The data given is
ϕAl=4.1eV
ϕAu=5.1eV
The energy is given by
hν=ϕ+KE⇒hν=ϕ+eV0
where V0 is the stopping potential and K=eV0
Thus, the above equation can be written as
eV0=hν−ϕ
⇒V0=ehν−eϕ
The equation is of the form of y=mx+c. The slope is
m=eh.
Since the Planck constant h and charge of an electron is a constant e, the slope does not depend on the work function. Hence the ratio of the slope of the plot of stopping potential and frequency for both the given metals is 1.