It is not correct saying that no force is applied. A photon carries momentum see PE here so on reflection there is momentum transfer. This is the idea behind laser propulsion discussed here.
Concerning the speed it is even more complicated. The fact that light gets reflected usually requires an abrupt change in the index of refraction. To get reflected, hence, the photon interacts with the medium that reflects it. So it penetrates the medium to some extend. This takes time especially as the speed of light is slower in the medium, due to the interaction of the electric field with the surrounding matter. The speed of light in a dielectric medium is often subject of discussions as shown here. Unfortunately the accepted answer in that discussion is not really satisfying as the slowing down does not require loss of energy nor is it stochastic. It is better to look at the overall electromagnetic wave and its interaction with a surrounding that can give an electromagnetic response.
In any case, thinking about light (particle, wave or what) and, furthermore, its interaction with matter is sort of mind boggling.
Edit 1 on loosing speed:
Mostly one would consider very large objects so they are considered static and the photon is outgoing with the same momentum. If the reflecting object is very small, one might consider a movement and a reduced momentum. This would, however not reduce the speed of the reflected photon buts its momentum and, therefore, its energy. This, hence, changes the wavelength. This process would be somewhat like inelastic scattering. Still, one has to consider conservation of momentum and energy (unless concerned with general relativity; then energy conservation goes out the window).
Edit 2 on energy and momentum transfer:
Considering one of the questions in the comments if the energy and momentum transfer has something to do with solar energy or could be used as alternative, here some simple estimates for a photon with wavelength $\lambda = 560\;\mathrm{nm}$ (somewhat yellowish).
The energy of the photon is $E= h c/ \lambda= 2.2\;\mathrm{eV}=3.54\;10^{-19}\;\mathrm{J}$. The momentum is $p=h/\lambda$ and the transfer is approximately $\Delta p= 2p=2.37\;10^{-27}\;\mathrm{kg\;m/s}$. A mirror of mass $m=1\;\mathrm{kg}$ responsible for that momentum transfer, will have kinetic energy $p^2 /(2 m)=2.8\; 10^{-54}\;\mathrm{J}$. The required change in wavelength of the photon providing this energy can be approximated by (the approximation assumes that the change is small, which it is as we see now) $\Delta \lambda \approx \Delta E \lambda^2/(h c)=2h/(mc)=4.42\; 10^{-33}\;\mathrm{nm}$
Note that you get the same if you interpret this as relativistic dopplereffect, i.e. $\Delta \lambda \approx \lambda v/c$, where $v$ is the speed of the mirror due to the momentum transfer of the photon. (The approximation holds for $v \ll c$). Also note that the absolute change in wavelength, for large wavelength i.e. small photon energies, is independent of the wavelength; so $\Delta \lambda/\lambda \rightarrow 0$ for $\lambda \rightarrow \infty$.
So a hypothetical change in wavelength is below anything one might hope to measure (33 digits behind the comma). Furthermore the energy gain in the mirror is ridiculously small compared to the energy one can get by absorption. Nevertheless, the momentum transfer is not that bad, as one can easily "fire" $10^{27}$ photons to get a reasonable overall effect. Efficiency seems poor though, so make a quick search on photon propulsion for why and how it is done in the first place.
