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Physics Lesson 19.3.4 - Interpretation of Light Pressure

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Welcome to our Physics lesson on Interpretation of Light Pressure, this is the fourth lesson of our suite of physics lessons covering the topic of The Compton Effect and Pressure of Light, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Interpretation of Light Pressure

The phenomenon of light pressure can be explained correctly through both the wave and particle approach (classical and modern theory of light). This fact is a demonstration of the dual nature of light. Hence, we can say that light waves are made up by light particles (photons).

The particle interpretation of light pressure takes into consideration the collision of photons with the surface of objects. In every collision, small forces caused by the incident photons act on the object's surface, the resultant of which, gives the pressure of light on that surface.

The figure below shows a beam of parallel light waves incident to the surface of an object at an angle θ to the normal line.

Physics Tutorials: This image provides visual information for the physics tutorial The Compton Effect and Pressure of Light

Let's denote by I the intensity of the incident light and by A the area of the zone when the light falls at the angle θ to the normal line. Light intensity represents the light energy incident on the unit area in every second. If ΔE is the light energy incident on the area A0 during a very short time interval Δτ, we obtain

∆E = I ∙ A0 ∙ ∆τ

On the other hand, based on the particle definition of light, it is clear that this energy ΔE is transported from the source to the surface by a number of photons ΔN, which are at any instant inside the imaginary cylinder of light where the A0 is the base of cylinder and c · Δτ its height. If we write by n the concentration of photons (i.e. the number of photons in the unit of volume), we have for the number ΔN of photons incident on the given surface during the time interval Δτ:

∆N = n ∙ A0 ∙ c ∙ ∆τ

For simplicity, we will consider only the case of monochromatic light, i.e. when all incident photons have the same frequency. Obviously, the energy of each of them is

Eph = h ∙ f

while the total energy ΔE transported by (all) ΔN photons is

ΔE = (h ∙ f) ∙ ∆N
= (h ∙ f) ∙ n ∙ A0 ∙ c ∙ ∆τ

Comparing the two expressions found for the energy of incident light (photons), we obtain the following relationship between the intensity of light beam and energy of incident photons:

I ∙ A0 ∙ ∆τ = (h ∙ f) ∙ n ∙ A0 ∙ c ∙ ∆τ

or

I = (h ∙ f) ∙ n ∙ c

The quantity (h ∙ f) ∙ n represents the volume density of photons energy and it is denoted by w. Hence, we obtain for the pressure of light

P = I/c ∙ (1 + r) ∙ cos2 θ
= w ∙ (1 + r) ∙ cos2 θ

Example 2

A monochromatic light beam of intensity 8 × 10-2 W/m2 and wavelength equal to 500 nm is incident at 37° to the normal line on a metal surface of reflection coefficient r = 0.9. Calculate:

  1. Pressure of the light beam
  2. The number of incident photons per unit of volume
  3. The number of photons incident on a 0.2 m2 section of the metal surface during 2 seconds

Solution 2

Clues:

I = 8 × 10-2 W/m2
λ = 500 nm = 5 × 10-7 m
θ = 37° (cos 37° = 0.8)
r = 0.9
(c = 3 × 108 m/s)
(h = 6.626 × 10-34 J · s)

  1. P = ?
    Using the equation of light pressure
    P = I/c ∙ (1 + r) ∙ cos2 θ
    we obtain after substitutions:
    P = (8 × 10-2 W/m2 )/(3 × 108 m/s) ∙ (1 + 0.9) ∙ (0.8)2
    = 3.243 × 10-10 Pa
  2. n = ?
    The concentration of photons (the number of photons per unit volume) giving that I = w · c, is
    n = w/h ∙ f
    = I/c ∙ h ∙ f
    = I ∙ λ/h ∙ c2
    = (8 × 10-2 W/m2 ) ∙ (5 × 10-7 m)/(6.626 × 10-34 J ∙ s) ∙ (3 × 108 m/s)2
    = 6.7 × 108 photons/m3
  3. ΔN = ?
    The number of photons incident on the given surface (A0 = 0.2 m2) during the time interval Δτ = 2 s, is
    ∆N = n ∙ A0 ∙ c ∙ ∆τ
    = (6.7 × 108 photons/m3 ) ∙ (0.2 m2 ) ∙ (3 × 108 m/s) ∙ (2 s)
    = 8.04 × 108 photons

You have reached the end of Physics lesson 19.3.4 Interpretation of Light Pressure. There are 5 lessons in this physics tutorial covering The Compton Effect and Pressure of Light, you can access all the lessons from this tutorial below.

More The Compton Effect and Pressure of Light Lessons and Learning Resources

Modern Physics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial		Revision
Notes		Revision
Questions
19.3The Compton Effect and Pressure of Light
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
19.3.1What is the Compton Effect?
19.3.2Quantum Interpretation of Compton Scattering
19.3.3Pressure of Light
19.3.4Interpretation of Light Pressure
19.3.5Wave Interpretation of Light

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