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 < 0 $ then: $$ e^{-x} = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \dfrac{x^4}{3!} + ... $$
now if also $|x| << 1$ 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 $2.303A << 1$ 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 $\ce{c}$.
