Photoelectric effect experiments are performed using three different metal plates p,q and r having work functions  ${\varphi }_p=2.0eV,{\varphi }_q=2.5eV,{\varphi }_r=3.0eV$ respectively. A light beam containing wavelengths of 550nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is[Take hc= 1240eV nm] 

For $550nm$, $E_1=\frac{1240}{550}=2.25eV$

For $450nm$, $E_2=\frac{1240}{450}=2.75eV$

For $350nm$, $E_3=\frac{1240}{350}=3.54eV$

${\varphi }_p<E_1,E_2,E_3$ ((all wavelengths can eject photoelectrons from p)

$E_1<{\varphi }_q<E_2, E_3$(only $E_2,E_3$can eject photoelectrons from q)

$E_1,E_2<{\varphi }_r<E_3$(only  $E_3$ can eject photoelectrons from r)

All wavelengths have equal intensity (say I)

Now, photocurrent is directly proportional to intensity of incident radiation that can cause ejection of photoelectrons

So, $i_p:i_q:i_r=3I:2I:l$

For p: $3.54=2+e{V}_p\Rightarrow V_p=1.54eV$
For q: $3.54=2.5+e{V}_q\Rightarrow V_q=1.04eV$

For r: $3.54=3.0+e{V}_r\Rightarrow V_r=0.54eV$