A photodiode forms part of a light meter used for checking light levels in an office - AQA - A-Level Physics - Question 4 - 2017 - Paper 8

The operational amplifier is in comparator mode.

To estimate the light level that causes a reverse current of 0.10 mA, we can refer to Figure 11. The corresponding light level for a reverse current of 0.10 mA is approximately 1000 lux, with a tolerance of ±10%.

Using Ohm's law, we can calculate the voltage at point X:

$V_X = I imes R$

Given that the current ($I$) through the photodiode is 0.10 mA (0.0001 A) and the resistance ($R$) connected at point Y is 20 kΩ (20000 Ω), we have:

$V_X = 0.0001 A imes 20000 Ω = 2 V$

To determine if the LED is ON or OFF, we rule that if $V_{in} > V_{out}$, then $V_{out} = 5 V$ (high). Since the voltage at point Y is 1.75 V and the voltage at point X is 2 V, the LED will be OFF because the operational amplifier will not trigger the output.