Proportionality between number of absorbed photons and optical density
Yes, your statement that "the number of absorbed photons only is proportional to the optical density at low optical densities" is correct. (I'll point out that optical density is a depreciated term for what is now referred to as absorbance.)     

First let's define some terms.


  T: Transmittance, i.e. the fraction of photons not absorbed. So a
     transmittance of 0.90 would mean that 90% of the photon pass through
     the sample.
  
  A: Absorbance of sample which is given by the Beer–Lambert law: $\ce{A = -log_{10}(T)}$ or: $\ce{T = 10^{-A}}$


Now a bit of math swizzling...

For exponentials of $e$ there is a nice series expansion. $$ e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{3!} + ... $$ and if $x &lt; 0 $ then: $$ e^{-x} = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \dfrac{x^4}{3!} + ... $$
now if also $|x| &lt;&lt; 1$ then the higher order terms can be ignored and the function becomes linear where:
$$ e^{-x} \approx 1 - x $$

Now for a bit more math voodoo... Let's let $$10^x = e^y$$ so: $$y = x\mathrm{ln}(10) \approx 2.303x$$ 

So now substituting $\ce{e^{-2.303A}}$ for $\ce{10^{-A}}$ we get: $$\ce{T = e^{-2.303A}}$$ and thus only when $2.303A &lt;&lt; 1$ is the Beer-Lambert law linear. 

Now for the rest of the story let's define some more terms:


  $\epsilon$ is the molar attenuation coefficient of the attenuating species  in the sample;
  
  $c$ is the molar concentration of the attenuating species in the sample;
  
  $l$ is the path length of the beam of light through the material sample.


Now also accoding to the Beer-Lambert law: $\ce{A = \epsilon c l}$ so $\ce{A}$ is proportional to $\ce{c}$.