The curves (a), (b), (c) and (d) show the var | vedclass

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(B) In the photoelectric effect, the photoelectric current $(i)$ is directly proportional to the intensity of incident light $(I)$ for a fixed frequen

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(B) In the photoelectric effect, the photoelectric current $(i)$ is directly proportional to the intensity of incident light $(I)$ for a fixed frequency of light. Given that $I_1 > I_2,$ the saturation current for intensity $I_1$ must be greater than that for $I_2$ $(i_1 > i_2).$Furthermore, the stopping potential $(V_0)$ depends only on the frequency of the incident light and the work function of the metal surface, not on the intensity of the light. Therefore, for both intensities $I_1$ and $I_2,$ the stopping potential $(V_0)$ remains the same.Looking at the provided graphs, figure $(b)$ correctly shows that both curves originate from the same stopping potential $(V_0)$ on the negative $V$-axis, and the saturation current for $I_1$ is higher than for $I_2.$ Thus, the correct variation is shown in figure $(b).$

The curves $(a), (b), (c)$ and $(d)$ show the variation between the applied potential difference $(V)$ and the photoelectric current $(i),$ at two different intensities of light $(I_1 > I_2).$ In which figure is the correct variation shown?

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