## Lesson-23 Radiation between two infinite parallel plates and proof of Kirchhoff’s law of Radiation, Radiation Shield

Radiation between two infinite parallel plates and proof of Kirchhoff’s law of Radiation:

At a given temperature, the total emissive power of a any body is equal to its absorptivity multiplied by total emissive power of a perfect black body at that temperature.

Therefore        E = αEb

But the ratio of total emissive power of any body to the total emissive power of a black body at the same temperature is called the emissivity of the body and is numerically equal to absorptivity.

Consider two bodies C and D whose absorptivity are αc and αd as shown in Figure 1.

Considering the energy emitted by the body D.

(1) D emits the energy = Ed                                                                                                 (1)

(2) C absorbs energy = Ed. αc                                                                                               (2)

and reflects energy=Ed (1 - αc)                                                                                        (3)

(3) D absorbs energy = Ed αd (1 - αc)                                                                                   (4)

and reflects energy = Ed (1 - αd) (1 - αc)                                                                         (5)

(4) C absorbs energy  = Ed αc (1 - αc)  (1 - αd)                                                                      (6)

and reflects energy = Ed  (1 - αc)2  (1 - αd)                                                                       (7)

(5) D absorbs energy  = Ed. αd (1 - αc)2  (1 - αd)                                                                    (8)

and reflects energy = Ed  (1 - αc)2  (1 - αd)2                                                                      (9)

and so on upto  times.

Considering the energy emitted by the body C.

(1) C emits the energy = Ec                                                                                                    (10)

(2) D absorbs energy = Ec. αd                                                                                                 (11)

and reflects energy=Ec (1 – αd)                                                                                          (12)

(3) C absorbs energy = Ec αc (1 – αd)                                                                                      (13)

and reflects energy = Ec (1 – αc) (1 – αd)                                                                           (14)

(4) D absorbs energy  = Ec αd (1 – αd)  (1 – αc)                                                                        (15)

and reflects energy = Ec  (1 – αd)2  (1 – αc)                                                                         (16)

(5) C absorbs energy  = Ec. αc (1 – αd)2  (1 – αc)                                                                       (17)

and reflects energy = Ec  (1 – αd)2  (1 – αc)2                                                                        (18)

and so on upto infinite number of  times.

Considering equations (1), (4), (8), (11), (I5),   net energy lost by the body D

= Energy emittedby it – energy absorbedby it

q(dc) =Ed – [Ed αd (1 - αc) + Ed. αd (1 - αc)2  (1 - αd) + ….] – [Ec. αd + Ec αd (1 – αd)  (1 – αc) +….]

Assuming   (1 - αc) (1 – αd) = K

q(dc)  = Ed – Ed αd (1 - αc) [1 + K + K2  + …...] – Ec. αd [1 + K + K2  + ….]

But   1 + K + K2  + …... = (1 –K)-1

q(dc) = Ed – (1 –K)-1 [Ed αd (1 - αc) + Ec. αd ]

i) If originally both bodies are at same temperature

Then     q(dc)  = 0          from equation (18A)

Ed ac = Ec ad

Assuming C as black body

Subscription ‘b’ represents black body.

However, Ed / Eb is the emissivity of body ‘D’ according to definition of emissivity.

εd = αc This is the statement of Kirchoff’s law and hence it is proved.

ii) If both the bodies are at different temperatures

Using equation (19)

According to Stefan-Boltzman Law

Ed = εd σb T14

Ec = εc  σb T24

Substituting these values in equation (19), we get