Are composite bosons distinguishable from elementary bosons? I have learnt that an even number of fermions can behave like a boson, with a net integer spin.
Am I correct in thinking that this is only true on scales where the distance from the composite particle is much greater than the separation of the fermions in the composite particle, and that if we 'zoom in' we can no longer apply this notion?
 A: Yes, you are correct.  The two-particle bosonic wave function, such as $$\Psi\left(\vec{r}_{1},\vec{r}_{2}\right)=\psi_{1}\left(\vec{r}_{1}\right)\psi_{2}\left(\vec{r}_{2}\right)+\psi_{1}\left(\vec{r}_{2}\right)\psi_{2}\left(\vec{r}_{1}\right)$$ for a pair of composite bosons is not exact.  It is merely a good approximation on scales larger than the internal scales of the composites.
If each boson is made up of two different fermions (like a hydrogen atom or a meson), the internal wave function for one of them looks like $$\psi\left(\vec{r}\right)=\phi_{1}\left(\vec{r}+\Delta_{a}\vec{r}\right)\phi_{2}\left(\vec{r}+\Delta_{b}\vec{r}\right),$$ with the two constituent fermions located at $\vec{r}+\Delta_{a}\vec{r}$ and $\vec{r}-\Delta_{b}\vec{r}$, slightly offset from the boson's center of mass position $\vec{r}$.  If you plug this form into the form for $\Psi$, you do not get a function that is properly antisymmetric under the exchange of two identical fermions between the two bosons.  The discrepancy is very small, however, except where there is significant overlap between the wave functions of the two bosons, so that their constituent fermion wave function clouds actually interpenetrate.
