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(D) $1$. The saturation current is directly proportional to the intensity of the incident light. Since the intensity is reduced, the saturation curren

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$D$

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(D) $1$. The saturation current is directly proportional to the intensity of the incident light. Since the intensity is reduced, the saturation current must decrease, which means the new curve must lie below the original broken-line curve.$2$. The stopping potential is directly proportional to the frequency of the incident light $(eV_s = h
u - \phi)$. Since the frequency is increased, the magnitude of the stopping potential increases (it becomes more negative). This means the curve must intersect the negative potential axis at a point further to the left than the original curve.$3$. Comparing the options, curve $D$ has a lower saturation current and a larger magnitude of stopping potential compared to the original broken-line curve. Therefore, curve $D$ represents the new situation.

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