The current in the forward bias is known to be more `(~mA)` than the current in the reverse bias `(~muA)`. What is the reason then to operate the photodiodes in reverse bias?

Explanation

For an n-type semiconductor, the majority carrier density `(n_(e ))` is larger than the minority hole density `(n_(h))` (mean`n_(e ) gt gt n_(h))`.
On illumination, let the excess electrons and holes generated be `Delta n_(e )` and `Delta n_(h)` respectively, so
`n._(e )=n_(e )+Delta n_(e )`
`n._(n)=n_(h)+Delta n_(h)`
where `n._(e )` and `n._(h)` are the electron and hole concentration respectively due to intensity of light incident on p-n junction.
And `n_(e ) and n_(h)` are carriers concentration when there is no illumination.
But `Deltan_(e )=Delta n_(h) and n_(e ) gt gt n_(h)`, hence fractional change in majority carriers is `((Delta n_(e ))/(n_(e )))` and for minority carrier is `((Delta n_(h))/(n_(h)))` which is much less than majority carriers.
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.