The Fourier transform definition you are using is:
$\mathfrak{F}(h(t))(\nu) = \tilde{H}(\nu) =\int_{-\infty}^\infty e^{-2\pi\,i\,\nu\,t} \,h(t)\mathrm{d}t$
In this convention, the transform variable is in $\mathrm{Hz}$, as opposed to radians per second, when $t$ is in seconds. So, your independent variable $t$ needs to be in seconds, to get the independent frequency variable to be in hertz.
Wavelength is inversely proportional to frequency $c = \nu \lambda$, so the transformation between frequency and wavelength spectrums is nonlinear. If your pulse is narrowband, so that its frequency spread is much less than its carrier frequency, you can use $\nu = \frac{c}{\lambda} \Rightarrow \Delta \nu \approx -\frac{c}{\lambda^2} \Delta \lambda = -\frac{\nu}{\lambda} \Delta \lambda$. The scaling constant therefore is:
$\nu \approx \nu_0 - \frac{\nu_0}{\lambda_0} (\lambda - \lambda_0) = 3.33\times10^{14} \mathrm{Hz} - 3.7\times10^{11}(\lambda - 900\mathrm{nm}) = 6.67\times10^{14} - 3.7\times10^{11}\lambda $
With $\lambda$ in nanometres. So, to convert your Fourier transform with transform variable of frequency in Hz to one with transform variable of wavelength in nanometres, you plug in:
$\tilde{H}(\nu) = \tilde{H}(6.67\times10^{14} - 3.7\times10^{11}\lambda)$
Again, this formula assumes $\lambda$ is in nanomatres. Beware, though, if your source is broadband (pulse duration less than 0.1 picoseconds) on interpreting power spectrums calculated from the Fourier transform. If your power spectrum is $G(\nu) = |\tilde{H}(\nu)|^2$, and you are trying to interpret $G$ as a power per unit wavelength, you must bring in the Jacobian of the nonlinear transformation. Given $\nu = \frac{c}{\lambda}$, witness that the power in the interval spanned by $[\nu_1,\nu_2]$ is
$\int_{\nu_1}^{\nu_2} G(\nu) d\nu = -\int_{\frac{c}{\nu_1}}^{\frac{c}{\nu_2}} G(\frac{c}{\lambda}) \frac{c}{\lambda^2} d\lambda$
So to get a power spectrum in power per wavelength, you have to use $\left|\tilde{H}\left(\frac{c}{\lambda}\right)\right|^2 \frac{c}{\lambda^2} $, not simply $\left|\tilde{H}\left(\frac{c}{\lambda}\right)\right|^2$