The current in the forward bias is known to be more $(\sim mA)$ than the current in the reverse bias $(\sim \mu A)$. What is the reason then to operate the photodiodes in reverse bias?

(N/A) Consider the case of an $n$-type semiconductor. Obviously,the majority carrier density $(n)$ is considerably larger than the minority hole density $(p)$ (i.e.,$n \gg p$). On illumination,let the excess electrons and holes generated be $\Delta n$ and $\Delta p$,respectively:
$n^{\prime} = n + \Delta n$
$p^{\prime} = p + \Delta p$
Here,$n^{\prime}$ and $p^{\prime}$ are the electron and hole concentrations at any particular illumination,and $n$ and $p$ are carrier concentrations when there is no illumination. Remember $\Delta n = \Delta p$ and $n \gg p$.
Hence,the fractional change in the majority carriers (i.e.,$\Delta n / n$) would be much less than that in the minority carriers (i.e.,$\Delta p / p$). In general,we can state that the fractional change due to the photo-effects on the minority carrier-dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence,photodiodes are preferably used in the reverse bias condition for measuring light intensity.