dividing voltage of shunt Vref Lm4040
A photodiode produces a photo-current proportional to the incident light power hitting its active area but, it has to be the correct wavelength light (for the photodiode) or values measured will be highly confusing.

The circuit you have shown with a photo-diode built into a potential divider sat around a voltage reference is missing the point. You should consider using a transresistance amplifier to convert the photo-diode current to a voltage that then feeds into what measurement system you have. In simple terms it is one of these: -



So, for the TEMD1000 specified above, the data sheet will show you what current output you get for a given light incident power: -



Take into account the photo-diodes active area to convert incident power density to actual power received and you get photo-current and that photocurrent is amplified by the op-amp and resistor to give you the required output voltage per uW you require for your ADC input.