Skip to main content
Physics LibreTexts

1.14: Relations between Flux, Intensity, Exitance, Irradiance

  • Page ID
    8000
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    In this section I am going to ask, and answer, three questions.

    i. (See figure I.3 )

    alt
    \(\text{FIGURE I.3}\)

    A point source of light has an intensity that varies with direction as \(I(\theta , \phi\)). What is the radiant flux radiated into the hemisphere \(\theta < \pi /2\)? This is easy; we already answered it for a complete sphere in equation 1.6.3. For a hemisphere, the answer is

    \[\phi = \int_0^{2\pi} \int_0^{\pi/2} I(\theta, \phi) \sin \theta d \theta d \phi. \tag{1.14.1} \label{1.14.1}\]

    ii. At a certain point on an extended plane radiating surface, the radiance is \(L(\theta ,\phi\)). What is the emergent exitance \(M\) at that point?

    alt
    \(\text{FIGURE I.4}\)

    Consider an elemental area \(\delta A\) (see figure I.4). The intensity \(I( \theta , \phi\)) radiated in the direction \((\theta ,\phi )\) is the radiance times the projected area \(\cos \theta \ \delta A\). Therefore the radiant power or flux radiated by the element into the hemisphere is

    \[\delta \phi = \int_0^{2\pi} \int_0^{\pi/2} L(\theta, \phi) \cos \theta \sin \theta d\theta d \phi \delta A, \tag{1.14.2} \label{1.14.2}\]

    and therefore the exitance is

    \[M=\int_0^{2\pi} \int_0^{\pi/2} L(\theta, \phi) \cos \theta \sin \theta d\theta d\phi \tag{1.14.3} \label{1.14.3}\]

    iii. A point \(O\) is at the centre of the base of a hollow radiating hemisphere whose radiance in the direction \((\theta , \phi )\) is \(L(\theta , \phi)\). What is the irradiance at that point \(O\) ? (See figure I.5.)

    alt
    \(\text{FIGURE I.5}\)

    Consider an elemental area \(a^2 \ \sin \theta \ \delta \theta \ \delta\phi\) on the inside of the hemisphere at a point where the radiance is \(L(\theta ,\phi )\) (figure I.5). The intensity radiated towards \(O\) is the radiance times the area:

    \[\delta I (\theta, \phi) = L(\theta, \phi) a^2 \sin \theta \delta \theta \delta \phi \tag{1.14.4} \label{1.14.4}\]

    The irradiance at \(O\) from this elemental area is (see equation (1.10.1)

    \[\delta E = \frac{\delta I (\theta,\phi) \cos \theta}{a^2} = L (\theta, \phi) \cos \theta \sin \theta \delta \theta \delta \phi, \tag{1.14.5} \label{1.14.5}\]

    and so the irradiance at O from the entire hemisphere is

    \[E = \int_0^{2\pi} \int_0^{\pi/2} L(\theta, \phi) \cos \theta \sin \theta \delta \theta \delta \phi. \tag{1.14.6} \label{1.14.6}\]

    The same would apply for any shape of inverted bowl - or even an infinite plane radiating surface (see figure I.6.)

    alt
    \(\text{FIGURE I.6}\)


    This page titled 1.14: Relations between Flux, Intensity, Exitance, Irradiance is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

    • Was this article helpful?