We will use the following formulas to calculate the intensity of my lights. \begin{align} A_{\text{table}} = .35 \, m^2 \quad R_{\text{table}} &= .51 \,m \quad R_{\text{plants}} = .27 \, m \nonumber \\ \left<S\right> &=\frac{ \left<P\right>}{4 \pi r^{2}} \\ &\text{ and} \nonumber \\ \left<S\right> &= \frac{ \left<P\right>}{A} \end{align} Where (1) is the sphereical light intensity at distance r, and (2) is the intensity over an area A. We'll see that (2) is more suited for our needs. \begin{align} \left<S_{table}\right> = \frac{96 \, W}{4\pi\cdot.51^2 \, m} = 30 \,W/m^2 \quad\text{and}\quad \left<S_{plant}\right> = \frac{96 \, W}{4\pi\cdot.27^2 \, m} = 105 \,W/m^2 \end{align} As you can see, even a few centimeters make a huge difference in intensity. However, this result is WRONG, and there's a lot of confusion here. This result is is the light source was a "point" and we consider the resulting intensity to be on the surface of the sphere of light hitting this point. This is not the case with (my) grow lights. The area of the lights is roughly equal to the area of the table, so we actually have infinitely many tiny point sources illuminated infinitely many little sphereical areas on the table. eq (3) does actually apply here. \begin{align} \left<S\right> = \frac{\left<P\right>}{A_{\text{table}}} =\frac{ 96 \,W}{.35\,m^2} =275 \, W/m^2 \end{align} Therefore, the closest approximation should be 275 \,W/m^2