Upon excitation of an 'isotropic' sample (assuming an ultra short excitation pulse) the relationship between the fluorescence intensity detected at a time $t$ and through a polarization analyser oriented at an angle $\beta$ with respect to the electric polarization of the exciting beam is given by $$I(t,\beta )\propto N(t)[1+(3{\text{cos}}^2\beta -1)R(t)]$$ where $R(t)$ is the degree of alignment of the emitting transition dipole in the laboratory frame and $N(t)$ is the excited-state population, both at time $t$. For $\beta ={54.7}^{\text{⚬}}$ (the magic angle), the dipole-alignment contribution vanishes and $I(t,\beta ={54.7}^{\text{⚬}})\propto N(t)$.