- #1

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$$I=\int_{0}^{T_{eye}}e^2dt$$

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?