B.1.16—Luminosity and brightness

Brightness–luminosity relation

For isotropic emission without absorption,

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

where $$L$$ is total luminosity and $$d$$ is the source–observer distance.

Use the inverse-square pattern

At fixed luminosity, doubling distance makes apparent brightness one quarter as large. At fixed distance, doubling luminosity doubles apparent brightness.

Rearrange before calculating

$$L=4\pi d^2b$$

so

$$d=\sqrt{\frac{L}{4\pi b}}$$

Keep luminosity in watts, distance in metres and brightness in W m⁻².

Worked example from local Question Bank row 30016

Mars is about $1.5$ times farther from the Sun than Earth. If solar intensity at Earth is $1.36\times10^3\,\mathrm{W\,m^{-2}}$,

$$b_{Mars}=b_{Earth}\left(\frac{d_E}{d_M}\right)^2=(1.36\times10^3)\frac{1}{1.5^2}=6.04\times10^2\,\mathrm{W\,m^{-2}}$$

The same solar luminosity is spread over a sphere with larger radius.

Common trap

The factor is $$d^2$$, not $$d$$. Also distinguish a source’s total emitted power from the power received per square metre.