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Radiation | Thermal Radiation

Radiation :- The process of the transfer of heat from one place to another place without heating the intervening medium is called radiation.

For example, if you heat an object and place it in vacuum (empty space with no air or other matter), it will still lose heat—even though there’s nothing around it to carry the heat away. This happens because heat cannot escape through conduction (which needs solids/liquids) or convection (which needs fluids like air or water). Instead, the heat leaves as invisible light waves (infrared radiation). Radiation does NOT need any material (like air or water) to transfer heat. It is by radiation that the heat from the Sun reaches the Earth.

Properties of radiation

Heat radiation exhibits many of the same behaviors as visible light, indicating that it is a form of electromagnetic wave. Below are its key properties :

1. Heat radiation travels in straight lines : Like light, thermal radiation moves in straight lines. If an object obstructs the path, a shadow is formed at the detector.

Example: If you place a metal plate in front of a heat source and an infrared detector behind it, a shadow appears on the detector screen.

2. It can travel through vacuum : Unlike conduction and convection, radiation does not require any medium to propagate.

Example: The Sun’s heat reaches Earth through the vacuum of space via radiation.

3. It obeys the inverse square law : The intensity of radiation decreases with the square of the distance from the source.

$\displaystyle I\propto \frac{1}{{{r}^{2}}}$

Example: If you double the distance from a heat source, the radiation intensity becomes one-fourth.

4. It shows reflection, refraction, and interference : Heat radiation can be reflected, refracted, and can interfere just like visible light.

Example: Infrared rays can be reflected off a shiny metal surface or refracted through a special prism.

5. It can be polarized : Just like light, thermal radiation can be polarized using materials like a Nicol prism.

Example: When polarized thermal radiation passes through a second prism (analyzer) at a particular angle, its intensity can change or vanish.

6. Heat radiation lies in the infrared region : Thermal radiation consists mainly of infrared rays, which have wavelengths longer than visible light.

Example: Infrared cameras detect heat radiation from objects even in complete darkness.

7. Radiation causes heating of a medium : When radiation passes through a medium, part of it is absorbed depending on the medium’s absorptive power, causing a slight rise in temperature.

Example: Glass may warm up when exposed to infrared radiation due to partial absorption.

8. Special prisms are used for infrared spectrum : Common prisms like glass or quartz absorb infrared rays and are not suitable. Instead, special materials like : KCl prism, Rock salt prism, Fluorspar prism are used to disperse heat radiation into a spectrum.

9. Radiation intensity is measured using a bolometer : A bolometer is a sensitive instrument used to measure the intensity of heat radiation.

Example: Bolometers are used in infrared astronomy and in measuring thermal radiation from heated bodies.

10. Radiation exerts pressure : When thermal radiation is incident on a surface, it exerts a small force known as radiation pressure.

Example: This effect is extremely small but measurable and is considered in spacecraft propulsion using solar sails.

Conclusion

All these properties demonstrate that thermal radiation behaves very similarly to visible light and is a part of the electromagnetic wave spectrum, differing mainly in wavelength. Heat radiation (infrared rays) has longer wavelengths than visible light, and due to this, it is invisible to the naked eye but can be detected using specialized instruments.

Basic Fundamental definitions about Thermal Radiation

Following terms describe how energy is stored, absorbed, and emitted by materials in the form of thermal radiation :

(1). Energy Density (u) : It refers to the amount of heat energy present per unit volume at any point in the space surrounding the source. It can be defined in two ways :

(A). Spectral Energy Density (u_λ) : Spectral energy density is the radiant energy per unit volume per unit wavelength interval, i.e., between (λ – ½) and (λ + ½) at a specific wavelength (λ). It describes how energy is distributed across different wavelengths (or frequencies) in a radiation field. It depends upon the value of λ and the temperature of the heat source.

SI UNIT : J/m³ Å or J/m⁴

(B). Total Energy Density (u) : The radiation energy of whole wavelength (0 to ∞) present in unit volume at any point in space is defined as total energy density.

$\displaystyle u=\int\limits_{0}^{\infty }{{{u}_{\lambda }}}d\lambda$

SI UNIT : J/m³

(2). Absorptive Power or Absorptive Coefficient (a) : Absorptive power (also called the absorptive coefficient) refers to the ability of a surface or material to absorb a portion of the incident thermal radiation or energy falling on it. Absorptive power is discussed under two main types :

(A). Spectral Absorptive Power or Monochromatic Absorptive Coefficient (a_λ) : Spectral absorptive power is defined as the ratio of the amount of radiation absorbed by a surface at a particular wavelength , to the amount of radiation incident on it at the same wavelength (. It quantifies how efficiently a surface absorbs radiation of a specific wavelength.

$\displaystyle {{a}_{\lambda }}=\frac{{{Q}_{{{a}_{\lambda }}}}}{{{Q}_{\lambda }}}$

(B). Total Absorptive Power or Absorptive Coefficient () : Total absorptive power (denoted by ) is defined as the ratio of the total amount of radiation absorbed by a surface to the total amount of radiation incident upon it, considering all wavelengths of the radiation.

$\displaystyle a=\frac{{{Q}_{a}}}{Q}$

It represents how efficiently a surface absorbs the entire spectrum of incident thermal radiation, not just at a specific wavelength. Also at a given wavelength :

$\displaystyle a=\int\limits_{0}^{\infty }{{{a}_{\lambda }}}d\lambda$

a_λand both are unitless. For an ideal black body, and $= 1 . Because an ideal black body is a perfect absorber — it neither reflects nor transmits any part of the radiation.$

(3). Emissive power or Emittance (E) : Emissive power (also called Emittance) of a surface is defined as the amount of thermal radiation energy emitted per unit area per unit time by the surface.

$\displaystyle E=\frac{Energy\text{ e}mitted}{Area\times time}$

It measures how efficiently a surface gives off thermal radiation.

Types of Emissive Power :

(A). Spectral Emissive Power (e_λ) : It is the amount of radiation energy emitted per unit area, per unit time, and per unit wavelength interval at a particular wavelength . It tells how energy is distributed among different wavelengths.

$\displaystyle {{e}_{\lambda }}=\frac{dE}{d\lambda }$

For Example a hot body emits more energy in the infrared region than in the visible region — that variation is captured by .

SI UNIT : W/m² Å

(B). Total Emissive Power or Radiant Emittance (E) : It is the total energy emitted per unit area per unit time over all wavelengths.

$\displaystyle E=\int{dE}=\int\limits_{0}^{\infty }{{{e}_{\lambda }}d\lambda }$

SI UNIT : Watt/m²

Surface Behavior Under Incident Radiation: Reflection, Absorption, Transmission and Their Coefficients

When radiation is incident on the surface of a body, the following phenomena occur, dividing the radiation into three parts : (a) Reflection, (b) Absorption, and (c) Transmission.

From conservation of energy,

$\displaystyle Q={{Q}_{r}}+{{Q}_{a}}+{{Q}_{t}}$

$\displaystyle \Rightarrow \frac{Q}{Q}=\frac{{{Q}_{r}}}{Q}+\frac{{{Q}_{a}}}{Q}+\frac{{{Q}_{t}}}{Q}$

$\displaystyle \Rightarrow r+a+t=1$

Here,

r = Reflective Coefficient = Q_r/Q

a = Absorptive Coefficient = Q_a/Q

t = Transmittive Coefficient = Q_t/Q

If :-

(1). r = 1 , a = 0 and t = 0 ⇒ Perfect reflector

(2). a = 1 , r = 0 and t = 0 ⇒ Ideal absorber (Ideal black body)

(3). t = 1 , a = 0 and r = 0 ⇒ Perfect transmitter (diathermanous)

Reflection Power = $\displaystyle \left[ \frac{{{Q}_{r}}}{Q}\times 100 \right]%$

Absorption Power = $\displaystyle \left[ \frac{{{Q}_{a}}}{Q}\times 100 \right]%$

Transmission Power = $\displaystyle \left[ \frac{{{Q}_{t}}}{Q}\times 100 \right]%$

Example 1.

Heat energy is incident on the surface of a material at the rate of $1000 J mi n^{-1}$ . If the coefficient of absorption is $0.8$ and the coefficient of reflection is $0.1$ , then the amount of heat energy transmitted through the material in $5 min u t es$ is

(A) 100 J

(B) 500 J

(D) 900 J

Solution :

Here Q = $1000 J mi n^{-1}$

As $\displaystyle r+a+t=1$ , we get

$\displaystyle 0.1+0.8+t=1$

$\displaystyle \Rightarrow t=1-0.9=0.1$

Now,

$\displaystyle t=\frac{{{Q}_{t}}}{Q}\Rightarrow {{Q}_{t}}=Q\times t$

Hence heat energy transmitted through the material in $5 min u t es,$

$\displaystyle {{Q}_{t}}=\left[ 1000\times 0.1 \right]\times 5=500J$

Correct option is (B).

Example 2.

100 units of energy is incident on a surface. In this 20 units are of wavelength λ₁, 30 units are of wavelength λ₂ and rest 50 units are of other wavelengths. The total 60 units of energy is absorbed by the surface. In these 60 units, 5 units are of λ₁ and 25 units are of λ₂. Find a , $\displaystyle {{a}_{{{\lambda }_{1}}}}$ and $\displaystyle {{a}_{{{\lambda }_{2}}}}$ .

Solution :

Total absorptive power/absorptive coefficient (a),

$\displaystyle a=\frac{{{Q}_{a}}}{Q}=\frac{60}{100}=0.6$

Spectral absorptive power for λ₁,

$\displaystyle {{a}_{{{\lambda }_{1}}}}=\frac{{{Q}_{{{a}_{{{\lambda }_{1}}}}}}}{{{Q}_{\lambda }}}=\frac{5}{20}=0.25$

Spectral absorptive power for λ₂,

$\displaystyle {{a}_{{{\lambda }_{2}}}}=\frac{{{Q}_{{{a}_{{{\lambda }_{2}}}}}}}{{{Q}_{\lambda }}}=\frac{25}{30}=0.83$

Here $\displaystyle {{a}_{{{\lambda }_{2}}}}>{{a}_{{{\lambda }_{1}}}}$ (0.83 > 0.25), this indicates that the surface is good absorber of wavelength λ₂ .