How is the electric field of light related to its wavelength?

Light is an electromagnetic field. At the same time light is composed of photons; each one carries an energy given by $$E=\frac{hc}{\lambda}$$. The electromagnetic energy of light is expressed by its electric and magnetic fields. The question is what is the relation between the electric field of a given photon and its wavelength?

• You should know it maxwell!
– user153036
Commented Nov 23, 2018 at 14:13

In general, there's no relationship at all. The electric field is a function, which generally has the form $$E(x,y,z,t) = E_0 \hat{\mathbf e}_y \cos\left(\frac{2\pi}{\lambda}(x-ct)\right)$$ in the simplistic case of a plane wave. When you say things like
the part that really matters is the electric field amplitude, $$E_0$$, and this is completely independent of the wavelength.
When you go to a regime when the total energy in the beam is comparable to the photon energy $$E=h\nu=hc/\lambda$$, then the electric field becomes a quantum variable with an undefined value, but roughly speaking its RMS value does get quantized as a multiple of the photon energy (i.e. the photon energy multiplied by the number of photons in the beam) divided by something called the mode volume - but the upshot is that absent additional restrictions, its value can be pretty much anything.