Fringe Spacing Understanding

Provided everything remains constant, does the fringe separation, that is the distance between adjacent fringes become further apart for higher order maxima?

Consider the above diagram. From the diagram it appears that the distance between adjacent maxima is getting larger with distance from the central maxima. My intuition agrees with this. However, I have heard that the distance between adjacent fringes is constant. Assume this is true. How so?

Double slit interference pattern: Fringes are equally spaced and of equal widths.

The above statement seems highly unintuitive. Shouldn't the spacing between adjacent fringes get larger, as the wave spreads out. How do I satisfy myself with the above statement? Proof?

The equation you need to consider is that which gives the angle $$\theta_{\rm n}$$ for the $$n^{\rm th}$$ bright fringe, $$n\lambda = d \sin \theta_{\rm n}$$, where $$\lambda$$ is the wavelength and $$d$$ the separation of the slits.

The angular separation of adjacent bright fringes is $$\theta_{\rm n+1} -\theta_{\rm n}$$.

If the angle $$\theta_{\rm n}$$ is small then an approximation can be made $$\sin \theta_{\rm n}\approx \theta_{\rm n}$$.

So the angular separation of the fringes is approximately $$\frac{(n+1)\lambda}{d} - \frac{n\lambda}{d} = \frac \lambda d$$ which is constant.
The separation of the fringes on a screen a distance $$D$$ from the double slit is $$x_{\rm n+1}-x_{\rm n}$$ where $$x$$ is the distance from the central maximum and $$\theta \approx \frac x L$$.

Thus $$x_{\rm n+1} - x_n \approx \frac {\lambda L}{d}$$ - constant fringe separation.

However the sine function is not linear with respect to $$\theta$$ and decreases less per increase in $$\theta$$ as $$\theta$$ increases.
This means that the distance between adjacent maxima will get bigger as $$\theta$$ increases and effect you may have well observed when looking at a spectrum using a diffraction grating which in really much more than a double slit but with many more slits.