical object whose intrinsic luminosity L is known (or can be determined reliably). By comparing the known luminosity with the observed radiant flux F we can determine the distance d to the object.

Key Objective

Recall and use the inverse‑square law for radiant flux intensity

\$F = \frac{L}{4\pi d^{2}}\$

to calculate distances or fluxes for standard candles.

Derivation of the Inverse‑Square Law

  1. Consider a point source emitting isotropically with total power (luminosity) L (units: W).
  2. At a distance d the emitted energy is spread over the surface of a sphere of radius d.
  3. The surface area of a sphere is 4πd².
  4. Radiant flux F is defined as power per unit area, therefore

\$F = \frac{L}{\text{area}} = \frac{L}{4\pi d^{2}}\$

This relationship holds provided there is no absorption or scattering between source and observer.

Using the Law – Typical Problems

  • Given L and measured F, find the distance d.
  • Given d and measured F, determine the luminosity L (useful for classifying a new standard candle).
  • Compare two objects: if they have the same L, the ratio of their fluxes equals the inverse square of the ratio of their distances.

Worked Example

A Type Ia supernova has an absolute luminosity of \$L = 1.0\times10^{43}\,\text{W}\$. An observer measures a flux \$F = 2.5\times10^{-13}\,\text{W m}^{-2}\$. Find the distance to the supernova.

  1. Re‑arrange the inverse‑square law:
  2. \$d = \sqrt{\frac{L}{4\pi F}}\$

  3. Insert the numbers:
  4. \$d = \sqrt{\frac{1.0\times10^{43}}{4\pi (2.5\times10^{-13})}}\$

  5. Calculate:
  6. \$\$d \approx \sqrt{\frac{1.0\times10^{43}}{3.14\times10^{-12}}}

    \approx \sqrt{3.18\times10^{54}}

    \approx 5.6\times10^{27}\,\text{m}\$\$

  7. Convert to light‑years (1 ly ≈ \$9.46\times10^{15}\,\$m):
  8. \$d \approx \frac{5.6\times10^{27}}{9.46\times10^{15}} \approx 5.9\times10^{11}\,\text{ly}\$

Common Standard Candles

ObjectTypical Absolute Luminosity \$L\$ (W)Typical Wavelength RangeNotes
Cepheid \cdot ariable\$\sim10^{30}\$\$10^{31}\$VisiblePeriod–luminosity relation provides \$L\$.
RR Lyrae Star\$\sim10^{28}\$\$10^{29}\$VisibleNearly constant \$L\$, useful for globular clusters.
Type Ia Supernova\$\sim10^{43}\$Optical/UVStandardised peak luminosity after light‑curve correction.
Tip of the Red Giant Branch (TRGB)\$\sim10^{31}\$Near‑infraredSharp cutoff in luminosity of red giants.

Practice Questions

  1. A Cepheid variable has a known luminosity \$L = 5.0\times10^{30}\,\$W. If its observed flux is \$F = 1.2\times10^{-12}\,\$W m\$^{-2}\$, calculate its distance.
  2. Two identical Type Ia supernovae are observed. Supernova A has a flux \$FA = 4.0\times10^{-13}\,\$W m\$^{-2}\$, and Supernova B has \$FB = 1.0\times10^{-13}\,\$W m\$^{-2}\$. What is the ratio of their distances \$dB/dA\$?
  3. Explain why interstellar extinction would cause the inverse‑square law to underestimate the true distance if not corrected.

Suggested diagram: Sketch of a point source emitting isotropically, with a spherical surface of radius \$d\$ intersected by a detector measuring flux \$F\$.