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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?

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1 Answer 1

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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
(a) not taken the time average (giving a factor of 1/2) and
(b) swapped out $A_0^2$ for $I_0$ rather than using $2A_0^2=I_0$ as the document uses.

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  • $\begingroup$ Assuming you know about phase interference, is there any difference between time-averaging the second factor and using the different conversion factor for intensity and amplitude? I.e., what assumptions does MIT make that my text doesn't? $\endgroup$
    – genap
    Commented Apr 27, 2015 at 1:05
  • $\begingroup$ 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. $\endgroup$
    – Kyle Kanos
    Commented Apr 27, 2015 at 1:08

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