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

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(B) Given that the intensity of light $I_1 > I_2$. According to the laws of the photoelectric effect,the photoelectric current $(i)$ is directly propo

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(B) Given that the intensity of light $I_1 > I_2$. According to the laws of the photoelectric effect,the photoelectric current $(i)$ is directly proportional to the intensity of the incident light. Therefore,the saturation current for intensity $I_1$ will 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 nature of the material,not on the intensity of the light. Thus,for both intensities $I_1$ and $I_2$,the stopping potential $(V_0)$ must be the same. Looking at the given figures,Figure $(b)$ correctly shows that both curves have the same stopping potential $(V_0)$ on the negative $V$-axis,while the saturation current for $I_1$ is higher than for $I_2$.

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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