Concept:Each absorbed photon with \( hf \ge E_g \) generates one electron-hole pair. The reverse photocurrent is directly proportional to the photon arrival rate (light intensity).
Formula:$$I_{\text{photo}} = \mathcal{R} \cdot P_{\text{optical}} = \left(\frac{\eta_{\text{ext}} q}{h f}\right) P_{\text{optical}} \propto \Phi$$
Solution:- Higher light intensity delivers more photons per second to the depletion region.
- This generates more electron-hole pairs per second, producing a reverse photocurrent that is directly and linearly proportional to incident light intensity.
Why other options are incorrect:- Option B: Photocurrent increases with illumination; it does not decrease exponentially.
- Option C: Photodiodes are specifically designed to produce current proportional to light intensity.
- Option D: The response is linear, not an inverse-square relationship.
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