Missing factor of 1/2 in the definition of light intensity The light field can be written as $E=A \cos\theta$ where $\theta=kz+wt$ is the phase of light wave. 
What we measure or see is the intensity which is the square of the light field
$$I=E^2=A^2 \cos^2\theta$$
Can anyone explain   why most of the textbooks directly show $\ I=E^2=A^2 \cos^2\theta=A^2$ with the reason the phase is changing very fast.
I am confused  why the fast change results in  $\ I=A^2$ rather than $\ I=(1/2)A^2$.
 A: I would say: do not worry about the 1/2, the only thing you need to know
is that it is proportional to the square of the field. You probably do
not want to get really quantitative, so the factor 1/2 does not really
matter.
If you really want to get quantitative, then you should avoid the word
“intensity”, and be more specific about what you mean. Is
it the Poynting vector?
The irradiance?
The radiance?
The radiant intensity?
“Intensity” is pretty vague. From the context, it sounds like you mean
irradiance on a plane perpendicular to the Poynting vector. But when an
textbook author uses this term, it generally means that he/she does not
want to get into exact radiometric definitions. Only to convey the
general idea that your eye/detector only cares about the square of the
field.
BTW, you are right in that averaging cos2(θ) over θ yields 1/2.
A: You're correct in that the angular average $\langle \cos^2 \theta \rangle = 1/2$; the simplest way to see this is to take the average of the trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$, and use the fact that $\langle \cos^2 \theta \rangle = \langle \sin^2 \theta \rangle$. Thus, if your textbook defined the irradiance as $I\equiv \langle \mathbf E^2(\theta)\rangle$, where the electric field $\mathbf E(\theta) = \mathbf E_0 \cos \theta$, then the correct equation is:
$$ I= \frac12 \mathbf E_0^2 $$
However, as Sofia mentioned in the comments, it is also common to represent electromagnetic fields in a complex notation $\mathbf{\tilde E}(\theta) = \mathbf {\tilde E}_0 e^{i\theta}$ to simplify calculations, where the physical field is understood to be the real part $\mathbf{E(\theta)} = \text{Re}\big\{\mathbf{\tilde{E}}(\theta)\big\}$. In this notation, the irradiance is usually defined as $I \equiv \langle \mathbf{\tilde E}^*\!\!(\theta)\cdot\mathbf{\tilde E}(\theta) \rangle$, in which case you get:
$$ I = \left|\mathbf{\tilde E}_0\right|^2$$
A: I like to explain this in a simple way.This  example might give you some feeling about intensity.
consider a single bulb glowing it has some intensity,say "I" and you turn on the second bulb now the intensity has become twice than before i.e. "2I"
And more technically it is;  power transferred per unit area.
