# B Intensity of two waves

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1. Mar 17, 2016

### Cozma Alex

Lets consider two waves, the intensity of each wave is proportional to the square of their amplitude, now why the intensity of the sum of the two waves is not the sum of each intensity but is proportional to the square of the sum of the amplitudes? (i know it on the math side and i can demonstrate that but i don't get the physical meaning of that)
Whats the meaning of that? Why is the intensity of the result wave different from the sum of the intensity of the waves? Does this mean that there's a difference between the intensity of two waves and the intensity of the resultant wave, but why this happens?
There are points where the sum of the amplitude is zero then there is no intensity, and others where the intensity is even greather than the sum of the intensity of each wave, i can't get why, doesn't this violate the energy conservation?

Last edited: Mar 17, 2016
2. Mar 17, 2016

### blue_leaf77

The wave equation is a linear equation, if you have two or more waves, the will superpose and become a single wave of its own. In short, it's because waves can superpose.
How do you define the terms "two waves" and "resultant wave"?
The energy is conserved, but unfortunately at the moment I don't have enough free time to brainstorm a representative example for this issue. I will leave this part to another member.

3. Mar 17, 2016

### Staff: Mentor

It simply redistributes energy from the locations of the minima to the locations of the maxima. It doesn't create new energy at the maxima or destroy energy at the minima. For simple two-slit interference, if the original beams each would produce uniform intensity I0 on the observing screen, their combination produces instead of a uniform 2I0, the intensity pattern $$I = 4I_0 \cos^2 \left( \frac{\delta}{2} \right)$$ where $\delta$ is the phase difference between the two beams at position x on the screen. For Fraunhofer diffraction and small angles, $\delta$ is proportional to x (x = 0 being the center of the screen): $$\delta = 2 \pi \frac{d}{\lambda} \frac{x}{L}$$ (L = distance from slits to screen, d = spacing between slits)

Last edited: Mar 17, 2016