# Electric field intensity of light

Light is a transverse wave in which electric field and magnetic field components oscillate in perpendicular directions. Based on this I have two questions: Question 1 What would be the intensity of this electric field produced due to to a single photon? (Please mention the exact formula if possible)

Question 2: If two light waves of the same energy undergo complete destructive interference then does their electric field and magnetic field cancel out?

• You are asking two questions. The fields are added for interference, the intensity is proportional to the squared amplitude of the electric field. Jun 6, 2018 at 5:28
• But is there an exact value or even a approximation of the electric field intensity produced by a photon? Jun 6, 2018 at 13:09
• Are you asking about a single photon? That is in no way clear from the question. Jun 6, 2018 at 17:06
• Yes I am, I'll edit the question so that it is more clear. Jun 7, 2018 at 1:07
• Possible duplicate of Amplitude of an electromagnetic wave containing a single photon Jun 7, 2018 at 12:14

What would happen to these fields during constructive and destructive interference?

Here is a photo of a section of a soap film that shows constructive and destructive interference.

High-contrast magnified-image of coloured interference-pattern in soap-film, (original area is approximately 10mm x 15mm). The black "holes" are areas where the film is very thin, (~10nm), and there is near-total destructive-interference, (hence they appear black).

Here is what constructive interference looks like:

Wikipedia explains optical interference this way:

Because the frequency of light waves ($~10^{14}$ Hz) is too high to be detected by currently available detectors, it is possible to observe only the intensity of an optical interference pattern. The intensity of the light at a given point is proportional to the square of the average amplitude of the wave. This can be expressed mathematically as follows. The displacement of the two waves at a point $r$ is:

$$U_1 (\mathbf r,t) = A_1(\mathbf r) e^{i [\varphi_1 (\mathbf r) - \omega t]}$$

$$U_2 (\mathbf r,t) = A_2(\mathbf r) e^{i [\varphi_2 (\mathbf r) - \omega t]}$$

where A represents the magnitude of the displacement, φ represents the phase and ω represents the angular frequency.

The displacement of the summed waves is:

$$U (\mathbf r,t) = A_1(\mathbf r) e^{i [\varphi_1 (\mathbf r) - \omega t]}+A_2(\mathbf r) e^{i [\varphi_2 (\mathbf r) - \omega t]}$$

The intensity of the light at r is given by:

$$I(\mathbf r) = \int U (\mathbf r,t) U^* (\mathbf r,t) dt \propto A_1^2 (\mathbf r)+ A_2^2 (\mathbf r) + 2 A_1 (\mathbf r) A_2 (\mathbf r) \cos {[\varphi_1 (\mathbf r)-\varphi_2 (\mathbf r)]}$$

This can be expressed in terms of the intensities of the individual waves as:

$$I(\mathbf r) = I_1 (\mathbf r)+ I_2 (\mathbf r) + 2 \sqrt{ I_1 (\mathbf r) I_2 (\mathbf r)} \cos {[\varphi_1 (\mathbf r)-\varphi_2 (\mathbf r)]}$$

Thus, the interference pattern maps out the difference in phase between the two waves, with maxima occurring when the phase difference is a multiple of 2π. If the two beams are of equal intensity, the maxima are four times as bright as the individual beams, and the minima have zero intensity.

The two waves must have the same polarization to give rise to interference fringes since it is not possible for waves of different polarizations to cancel one another out or add together. Instead, when waves of different polarization are added together, they give rise to a wave of a different polarization state.

• Even if it is a direct quote form wikipedia: is it common to speak of displacement when talking about electromagnetic waves? Also the animation does not show constructive interference but a standing wave. Jun 6, 2018 at 17:09
• Red is the result of adding blue and green, all are plotted together along with the formula.
– Rob
Jun 6, 2018 at 19:46
• True, but it's still a standing wave instead of constructive interference. Jun 6, 2018 at 19:53