Kirchoff's Law of Thermal Radiation

If Emissive power =$ E$
Absorptive power = $A$
The emissive power of a black body is the emissive power for unit absorptive power(per unit absorptive power) (Black body absorbs all radiations)  = $\dfrac{E}{A}$

$E\propto e \times A$
roughness increases area and black colour increases $e$.

Coefficient of absorption is defined as the ration of amount of radiation a body absorbs to the amount of radiation that is incident on the body.
Incident radiation = $p$
Rejected radiation = $q$
Hence absorbed radiation must be = $p - q$
From the definition of absorptive power it follows that the power = $\dfrac{p-q}{p}$

In thermodynamics, Kirchhoff's law of thermal radiation refers to wavelength-specific radiative emission and absorption by a material body in thermodynamic equilibrium, including radiative exchange equilibrium.

Coefficient of absorption is defined as the ration of amount of radiation a body absorbs to the amount of radiation that is incident on the body.
Incident radiation = $X$
Absorbed radiation  = $Y$
Hence absorptive power = $\dfrac{Y}{X}$

the temperature diffrence should be equal for equilibrium
$(t-20)=20-(-20)$;

The emissivity of Platinum black is 0.98. According to Kirchhoff's law of radiation, absorptivity is equal to emissivity at a given temperature. Hence, absorptivity = 0.98. 

Good reflectors are bad emitters of radiation.
Good reflectors, are called so because they absorb very little radiation and reflect the rest.
The ratio of absorptive and emissive power is constant for a given temperature and wavelength, so a bad absorber of radiation is also a bad emitter.
Hence the assertion and reason are true and the reason is the correct explanation.

Given,

With the help of kirchhoff's law  we can say that $\varepsilon =\alpha $
${\varepsilon}= emissivity $
$\alpha=absorptivity $
so kirchhoff's law state that total emissivity of body is equal to total absorptivity.
so ${\lambda} _{1},{\lambda} _{2}, {\lambda} _{3}, {\lambda} _{4}$ will be absorbed.  

In steady state, heat received by any small element = heat given by it
So, heat is partially radiated and no net heat is absorbed.
We give an argument for this.
suppose that the heat is being given out by the element.
In this case, the element's temperature will decrease.
But, by definition , at steady state, all the temperatures remain constant.
So, this will violate the steady state condition.

According to Kirchhoff law, good absorbers are good emitters. Since black spot is good absorbers so it is also a good emitter & will brighter than plate.

An ideal black body is a good emitter and good absorber. Initially it
will absorb all the energy and will appear dark. As soon as it attains the temperature equal to that of the furnace, it will start emitting better than the rest of furnace and hence appears the brightest.

When a non black body is placed inside a hollow enclosure, the total radiation from the body is the sum of what it would emit in the open ( with $\epsilon<1$ ) and the part of the incident radiation from the walls reflected by it. The two add up to a black body radiation.

$P=\sigma e AT^4$

From the krichoff's law we know that absorptivity is equal to emissivity,so the body which emitt more wil absorbe more nor reflect more. So good reflector are poor emittor.
Now, by taking there ratio it validate krichoff's law because at any fixed or constant temperature both are equal.

Since the radiation is continuously falling on the black body, the quantity radiation absorbed per second will increase.