I need to calculate the fourier transforms of these functions
single slit: $$ f(x)=1 $$ with $ \frac{-b}{2}<x<\frac{b}{2} $
and $ f(x)=0 $ elsewhere
double slit: $$ f(x)=1 $$ with $ \frac{-(g+b)}{2}<x<\frac{-(g-b)}{2}, \frac{(g-b)}{2}<x<\frac{(g+b)}{2} $
and $ f(x)=0 $ elsewhere
where b gives the slit width and g the slit distance.
For the single slit, using the definition of a fourier transform, the result for $f(x)=1$ must be the delta function, with $x_0=0$, so $\delta(x)$, right? and still $0$ elsewhere.
for the double slit, again using the definition of fourier transform, is it correct to say that $$ F(k)=\delta(x+g), F(k)=\delta(x-g), $$ with $F(k)=0$ elsewhere.
Is this correct? Is there a more intuitive way of looking at the results of this? I Think i've done the maths correctly but
You are not correct. The Fourier transform is absolutely not a delta function, since $f(x)$ isn't constant everywhere.
There are several definitions of the Fourier transform. I don't know which one you use, so I'll use mine
$$ F(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi ik x} dx = \int_{-b/2}^{b/2}e^{-2\pi ik x} dx = \frac{e^{\pi ik b}-e^{\pi ik b}}{2\pi ik} = \frac{\sin(b\pi k)}{\pi k} = b \operatorname{sinc}(b\pi k) $$
For the double-slit, we invoke the Fourier shift theorem $$ \int_{-\infty}^{\infty} f(x-g)e^{-2\pi ikx}dx = \int_{-\infty}^{\infty} f(x)e^{-2\pi ik(x+g)} dx = e^{-2\pi ig k} F(k) $$
So the transform turns out to be $$ F(k) = \big( e^{2\pi i gk} + e^{-2\pi i gk}\big)b\operatorname{sinc}(b\pi k) = 2b\cos(2\pi gk)\operatorname{sinc}(b\pi k) $$