The problem with your derivation is that you distributed the photons over a 360$^o$ circle, so the photons only spread out in a two-dimensional circle. This means that the intensity of light drops off at a rate proportional to $1/r$ instead of $1/r^2$ (where $r$ is the distance from the center of the sun) like it does in a three-dimensional universe.

So, starting with $N$ photons emitted per second, the intensity of photons at a distance $r$ from the sun is given by $$I = \frac{N}{4\pi r^2}.$$ This comes from spreading out the photons over the surface of a sphere surrounding the sun.

The number of photons seen by your eye per second is just the intensity multiplied by the area of the iris of your eye: $$n = IA_{eye} = \frac{N}{4\pi r^2}A_{eye}.$$ You are looking for the distance beyond which you would see less than one photon per second: $$n = \frac{N}{4\pi r^2}A_{eye} < 1.$$ Solving for $r$ gives $$r > \sqrt\frac{NA_{eye}}{4\pi}$$ Plugging in your numbers gives $$r > \sqrt{\frac{(10^{45})\pi(0.005\,\textrm{m}/2)^2}{4\pi}} = 4\cdot10^{19} \,\textrm{m} \approx 4000\,\textrm{light-years}$$ This distance is still well within our own galaxy.

The problem with your derivation is that you distributed the photons over a 360° circle, so the photons only spread out in a two-dimensional circle. This means that the intensity of light drops off at a rate proportional to $1/r$ instead of $1/r^2$ (where $r$ is the distance from the center of the sun) like it does in a three-dimensional universe.

So, starting with $N$ photons emitted per second, the intensity of photons at a distance $r$ from the sun is given by $$I = \frac{N}{4\pi r^2}.$$ This comes from spreading out the photons over the surface of a sphere surrounding the sun.

The number of photons seen by your eye per second is just the intensity multiplied by the area of the iris of your eye: $$n = IA_\text{eye} = \frac{N}{4\pi r^2}A_\text{eye}.$$ You are looking for the distance beyond which you would see less than one photon per second: $$n = \frac{N}{4\pi r^2}A_\text{eye} \lt 1$$ Solving for $r$ gives $$r > \sqrt\frac{NA_\text{eye}}{4\pi}$$ Plugging in your numbers gives $$r > \sqrt{\frac{(10^{45})\pi(0.005\,\textrm{m}/2)^2}{4\pi}} = 4\cdot10^{19} \,\textrm{m} \approx 4000\,\textrm{light-years}$$ This distance is still well within our own galaxy.

The problem with your derivation is that you distributed the photons over a 360$^o$ circle, so the photons only spread out in a two-dimensional circle. This means that the intensity of light drops off at a rate proportional to $1/r$ instead of $1/r^2$ (where $r$ is the distance from the center of the sun) like it does in a three-dimensional universe.

So, starting with $N$ photons emitted per second, the intensity of photons at a distance $r$ from the sun is given by $$I = \frac{N}{4\pi r^2}.$$ This comes from spreading out the photons over the surface of a sphere surrounding the sun.

The number of photons seen by your eye per second is just the intensity multiplied by the area of the iris of your eye: $$n = IA_{eye} = \frac{N}{4\pi r^2}A_{eye}.$$ You are looking for the distance beyond which you would see less than one photon per second: $$n = \frac{N}{4\pi r^2}A_{eye} < 1.$$ Solving for $r$ gives $$r > \sqrt\frac{NA_{eye}}{4\pi}$$ Plugging in your numbers gives $$r > \sqrt{\frac{(10^{45})\pi(0.005\,\textrm{m}/2)^2}{4\pi}} = 4\cdot10^{19} \,\textrm{m} \approx 4.2\,\textrm{light-years}$$ This distance is just beyond the next closest star to our sun: Proxima Centauri.

The problem with your derivation is that you distributed the photons over a 360$^o$ circle, so the photons only spread out in a two-dimensional circle. This means that the intensity of light drops off at a rate proportional to $1/r$ instead of $1/r^2$ (where $r$ is the distance from the center of the sun) like it does in a three-dimensional universe.

So, starting with $N$ photons emitted per second, the intensity of photons at a distance $r$ from the sun is given by $$I = \frac{N}{4\pi r^2}.$$ This comes from spreading out the photons over the surface of a sphere surrounding the sun.

The number of photons seen by your eye per second is just the intensity multiplied by the area of the iris of your eye: $$n = IA_{eye} = \frac{N}{4\pi r^2}A_{eye}.$$ You are looking for the distance beyond which you would see less than one photon per second: $$n = \frac{N}{4\pi r^2}A_{eye} < 1.$$ Solving for $r$ gives $$r > \sqrt\frac{NA_{eye}}{4\pi}$$ Plugging in your numbers gives $$r > \sqrt{\frac{(10^{45})\pi(0.005\,\textrm{m}/2)^2}{4\pi}} = 4\cdot10^{19} \,\textrm{m} \approx 4000\,\textrm{light-years}$$ This distance is still well within our own galaxy.