Best Illumination

Let $h$ be the height of the light and $r$ the radius of the illuminated area. If $I$ is the intensity of illumination, we are told that $I\sim \frac{\sin\theta}{d^2} = \frac{h}{d^3} = \frac{h}{(h^2+r^2)^{^3\!/\!_2}}$ Therefore $I$ will be maximized when the right hand side of the above equation is maximized. Setting the derivative of that expression equal to zero yields $\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}h}\frac{h}{(h^2+r^2)^{^3\!/\!_2}} &= 0 \\ \frac{(h^2+r^2)^{^3\!/\!_2}-3h^2(h^2+r^2)^{^1\!/\!_2}}{(h^2+r^2)^3} &= 0 \\ (h^2+r^2)-3h^2 &= 0 && \color{#3D99F6} \text{Multiply through by }(h^2+r^2)^{^5\!/\!_2} \\ 2h^2 &= r^2 \\ h &= \frac{r}{\sqrt{2}} && \color{#3D99F6} h\text{ and }r\text{ are both positive} \end{aligned}$ All that remains is to find the ratio of the height to the radius: $\frac{h}{r} = \frac{1}{\sqrt{2}} \approx \boxed{0.7071}$