1. In a photocell, when the anode voltage is kept fixed, the plate current depends mainly on the number of photoelectrons emitted per second that are collected by the anode.

2. As the wavelength $\lambda$ of incident light is changed:

   • For $\lambda > \lambda_0$ (where $\lambda_0$ is the threshold wavelength), the photon energy is $E = \frac{hc}{\lambda} < \phi$ so no photoelectric emission takes place. Hence, $I = 0$

3. For $\lambda < \lambda_0$, photoemission occurs. If the intensity of light is taken as fixed, then as wavelength decreases, energy per photon increases: $E = \frac{hc}{\lambda}$ so the number of incident photons per second decreases. Since photocurrent is proportional to the number of emitted electrons per second, the photocurrent decreases as $\lambda$ decreases.

4. Therefore, the graph of plate current $I$ versus wavelength $\lambda$ should be:

   • zero for $\lambda > \lambda_0$,
   • nonzero for $\lambda < \lambda_0$,
   • increasing with increasing $\lambda$ up to the threshold,
   • and then suddenly dropping to zero at $\lambda = \lambda_0$.

5. Among the given options, this corresponds to Option D.

Thus, the correct choice is: $\boxed{D}$