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(C) $1$. The saturation photoelectric current is directly proportional to the intensity of the incident radiation. Since the intensity is reduced, the

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$C$

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(C) $1$. The saturation photoelectric current is directly proportional to the intensity of the incident radiation. Since the intensity is reduced, the saturation current for the new graph must be lower than that of the original (broken) line.$2$. The stopping potential is directly proportional to the frequency of the incident radiation (according to Einstein's photoelectric equation: $K_{max} = h
u - \phi = eV_s$). Since the frequency is increased, the magnitude of the stopping potential must increase (i.e., it becomes more negative).$3$. Comparing the given curves, curve $C$ has a lower saturation current than the broken line and a larger magnitude of stopping potential (shifted further to the left). Therefore, curve $C$ represents the new situation.

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