Newton's third law in relation to a mirror reflecting light I am not too versed in the physics realms other than a couple courses from college, but my question mainly pertains to newton's third law of motion.
If for every action there is an equal and opposite reaction, why does the act of a mirror reflecting (or even partially deflecting) photons moving at the speed of light not cause an infinite reaction force in the opposite direction of the reflection?
I assume it has to do with photons being considered mass-less particles and newton's laws failing at quantum levels but I guess I hoped there was a better explanation than this.
 A: As you said, photons are massless. So getting a photon to travel at the speed of light (which, of course, it always is travelling at) does not take "infinite energy". This idea only applies to objects with mass.
However, photons do have momentum, which is defined by the equation:
$Momentum$= $Energy$ / $c$ 
with c being the speed of light. Of course, the energy of a typical photon is relatively very small and the speed of light is very large. Therefore, compared to usual objects in motion, a photon will have an extremely small momentum, which is why it has no large scale effect on something such as a mirror.
A: The energy of a photon is not infinite, but is given by $$E=hf=\dfrac{hc}{\lambda},$$
where $h$ is Planck's constant ($6.626\times 10^{-34}$ joule$\cdot$seconds), $f$ is the frequency of the photon, $c$ is the speed of light, and $\lambda$ is the wavelength of light.
The momentum of a photon is related to the energy by $E=pc$, where $p$ is the momentum.  Combining these relationships, we see that the photon momentum can be calculated by $$p=\dfrac{h}{\lambda}.$$
If a photon is reflected from a mirror, we can estimate that the momentum of the photon changes to $p$ in the opposite direction. Ignoring outside itneractions, conservation of momentum tells us (the mirror is initially at rest) $$ p + 0 = -p + p_{\mathrm{mirror}}.$$
This tells us the momentum of the mirror after the reflection should be (if the mirror is not attached to anything else, like a table with friction)$$p_{\mathrm{mirror}}=2\dfrac{h}{\lambda}.$$
Let's assume some numbers: a high energy visible photon has $\lambda = 4\times10^{-7}$ meters, and assume a very small, low-mass mirror with $m=1$ nanogram = $1\times 10^{-12}$ kilograms. This would result in a recoil speed for the mirror of about 3 femtometers per second.
I believe it's safe to assume that the resulting energy would easily be absorbed by frictional forces from objects to which the table is attached.
A: i think that photons suffer elastic collisions as they completely reflect back with same velocit that is c therefore :
                              1=(v2-u2)/(u1-v1)= coefficient of restitution which                                                                                 must be equal to 1 for elastic collissions.(u1,v1,u2,v2 are initial and final velocities of photon and mirror respectively
therefore         u1+u2=v1+v2____________A

and according to
                  p+0=-p+pm  (pm=momentum of mirror)___________B

        NOW compare equation A and B We get:
                         u2=0,v1=u1-u and v2=vm                                                          (vm=velocity of mirror) and u is the infinitesimal velocity change of photons reflecting back from mirror .

therefore 
therefore
*                                **u1=u1-u+vm **
                                    vm=u,which is at the order of fermimeters
  and hence very small and get absorbed by the mirror due to some frictional forces                                                       

