Major Formulas needed to understand the Topic Radiation are listed in an organized manner here. Take the help of the Radiation Formulas List and know the logic behind the concepts easily. Try to apply the Formulae in Radiation Cheat Sheet during your homework and arrive at the solutions quickly. Radiation Formulae covers everything like Spectral emissive power, Emissive power, Thermal radiation, Reflection, absorption and transmission coefficients, etc. For any other help on related concepts, you can have a quick look at our **Physics Formulas** existing.

**1. Radiation**

Process of transfer of energy by electromagnetic waves.

**2. Thermal radiation**

Electromagnetic waves corresponding to infra-red region.

**3. Reflection, absorption and transmission coefficients**

r = \(\frac{Q_{r}}{Q}\), a = \(\frac{Q_{a}}{Q}\) and t = \(\frac{Q_{t}}{Q}\)

r + a + t = 1

**4. Emissive power**

Total amount of radiations emitted per second per unit area.

E = Q/A.t

**5. Spectral emissive power**

Ratio of amount of radiations emitted in a given range of wavelength λ to (λ + dλ) per second per unit area to the wavelength spread i.e.,

e_{λ} = \(\frac{d Q_{\lambda}}{d \lambda}\)

and

e = \(\int_{0}^{\infty}\)e_{λ}dλ

**6. Emissivity**

Ratio of emissive power of a given surface to emisive power of a black body.

**7. Spectral absorptive power**

In a given range of wavelength, d-f. is ratio of amount of radiations absorbed to amount of radiations incident.

**8. Perfectly black body**

Which absorbs all incident radiation of all wavelengths.

a_{λ} = 1

and r = t = 0

Ferri’s ideal Black body:

**9. Kirchoff’s law**

\(\frac{\mathrm{e}_{\lambda}}{\mathrm{a}_{\lambda}}\) = constant = Emissive power of a black body.

**10. Stefan’s law**

σ is Stefan’s constant = 5.67 × 10^{-8} W/m^{2} – K^{4}

Net loss of energy per second per unit area = σ (T^{4} – T_{0}^{4})

For a black surface of area A the net rate of loss of heat

–\(\frac{d Q}{d t}\) = σA (T^{4} – T_{0}^{4}) J/s

For a surface of emissivity e

–\(\frac{d Q}{d t}\) = σAe (T^{4} – T_{0}^{4}) J/s

Rate of fall of temperature

\(-\frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{\sigma \mathrm{Ae}\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right)}{\mathrm{msJ}}{ }^{\circ} \mathrm{C} / \mathrm{s}\)

**11. Newton’s law of cooling**

Rate of cooling ∝ Excess of temperature

–\(\frac{d Q}{d t}\) = K (θ – θ_{0}), K is cooling constant.

or \(\frac{d \theta}{d t}=\frac{K}{m s}\)(θ – θ_{0}) = k'(θ – θ_{0})

Time to cool from θ_{1} to θ_{2}

t = \(\frac{2.303}{\mathrm{K}^{\prime}}\)log_{10}\(\left(\frac{\theta_{1}-\theta_{0}}{\theta_{2}-\theta_{0}}\right)\)

If variation of 0 from θ_{1} to θ_{2} can be treated as linear then

\(\frac{\Delta \theta}{\Delta t}=\frac{\theta_{1}-\theta_{2}}{t}=K^{\prime}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]\)

Specific heat of liquids

\(\frac{\left(\mathrm{W}+\mathrm{m}_{\mathrm{L}} \mathrm{s}_{\mathrm{L}}\right)\left(\theta_{1}-\theta_{2}\right)}{\mathrm{t}_{1}}=\frac{\left(\mathrm{W}+\mathrm{m}_{\mathrm{w}} \mathrm{s}_{\mathrm{w}}\right)\left(\theta_{1}-\theta_{2}\right)}{\mathrm{t}_{2}}\)

**12. Spectral distribution of radiant energy**

E_{λ} – λ curve has a maximum at λ = λ_{m}.

Area between E_{λ} – λ, curve and λ axis is proportional to T^{4}.

**13. Wien’s displacement law**

or λ_{m} ∝ \(\frac{1}{T}\)

or λ_{m} T = b (Wien’s constant)

b = 2.93 × 10^{-3}m-K

or v_{m} ∝ T

v_{m} = b’T, b’= \(\frac{c}{b}\)

**14. Laws of distribution**

Wien’s law:

E_{λ} dλ = \(\frac{A}{\lambda^{5}}\) e^{-a/λT} dλ.

Rayleigh – Jean’s law:

E_{λ} dλ = \(\frac{8 \pi \mathrm{kT}}{\lambda^{4}}\) dλ

Planck’s law:

E_{λ} dλ = \(\frac{8 \pi \mathrm{hc}}{\lambda^{5}} \frac{1}{\left[\mathrm{e}^{\mathrm{hc} / \lambda \mathrm{kT}}-1\right]}\) dλ

Planck’s law reduces to Wein’s law at short wavelengths and to Rayeigh- Jeans law at long wavelengths.

**15. Solar constant**

S = Radiant energy per minute per cm^{2} by earth at the mean distance from sun

= 1.94 cal/cm^{2}-mt

Solar constant in MKS units

S = 1.38 × 10^{3} W/m^{3}

Temperature of sun

T = \(\left(\frac{\mathrm{S}}{\sigma}\right)^{1 / 4} \cdot\left(\frac{\mathrm{d}}{\mathrm{R}_{\mathrm{s}}}\right)^{1 / 2}\) = 5800 K

Equilibrium temperature of a planet T_{p} is inversely proportional to the square root of its distance upon the sun i.e., T_{p} ∝ \(\frac{1}{\sqrt{d}}\).

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