# On Superposition In Diffraction

I am a high school student learning Physics. I am currently learning about diffraction of waves, be it water waves, light waves each, in systems where there are 2 slits for it to diffract through.

1. Where are the peaks and where are the troughs of the waves in the diagrams? Are the blank spaces the troughs and the red lines the peaks?

2. I understand that constructive interference occurs when the peaks of 2 waves align, and destructive interference when the peak of the wave hits the trough of another wave. But why in the diagram above does constructive interference occur in the center (represented by bright)? And why does destructive interference occur on the 2 sides of the center (represented by dark), followed by another 2 zones of constructive interference (represented by bright)?

Can all this be explained at the level of a high school student, and without Calculus based mechanics?

We assume the screen is far away compared to the slit separation (which is the standard assumption). Let the slit separation be $d$, and look at a point on the screen A (it might be helpful to draw it out). Let the line connecting point A to the center of the slits make an angle $\theta$ with the line connecting the center of the screen to the center of the slits. Connecting each of the slits to point A and using similar triangles, for constructive interference, We have $$d\, sin \theta=n\lambda,\;\; n \in \mathbb Z^+$$ where $\lambda$ is the wavelength. Let me explain this formula. $d\, sin \theta$ is the difference in distance traveled by light from the 2 slits to point A. $n \in \mathbb Z^+$ because the two coherent beams must differ by whole number of wavelengths for the wavefronts to meet again. Similarly, for destructive interference, we have $$d\, sin \theta=(n + \frac{1}{2})\lambda,\;\; n \in \mathbb Z^+$$ because their distances must differ by half a wavelength for their amplitudes to cancel out. This can be used to further derive the fringe widths.