Can you convert light to radio waves

Let us start with the basics, because there seems to be a lot of confusion in the comments.

Fact 1: A light ray or wave (not a single photon) is characterized by its wavelength while its propagation speed depends on the medium. For vacuum it is always the speed of light $c$. It is in that case related to the equation: $$lambdanu = c$$ where $nu$ is the frequency, $lambda$ the wavelength.

Fact 2: Light IS an electromagnetic wave.

Fact 3: Radio waves ARE also an electromagnetic wave with a specific range of frequencies (wavelengths).

So your question translates to: can we convert electromagnetic radiation with a certain wavelength into electromagnetic radiation with a different wavelength.

The answer is sure you can, at a theoretical level, as long as you respect energy conservation or you put in some energy into the system.

If you want to compute what the difference in energy that might be, then you have to specify the initial wavelength of the electromagnetic radiation you are using.

Say we start with visible light, say yellow light: $lambda = 580$ nm which corresponds to $5.16times 10^{14}$ Hz.

Energy in light is carried by photons which follow : $$E = hbar nu, $$ where $hbar$ is the Planck constant. So the energy of one photon in a yellow light wave has $$E_y=5.4times 10^{-20} ,text{J}$$

Let us compare that to a radio wave, which has typically something close $lambda=3times 10^5$ m or equivalently a frequency of about 1000 Hz.

So a photon of a radio wave will have $$E_r=hbar nu = 1.0545718 × 10^{-31} ,text{J}$$ of energy.

So the energy fraction between them is of the following order of magnitude: $$frac{E_y}{E_r} sim 10^{11}$$

That is a radio wave photon has $10^{-9}$ % the energy of a visible yellow photon. So you can see how much more energy per photon is carried by visible light!!! Most of the energy has to be taken away (if you were to do this photon by photon). If you were able to absorb a photon and then release a photon with a smaller energy, you would be "converting" visible light into radio waves. Processes such as fluorescence comes to my mind.

This however is not the easiest way to produce radio waves, since we haven't spoken about intensity and relevant energy scales in daily life. We normally produce radio waves using antennas which consist of accelerating electric charges, so as some comments indicate this is what your phone is doing. If you were to power your phone with visible light, let us say the sun, you would first capture the energy in the phones battery, which is later use to power the antenna and producing radio waves.

In theory, you can use parametric down-conversion to convert high frequency photons into low-frequency photons. There are three big problems here: first, efficiency is very low (on the order of a million), second, you would need to get a million-fold reduction in frequency which likely requires passing through several stages (with losses each time), and third, I am not even sure if we know materials for the process in the intermediate IR and teraherz range to do the conversion.