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(C) The photoelectric effect depends on the intensity and frequency of incident light.Since the incident lights have the same wavelength,their frequen

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(C) The photoelectric effect depends on the intensity and frequency of incident light.Since the incident lights have the same wavelength,their frequencies are identical. Therefore,the maximum kinetic energy of the emitted photoelectrons is the same,which means the stopping potential $(V_0)$ remains the same for both intensities.Since the intensity $I_2 > I_1$,the number of photoelectrons emitted per unit time is greater for $I_2$ than for $I_1$. Consequently,the saturation current for $I_2$ will be greater than that for $I_1$.Comparing this with the given figures,Figure $C$ correctly shows that both curves start from the same stopping potential $(-V_0)$ on the negative $V$-axis,and the saturation current for $I_2$ is higher than for $I_1$.Thus,the correct figure is $C$.

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 the same wavelength?

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