# I Derivation about the wave interference

#### christang_1023

Summary
There are two coherent waves, and I was trying to derive the interference of them but failed, only to get $I=\frac{1}{2}(a_1^2+a_2^2)T_{eye}$.
Starting from the simple case, there is a single wave $e=a\cos(2\pi ft+\frac{2\pi}{\lambda}x+\phi_0)$, and integrate in such a way, where $T_{eye}$ stands for the response time of human eyes' response time towards energy change:
$$I=\int_{0}^{T_{eye}}e^2dt$$
The calculation includes triangular identity and approximation(i.e. there is a part $\frac{\sin(a)-sin(b)}{4\pi f}\approx 0$, due to $f>>2$). The result of the integration is $I\approx \frac{1}{2}a^2T_{eye}$, showing that we can only observe the average energy change of a wave.

Then, I consider a more complex case in which there are two waves, $e_1=a_1\cos(2\pi ft+\frac{2\pi}{\lambda}x+\phi_0),e_2=a_2\cos(2\pi ft)$.
$$I=\int_{0}^{T_{eye}}(e_1+e_2)^2dt$$,
After expanding it, $I=\frac{1}{2}(a_1^2+a_2^2)T_{eye}+\int_{0}^{T_{eye}}2(e_1e_2)dt$. Concerning the integration, I use the same approximation mentioned above, such that $\int_{0}^{T_{eye}}2(e_1e_2)dt\approx 0$.

Finally, I get $I=\frac{1}{2}(a_1^2+a_2^2)T_{eye}$, which doesn't indicate the sign of bright and dark pattern.

Is there anything wrong with my approximation or integration?

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#### Ibix

You seem to have a wave travelling in the x direction and a wave that isn't travelling. Is this a circumstance you would expect to lead to visible interference? Or, indeed, one you would expect to exist at all?

Under what circumstances would you expect detectable interference?

#### sophiecentaur

Gold Member
You seem to have a wave travelling in the x direction and a wave that isn't travelling.
Yes. this is a strange model.
The value of the second 'oscillation' is independent of position so can it be a wave?

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