The figure (1) shows the variation of photo current with anode potential for a photo-sensitive surface for three different radiations. Let ${\mathrm{I}}_{\mathrm{a}},{\mathrm{I}}_{\mathrm{b}}\text{ and }{\mathrm{I}}_{\mathrm{c}}$ be the  intensities and $f_a,f_b\text{ and }{f}_c$ be the frequencies for the curves a.b and c respectively
 

It is obvious from the figure (1) that tho stopping potential for curve (a) and (b) is the same. Therefore ${\mathrm{f}}_{\mathrm{a}}={\mathrm{f}}_{\mathrm{b}}$ This is because, the stopping potential V₀ is frequency dependent $\left(\mathrm{h}\vee -{\mathrm{W}}_0={\mathrm{eV}}_0\right)$. Here W₀  is the work function of the metal. We also know that saturation current is proportional to intensity of the incident radiation. So ${\mathrm{I}}_{\mathrm{a}}<{\mathrm{I}}_{\mathrm{b}}\text{ i.e., }{\mathrm{I}}_{\mathrm{a}}\ne {\mathrm{I}}_{\mathrm{b}}$