Special distribution of radiations emitted from a surface

• Fig. shows a small black surface of area dA (emitter) emitting radiation in different directions.
• A black body radiation collector through which the radiation pass is located at an anglular position characterized by zenith angle θ towards the surface normal and angle $\phi $ of a spherical coordinate system.
• Further the collector subtends a solid angle dw when viewed from a point on the emitter.

• Let us now consider radiation from the elementary area dA1 at the centre of a sphere as shown in fig.
• Suppose this radiation is absorbed by a second elemental area dA2, a portion of the hemispherical surface.
• The projected area of dA1 on a plane perpendicular to the line joining $dA1 and dA2=dA1cos\theta$.

The solid angle subtended by

dA2 = dA2r2

∴The intensity of radiation,

$I = dQ1 - 2dA1cos\theta × dA2r2$ …(1)

where dQ1-2 is the rate of radiation heat transfer from dA1 to dA2.

It is evident from the fig. that,

$dA2 = r.d\theta(r sin\theta d\phi) \\ dA2 = r2sin\theta\theta.d\theta. dϕ$ …(2).

From eqn.(1) and (2), we obtain,

$dQ1-2 = I.dA1.sin \theta.cos \theta.d\theta.d\phi$

The total radiation through the hemisphere is given by,

$Q=I dA1\theta=0\theta= \pi2 \phi=0\phi=2\pi sin \theta.cos\theta.d\theta.d\phi \\ =2\pi I dA1 \theta=0\theta=\pi2 sin \theta.cos\theta.d\theta \\ =\pi I dA1 \theta=0\theta=\pi22sin \theta.cos\theta.d\theta \\ =\pi I dA1 \theta=0\theta=\pi2sin 2\theta.d\theta \\ Q=\pi I dA1 \\ Also, Q=E dA1 \\ ∴ E dA1=\pi I dA1 \\ E=\pi I$

i.e. ,The total emissive power of a diffuse surface is equal to $\pi$ times its intensity of radiation.