Theory of Intelligence
12TH•Physics•Space And Environment•Radiant Flux Intensity

21.2 Radiant Flux Intensity

Inverse-square law $F = L/4\\pi r^{2}$ and Example 21.2.

21.2 Radiant Flux Intensity

Radiant flux intensity is the amount of electromagnetic energy per unit area per unit time emitted from a star that reaches the Earth's surface. It is a small fraction of the total value of luminosity. In other words, radiant flux intensity is the luminosity per unit area measured on the surface of Earth. Radiant flux intensity from a star at a distance dd is expressed as:

Radiant flux intensity F=L4πr2…(21.2)\text{Radiant flux intensity } F = \frac{L}{4\pi r^{2}} \quad \ldots (21.2)

where LL is the luminosity and 4πr24\pi r^{2} is the surface area of star. As the luminosity of a star remains constant, therefore, radiant flux intensity follows inverse square law with distance as demonstrated in Fig. 21.1. According to this law, when the distance from centre of a star is doubled (from rr to 2r2r), intensity of the radiant flux reduces to one-fourth of its original value. Similarly, increasing the distance to three times its initial length (3r3r), decreases the flux by a factor of nine, and this pattern continues accordingly.

Tidbit!

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If Earth were twice as far from the Sun, the sunlight reaching it would be only one-fourth as intense. Such low energy might have made the evolution of life impossible.

(Fig. 21.1: Schematic illustration of the inverse-square law for radiant flux intensity)

Brain Teaser

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Two stars appear equally bright in the night sky. However, one is 1010 times farther away than the other. Which star is more luminous, and by how much?

Example 21.2

The luminosity of a star is 4.50×1026 W4.50 \times 10^{26}\,\mathrm{W}. It is located at a distance of 1.40×1017 m1.40 \times 10^{17}\,\mathrm{m} from a planet. Calculate the radiant flux intensity of the star at the surface of that planet.

Solution

Luminosity of the star L=4.50×1026 WL = 4.50 \times 10^{26}\,\mathrm{W} Distance between Earth and the star r=1.40×1017 mr = 1.40 \times 10^{17}\,\mathrm{m} Radiant flux intensity F=?F = ? Using the formula; F=L/4πr2F = L / 4\pi r^{2} =4.50×1026 W/(4×3.14(1.40×1017 m)2)= 4.50 \times 10^{26}\,\mathrm{W} / (4 \times 3.14(1.40 \times 10^{17}\,\mathrm{m})^{2}) F=1.83×10−9 W m−2F = 1.83 \times 10^{-9}\,\mathrm{W\,m^{-2}}

Do you know?

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The idea of standard candles dates back to the 1910s, when Henrietta Swan Leavitt discovered that Cepheid variable stars pulse in a predictable way. Her work laid the foundation for measuring cosmic distances and the scale of the universe. Without standard candles, we would not even know how big the universe is.

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