Explanation: Photodiodes are semiconductor devices specifically designed to convert light energy into electrical current. When photons of light strike the depletion region of the photodiode, they impart energy to electrons, creating electron-hole pairs. These charge carriers are then swept across the p-n junction by the diode's internal electric field (or an external reverse bias, if operated in photoconductive mode). This movement of charge constitutes an electric current. Therefore, if more light strikes the photodiode, more photons are absorbed, generating a greater number of electron-hole pairs and resulting in a larger diode current. Options A and B are incorrect because they contradict the fundamental operating principle of a photodiode, which is to produce more current in response to increased light. Option D is irrelevant to the effect of light intensity on current; while proper polarity is crucial for operation, the question assumes a functional setup and focuses on the *change* in current due to varying light.