Which figure shows the correct variation of applied potential difference (V) with photoelectric current (I) at two different intensities of light $$(I_1 < I_2)$$ of same wavelengths:

$$\text{Einstein's photoelectric equation: } eV_0 = \frac{hc}{\lambda} - \phi \implies V_0 = \frac{hc}{e\lambda} - \frac{\phi}{e}$$

Since the wavelength $$\lambda$$ and work function $$\phi$$ remain identical for both cases, their stopping potential $$-V_0$$ must be exactly the same

$$\text{The saturation photoelectric current is directly proportional to intensity: } i \propto I \implies \text{Since } I_1 < I_2\text{, the curve for } I_2 \text{ reaches a higher saturation current Level.}$$

Therefore, both curves must originate from the same stopping potential point $$-V_0$$ on the potential axis and separate as current rises