Let $r$ be the resistance of the ammeter. To double the measurement range of the meter, a shunt with resistance $r$ must be connected in parallel with the meter so that half of the measured current is diverted to the shunt. The meter and the shunt together has resistance $r/2$. Suppose that the photocell is a voltage generator $V$ and is connected to the load $R$. Without the shunt, the measured current is $V/(R+r)$. With the shunt, the measured current becomes $V/(R+r/2)$.