If I point a torch up, would the photons hit anything? If I point my torch up, the photons leave the atmosphere. They might hit the sun, moon or some clouds. If they miss all this, what would they likely hit - if anything?
 A: It depends exactly which way you are pointing your torch, but unless you are unlucky and point at the Galactic plane, then the photons will probably travel to cosmological distances.
A typical value for the extinction of visible light in the atmosphere is about 0.3 magnitudes (astronomer's units!) if you point your torch straight up; there is more extinction for blue photons, less for red photons. What this means is that about 75% of the photons make it through the atmosphere and into space.
As it gets further from the Earth, the light is very unlikely to be intercepted by another object in the Solar System. The two objects with the biggest solid angle are the Sun and the Moon, but these fill a total of only 3 millionths of the available solid angle. The contribution from planets and asteroids is completely negligible.
After it leaves the Solar System, some of the light will interact with the interstellar medium (ISM) and be absorbed/scattered. This will be the fate of almost all the light which is emitted towards the plane of our Galaxy, which contains sufficient molecular gas and dust to block visible light travelling through it for any distance. We know this happens because we can "see" dark clouds in the Milky Way, that can be penetrated by longer wavelength radiation to reveal all the billions of Sun-like stars that lie behind them. Roughly speaking, about half the visible light will be absorbed every 1000 light years when travelling in the Galactic plane, so it is essentially all absorbed within a few thousand light years.
But the Galactic plane is only a few hundred light years thick and so the band that is capable of absorbing the photons covers only about $\pm 10^{\circ}$ around the Galactic equator and thus only about 15% of sky. So the chances are good that the light from your torch will escape both the Solar System and the Galaxy.
The equivalent extinction number for the intergalactic medium is that light travels many billions of light years with little chance of being absorbed (see Zu et al. 2010). This means that most of the light from your torch will travel to cosmological distances (billions of light years) over the course of the next billions of years. Indeed light emitted from the Sun shortly after its birth has already travelled 4.5 billion light years. We know this has happened and will happen, because we can observe galaxies (the light from which is nothing more than the summation of light from many stars like the Sun) that are 4.5 billion (and much more) light years away. In fact it is more likely, because the universe has expanded significantly over the last 4.5 billion years and so any intergalactic dust that might cause some absorption has become even more rarefied.
As the torchlight travels towards cosmological distances in the distant future, its wavelength is "stretched" by the expansion of the universe, becoming redder and redder, moving into the infrared, microwave and ultimately radio part of the spectrum. We know this happens because distant galaxies have redshifted spectra and we can observe the cosmic microwave background (that began its journey 13.7 billion years ago as red/infrared light). At wavelengths larger than visible light the radiation is even less likely to be absorbed and even less so because the average density of the universe decreases. If the universe keeps expanding, then its density will continue to decrease, and there is little to stop the radiation from the Sun travelling on forever with a wavelength that increases as the scale factor, $a$, of the universe and an energy density that scales as $a^{-4}$.
A: Mean free path of a photon in material is :
$$ \ell_{\gamma} =(\mu_m \rho )^{-1} \tag 1,$$
where $\mu_m$ is mass attenuation coefficient and $\rho$ - material density. Interstellar space contains mostly hydrogen atoms with density of $1~ \text{atom}/m^3$, so this gives interstellar space density $\rho = 10^{−24}~g  / m^3$. Hydrogen has mass absorption coefficient about $\mu_m = 0.4~cm^2/g$.
Plugging this into (1) gives that photon who reaches interstellar space will travel about
$$ \ell_{\gamma} \approx 3~Tly = 10^{12}~\text{light years} \tag 2$$
until it is captured by a lonely Hydrogen atom in outer-space. But this free path distance is amazingly huge, it's about $\approx 28$ diameters of observable universe. So basically it means that photon is free to go anywhere. Of course some of them will be bumping into stars, dust, gas particles, planets (and into creatures eyes living on that planet if any), however probability of this to happen for an ordinary photon is very small, so most photons will be just wandering in empty space for no good.
A: If you were to miss all planets, debris, nebulas, starts, etc. the photons would continue on until they did one of two things:


*

*Reach the edge of the universe (which there is not, but is at the same time) and then we don't know what would happen from there, as we don't know what is outside of the universe.

*The more likely scenario is that the photons would eventually reach almost an equilibrium of velocity to the expansion velocity of the universe and essentially stop. 


Take a look here to see that there are theoretical "sides" of the universe that are moving apart faster than the speed of light but also there are theoretical stars and galaxies that we cannot see because they are moving to quickly away from us.
