# Disparity between two texts on intensity and phase interference - which is right?

So, I'm learning phase interference.

Imagine we have two waves.

$$E_1 = A_0sin(wt)$$

and

$$E_2 = A_0sin(wt+\phi)$$

With

$$\phi = \frac{2\pi}{\lambda}dsin(\theta)$$

Which is the path difference.

So, if we add the two together, we get

$$E_t = E_1 + E_2 = A_0sin(wt) + A_0sin(wt+\phi)$$

Which can be simplified to

$$2A_0cos(\frac{\phi}{2})sin(wt + \frac{\phi}{2})$$

And as the intensity is proportional to the square of the amplitude, we can thus say that

$$I = 4I_0cos^2(\frac{\phi}{2})$$

Which, when plugging in $\phi$, gives us

$$I = 4I_0cos^2(\frac{\pi}{\lambda}dsin(\theta))$$

However, a guide on the matter published by MIT states that the answer is actually

$$I = I_0cos^2(\frac{\pi}{\lambda}dsin(\theta))$$

Where did the 4 go?

You have that $$E=2A_0\cos\left(\frac{\phi}{2}\right)\sin\left(\omega t + \frac{\phi}{2}\right)$$ which is correct. To get the intensity, you then square and time average this: \begin{align} I=\langle E^2\rangle&=4A_0^2\cos^2\left(\frac{\phi}{2}\right)\left\langle\sin^2\left(\omega t + \frac{\phi}{2}\right)\right\rangle\\ &=4A_0^2\cos^2\left(\frac{\phi}{2}\right)\cdot\frac12\\ &=2A_0^2\cos^2\left(\frac{\phi}{2}\right)\\ &=I_0\cos^2\left(\frac{\phi}{2}\right) \end{align} Which is not the same as your relation because you've
(b) swapped out $A_0^2$ for $I_0$ rather than using $2A_0^2=I_0$ as the document uses.
• Well formally you are time-averaging the whole thing: $I=\langle E^2\rangle$. It is just that only the sine term has any dependence on time. Apr 27, 2015 at 1:08