10
$\begingroup$

Referring to this question How can I measure the amplitude of a light wave? I'm curious about what is a amplitude of a light wave. Especially for light from a thermic source.

$\endgroup$
0
6
$\begingroup$

A light wave is a traveling disturbance in the electric and magnetic fields. In the far-field these two components are coupled and in-phase so that one can define the amplitude entirely in terms of either the electric field strength or the magnetic field strength; by convention we give the electric field strength.

So, in SI, the amplitude of a light wave is some number of Newtons per Coulomb (or equivalently of Volts per meter) and it represents the magnitude of the largest change in the electric field strength from the ambient value that would be measured as the wave passes.

It is not practical to make such measurements in the optical range (the frequencies are around $10^{14}$$10^{15}\,\mathrm{Hz}$ after all), but this can be (and has been) measured explicitly in the radio regime.

The electric field of course is a vector, but again in the far-field and in free-space it is always perpendicular to the line of travel so the wave is transverse. We define the direction of polrization as agreeing with the direction of the electric field oscillation (the ambiguity of direction on either side of zero is of no consequence).

$\endgroup$
2
$\begingroup$

You cannot immobilize light, but you can absorb it, convert it to heat. A first step in measuring light intensity (watts/square meter) is to make a black body to absorb the light. Then, by conservation of energy, the heating of that black body tells you how much light is being absorbed, in watts. The illuminated area of the black body is the "square meters" part.

There are many other kinds of light sensors (electronic, acoustic, chemical) suitable for special cases, but they require different calibration curves for different light sources. Color, beam divergence, polarization, and other variables other than light intensity will affect those sensors.

$\endgroup$
0
$\begingroup$

So an amplitude in terms of an electromagnetic wave is related to its intensity. So technically, the more intense a light wave is at a point, the more its amplitude. What is the intensity Intensity is inversely proportional to the A^2 of a wave. If you know the intensity, then the amplitude can be found using this formula I=1/2*pvw*A^2 where p is density, v, w is angular velocity, v is speed of wave. Hence, if you measure the intensity in a a unit area, (W m^-2), you can find the amplitude inversely solving.

$\endgroup$
0
$\begingroup$

It is mentioned that the question has been derived from another question: “How can I measure the amplitude of a light wave?” Let me derive another question from the present question: “What is the unit for amplitude?” Everyone knows the answer for this derived question. To check validity of a formula in physics, the first step is to verify units on both sides. Apart from this general discussion, I like to give an answer to the present question.

Light rays are emitted from a tungsten bulb, when current passes through it, and in this case the Maxwell equations are satisfied. So, light rays are special cases of general solutions of Maxwell equations. Solutions of Maxwell equations provide orthogonal oscillations of magnetic field and electric field; not radiation (not propagation). (I am a mathematics teacher). In this specific situation, electric field creates light rays. One used to associate a number/parameter in solution of Maxwell equations as amplitude. That number/parameter of that solution is called the amplitude of the corresponding light rays emitted. When light rays are emitted from other sources, an imaginary model should be developed for which a reference is not to be provided (to avoid “self-promotion”). To my knowledge: there is no device to measure amplitude in length for light rays.

P.S. Amplitude in the question refers to length/height.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.