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Radiation: Explanation, Emissive and Absorptive Power, Emissivity, Black Body

Radiation: Explanation, Emissive and Absorptive Power, Emissivity, Black Body

The Sun is one of the most important sources of energy for the Earth. This energy is exploited by solar panels to generate electricity. These are generally installed to get maximum solar energy. We generally use a term called intensity to measure this energy which we will discuss hereafter. You would have noticed that the solar panels are installed at an angle. They are not just put flat (horizontal) on the ground or on the roof. Why is this done? This is done so as to extract the highest intensity of light throughout a day cycle. Let’s discuss intensity in this article.

Table of contents

  • Intensity of radiation
  • Radiation falling perpendicular to a surface
  • Radiation falling at an angle on a surface
  • Practice problems
  • FAQs

Intensity of radiation

Intensity of radiation can be understood as the radiation energy flux over an area under consideration. So it is the amount of radiation energy falling on a unit area perpendicular to it over a unit time interval.

Radiation falling perpendicular to a surface

The energy crossing per unit area perpendicular to the direction of propagation per unit time is known as Intensity of light(𝐼).

The unit of intensity is W/m2

Please enter alt text

Where n= number of photons incident on the area, 𝐴 per second

λ=  wavelength of light

Momentum of photon,

Where λ= wavelength, Speed of light ,  frequency

Planck constant

Energy contained in one photon,

Substituting the values of planck constant and speed of light,

Energy incident per second,

Therefore, intensity 

Radiation falling at an angle on a surface

Where n= number of photons incident on the area A per sec

λ=  wavelength of light

In this case we have to consider an area which is perpendicular to the direction of propagation. So, we have to take the projection of area which is perpendicular to the direction of propagation and here it is A cos θ.

Intensity of light, 

If intensity has to be increased without changing the wavelength, the number of photons incident per second must increase.

Properties of photon

  • The rest mass of photons is zero.
  • Photons do not carry any charge.
  • The number of photons may not be conserved. They can be destroyed or created due to radiation.
  • Photons travel with the speed of light.
  • Energy of a photon, 𝐸=ℎ𝑣 (where h is Planck constant and 𝑣 is frequency).
  • Though a photon doesn’t have any mass, its momentum is given by,  

Practice problems

Q. Estimate how many visible light photons a 120 W light bulb emit per second. Efficiency of the bulb can be assumed to be about 5%. (λ =500 nm h=6.63×10−34 𝐽𝑠).

A. Energy emitted per second is,

If n is the total number of photons emitted per second,

Q. Monochromatic light is incident on a surface area 4 cm2 . If the intensity of light is 150 mW/m2 and the wavelength is 3000 Å, then find the rate at which photons will be striking the target.

A. Given,

Area of the surface, A = 4 cm2

Intensity of incident light, I = 150 mW/m2

Total energy incident on the surface in one second,

Energy of the incident photon,

Number of photons,

Q. A parallel beam of monochromatic light having wavelength 663 nm is incident at an angle of 60o on a totally reflecting surface having area 1 m2. What will be the intensity of light if the number of photons falling on the surface per second is 1019?

A. Intensity of light,

Q. At a particular time of day, the earth's surface receives 1.4× 103 W/m2 sunlight from the sun. Assuming the light to have monochromatic wavelength of 500 nm and the average distance between the sun and the earth to be 1.5× 1011 m, calculate the number of photons in each m3 near the earth’s surface at the same instant.


Number of photons received per unit area of the earth's surface,

Distance covered by photons in

Thus photons contained in will fall on each  of the earth’s surface.

Therefore by unitary method, photons in  volume, 


Q. What are photons?

A. These are packets of energy that constitute the light.

Q. Do photons exert any force on the surface they are falling upon? If yes, how can we find that force?

A. Yes, the photons falling on a surface exert force and this force can be found by calculating the rate of change of momentum of photons. The force acting per unit area is known as radiation pressure.

Q. Do photons get affected by electric fields and magnetic fields?

A. No. Since they do not carry any charge, they are unaffected by any electric and magnetic fields .

Q. What kind of collision occurs between the photons of light falling on a perfectly reflecting surface?

A. Elastic collision occurs between the photons and the surface for a perfectly reflecting surface.

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