The stopping potential when a metal surface is illuminated by light of wavelength $\lambda$ is 15 V . The stopping potential when the same surface is illuminated by light of wavelength $4 \lambda$ is 3 V . The ratio of threshold wavelength to the initial incident wavelength $\lambda$ is:

From Einstein's photoelectric equation, the stopping potential $V_s$ is given by $eV_s = \dfrac{hc}{\lambda} - \Phi$, where $\Phi$ is the work function of the metal and $\lambda_0$ is the threshold wavelength such that $\Phi = \dfrac{hc}{\lambda_0}$.

For the first case, $eV_1 = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0} = 15e$. Thus, $\dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0} = 15e$ (Equation 1).

For the second case, $eV_2 = \dfrac{hc}{4\lambda} - \dfrac{hc}{\lambda_0} = 3e$. Thus, $\dfrac{hc}{4\lambda} - \dfrac{hc}{\lambda_0} = 3e$ (Equation 2).

Subtracting Equation 2 from Equation 1:

$\left( \dfrac{hc}{\lambda} - \dfrac{hc}{4\lambda} \right) = 15e - 3e = 12e$

$\dfrac{3hc}{4\lambda} = 12e \Rightarrow \dfrac{hc}{\lambda} = 16e$.

Substituting $\dfrac{hc}{\lambda} = 16e$ into Equation 1:

$16e - \dfrac{hc}{\lambda_0} = 15e \Rightarrow \dfrac{hc}{\lambda_0} = e$.

Since $\dfrac{hc}{\lambda} = 16e$ and $\dfrac{hc}{\lambda_0} = e$, we have $\dfrac{hc}{\lambda} = 16 \left( \dfrac{hc}{\lambda_0} \right)$, which implies $\dfrac{1}{\lambda} = \dfrac{16}{\lambda_0}$.

Therefore, $\dfrac{\lambda_0}{\lambda} = 16$, or the ratio of threshold wavelength to the initial wavelength is $16:1$.

Answer: $16: 1$