Full Width Half Max question??

May 2010
I have a plotted a graph of:
\(\displaystyle \frac{I(t)}{|E_p|^2}=e^{(-2\alpha t^2)}\)

Now I understand that the maximum is when t=0, however what about the full width half max?? Why is the half height 1/e??

Can someone help me obtain the full width half max of this?

Last edited:


MHF Helper
Apr 2005
I had to google "full width half max". Apparently, it is the distance between \(\displaystyle x_1\) and \(\displaystyle x_2\) for points such that \(\displaystyle f(x_1)= f(x-2) = \) 1/2 the maximum value of the function.

You have correctly calculated that the maximum value of this function occurs at x= 0 and, so, is e. But the "half max" is not 1/e, it is, as you should guess, half of e: e/2.

You need to solve \(\displaystyle f(x)= e^{-2\alpha x^2}= \frac{e}{2}\). Since f is symmetric about x= 0, the "full width at half maximum" is twice that x value.