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I don't understand why we use this formula:

enter image description here

Instead of this formula:

enter image description here

I mean, why we ignore $\omega\cdot t$ part.

Artur Owczarek
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2 Answers2

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I mean, why we ignore $\omega\cdot t$ part.

If you use the real representation

$$E\left(x,t\right)=\dfrac{1}{2}E_{0}\cos\left(kL_{1}-\omega t\right)+\dfrac{1}{2}E_{0}\cos\left(kL_{2}-\omega t\right)=E_{0}\cos\left(k\dfrac{L_{1}+L_{2}}{2}-\omega t\right)\cos\left(k\dfrac{L_{1}-L_{2}}{2}\right)$$

The oscillations in time of an electric field in the optical domain are so fast, our eyes and also electronic detectors just can't catch them. What you actually see is the time average of the intensity $I=\left<\left|E\left(x,t\right)\right|^{2}\right>_{t}$. Since $\left<\cos^{2}\omega t\right>=\frac{1}{2}$ over one period of oscillation, you get

$$I=\left<\left|E\left(x,t\right)\right|^{2}\right>_{t}=\dfrac{1}{2}\left|E_{0}\right|^{2}\cos^{2}\left(\dfrac{k\Delta L}{2}\right)=\dfrac{1}{4}\left|E_{0}\right|^{2}\left(1+\cos\left(k\Delta L\right)\right)$$

As you can see, including time just gives you an additional numerical factor and doesn't change the spatial interference pattern. Therefore, in most calculations you just ignore it.

eranreches
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  • I have another question. Can you tell me why amplitude of splitted wave is $\frac{1}{2}E_{0}$? I thought intensity is divided in half not amplitude of wave. Am I right? – Artur Owczarek Jun 17 '19 at 12:33
  • @ArturOwczarek Of course, I just used it because it is convenient the resulting sum of the two has amplitude $E_{0}$ without a factor. – eranreches Jun 17 '19 at 12:36
  • So in the quoted fragment of book is mistake? It should be $\frac{1}{\sqrt{2}}\varepsilon_{0}$? – Artur Owczarek Jun 17 '19 at 13:06
  • I think they use $\varepsilon=\frac{1}{2}\varepsilon_{0}e^{ikx}$ because then $\varepsilon+\varepsilon^{\ast}=\varepsilon_{0}\cos kx$. – eranreches Jun 17 '19 at 13:13
  • But at first, it is written that input beam has amplitude $\varepsilon_{0}$. Later it is only shown the conditions when field maxima and minima occurs. There is no use $\varepsilon+\varepsilon^{*}=\varepsilon_{0}cos{kx}$. So I think there is mistake, but I'm not sure and I ask you. :) – Artur Owczarek Jun 17 '19 at 13:33
  • If you assume the laser beam before the beam splitter has an amplitude $\varepsilon_{0}$, then after the beam splitter the amplitude is $\frac{1}{\sqrt{2}}\varepsilon_{0}$. Now if $\varepsilon$ is the complex field, it depends if you decide on writing $E={\rm Re}\varepsilon$ or $E=\varepsilon+\varepsilon^{\ast}$. – eranreches Jun 17 '19 at 13:37
  • So in the first case the original beam is $\varepsilon=\varepsilon_{0}e^{ikx}$ and after the beam splitter $\varepsilon=\frac{1}{\sqrt{2}}\varepsilon_{0}e^{ikx}$, while in the second case the original beam is $\varepsilon=\frac{1}{2}\varepsilon_{0}e^{ikx}$ and after the beam splitter $\varepsilon=\frac{1}{2\sqrt{2}}\varepsilon_{0}e^{ikx}$. In both cases before the beam splitter $E=\varepsilon_{0}\cos\left(kx\right)$ and after $E=\frac{1}{\sqrt{2}}\varepsilon_{0}\cos\left(kx\right)$ – eranreches Jun 17 '19 at 13:39
  • Ok, thanks very much! :) – Artur Owczarek Jun 17 '19 at 13:42
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I mean, why we ignore $\omega\cdot t$ part.

So why did you not ask this question when the double slit arrangement was being discussed and the positions of maxima and minima were found?

In both cases you need to find the resultant at a certain position of two superposing waves which have the same frequency, $\omega$, but differ in phase (and amplitude).

The resultant wave at the position in question will have the same frequency as the two superposing waves and an amplitude which depends on the phases (and amplitudes) of the superposing waves.

This is equivalent to saying that $e^{-i\omega t}$ is common to each term in the equation $\mathcal E^{\rm out}= \mathcal E_1 + \mathcal E_2$ and can be cancelled out.

Farcher
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