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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?

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  • $\begingroup$ You should know it maxwell! $\endgroup$ – santimirandarp Nov 23 '18 at 14:13
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what is the relation between the electric field of a given light and its wavelength?

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 electromagnetic energy of light is expressed by its electric and magnetic fields

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.

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