The FWHM of the Image Quality profile

If we suppose the seeing has a gaussian profile:

$$I(x)=\exp(-\frac{x^{2}}{\sigma^{2}})$$

$$\int I(x) dx=\sqrt{\pi}\sigma$$

$$FWHM=1.665\sigma=2\sqrt{\ln(2)}\sigma$$

$$erf(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}\exp(-t^{2}) dt=\frac{... ...rac{x^{2}}{1!3}+\frac{x^{4}}{2!5}-\frac{x^{6}}{3!7}+...\right)$$

Where we have taken in count that

$$erf(\infty)=1.0$$

defining:

$$J(x)=\exp(-\frac{x^{2}}{2\sigma^{2}})$$

In this case we have:

$$FWHM=2.35482\sigma$$

$$\int J(x) dx=\sqrt{2\pi}\sigma$$

$$G(x)=\frac{1}{\sqrt{\pi}\sigma}\exp(-\frac{x^{2}}{2\sigma^2})$$

$$\int G(x)=1$$

For example, for a gaussian which has a maximum value of I₀=141.55 and a FWHM of 840 Å. The integral is:

$$I=141.55\times\frac{840}{1.665}\sqrt{\pi}=126575.56$$ (2.8)

or

$$I=141.55\times\frac{840}{2.355}\sqrt{2\pi}$$ (2.9)

$$I_{tot}=I_{max}\cdot FWHM\cdot 1.0645$$ (2.10)

So the FWHM of the spatial profile can be calculated if one knows:

• The spatial profile distribution or
• I_tot, I_max