Physics, asked by PyushSharmaz1528, 10 months ago

Emissive power of black body is times the intensity of radiation

Answers

Answered by divyaman44
3

hey mate here is your answer

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 ϕ" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">ϕϕ 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 dA1anddA2=dA1cosθ" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">dA1anddA2=dA1cosθdA1anddA2=dA1cosθ.

The solid angle subtended by

dA2 = dA2r2

∴The intensity of radiation,

I=dQ1−2dA1cosθ×dA2r2" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">I=dQ1−2dA1cosθ×dA2r2I=dQ1−2dA1cosθ×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.dθ(rsinθdϕ)dA2=r2sinθθ.dθ.dϕ" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">dA2=r.dθ(rsinθdϕ)dA2=r2sinθθ.dθ.dϕdA2=r.dθ(rsinθdϕ)dA2=r2sinθθ.dθ.dϕ …(2).

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

dQ1−2=I.dA1.sinθ.cosθ.dθ.dϕ" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">dQ1−2=I.dA1.sinθ.cosθ.dθ.dϕdQ1−2=I.dA1.sinθ.cosθ.dθ.dϕ

The total radiation through the hemisphere is given by,

Q=IdA1θ=0θ=π2ϕ=0ϕ=2πsinθ.cosθ.dθ.dϕ=2πIdA1θ=0θ=π2sinθ.cosθ.dθ=πIdA1θ=0θ=π22sinθ.cosθ.dθ=πIdA1θ=0θ=π2sin2θ.dθQ=πIdA1Also,Q=EdA1∴EdA1=πIdA1E=πI" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">Q=IdA1θ=0θ=π2ϕ=0ϕ=2πsinθ.cosθ.dθ.dϕ=2πIdA1θ=0θ=π2sinθ.cosθ.dθ=πIdA1θ=0θ=π22sinθ.cosθ.dθ=πIdA1θ=0θ=π2sin2θ.dθQ=πIdA1Also,Q=EdA1∴EdA1=πIdA1E=πIQ=IdA1θ=0θ=π2ϕ=0

it will surely help you.

Answered by Anonymous
0

All objects with a temperature above absolute zero (0 K, -273.15 oC) emit energy in the form of electromagnetic radiation. A blackbody is a theoretical or model body which absorbs all radiation falling on it, reflecting or transmitting none.

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