Force exerted by light on a moving mirror 
Consider a light with energy density E shining uniformly over a mirror. The mirror has an area A. The mirror is moving at with a velocity β. Calculate the force that the photons exert on the mirror.

My Attempt:
Energy, under Lorentz transformation would be transformed as V E' = γ(V E - βp) = γ(V E - βVE/c) = γ V E(1 - β/c)
So we have V E' = γ V E(1 - β/c)
And so energy V E' = V γE(1 - β/c)
Then, change in momentum Δp' = 2 * V E' / c = 2 (γVE(1 - v/c)) / c 
So force F =  Δp' / Δt = 2 (γVE(1 - v/c)) / (c Δt) 
But Δt = Δx / c and V = A Δx and so F = 2 (γ A Δx E(1 - v/c)) / (c (Δx / c)) = 2 γ A E(1 - v/c))
Hence force = 2 γ A E(1 - v/c)).
Question:
This does not account the fact that the length $dx$ also shrinks down and now that the mirror is moving, less number of photons will be hitting the mirror. How do I take that into account?
 A: Perhaps a slightly different way of thinking about this would be in terms of the Poynting vector.
The force exerted is given by 
$$F =  \frac{1}{c} \int {\bf S} \cdot d{\bf A}, $$
where ${\bf S}$ is the Poynting vector and in your terms $S = c E$. In this case I assume everything is at normal angles, so no need to worry about that.
The Poynting vector is transformed relativistically (in this case) as follows:
$$ {\bf S'} = \gamma^2 (1 + \beta)^2 {\bf S} $$
Where $\beta = v/c$ and defined in the sense that the prime frame is moving with velocity $v$ with respect to the "stationary" frame. Using this, I get quite a different answer to you. 
When you are calculating the force, should you not be using a time interval of $\Delta x^{'}/c$?
NB: The considerations above apply to planes waves normally incident upon the mirror. If the mirror is travelling through an isotropic (e.g. blackbody) radiation field then the calculations become considerably more complicated because the photons do not strike the mirror normally.
