The curves (1),(2),(3),and (4) show the varia | 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 frequenc

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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 incident radiation.Given that $I_1 > I_2$,the saturation current for intensity $I_1$ will be greater than that for intensity $I_2$,i.e.,$i_1 > i_2$.However,the stopping potential $(V_0)$ depends only on the frequency of the incident light and the nature of the material of the photosensitive surface,not on the intensity of the light.Therefore,both curves must intersect the voltage axis at the same stopping potential $(V_0)$.Comparing this with the given options,the figure that shows different saturation currents but the same stopping potential is represented by the curve where both lines start from the same point $-V_0$ on the voltage axis.

The curves $(1)$,$(2)$,$(3)$,and $(4)$ 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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