# Formula Double slit experiment (approximation formula) Path difference    Slit distance    Fringe separation    Screen distance

## Path difference

Unit
The path difference tells by how many wavelengths $$\lambda$$ the light wave through the upper slit, differs from the light wave through the lower slit. In the case of constructive interference, the difference is a multiple of the wavelength: $\Delta s = m \, \lambda$With destructive interference, the path difference is: $\Delta s = \left( m-\frac{1}{2} \right) \, \lambda$Here $$m$$ is an integer 1, 2, 3, ... It numbers the light and dark stripes, respectively. For the first maximum / minimum is $$m = 1$$. For the fifth maximum / minimum is $$m = 5$$.

## Slit distance

Unit
Distance between the two slits (measured from the center of the slit). The further apart the two slits are, the closer the interference fringes are to each other.

## Fringe separation

Unit
Distance of the main maximum (which is exactly in the center of the screen) to a bright or dark interference fringe. If the condition for constructive interference is used for the path difference $$\Delta s$$, then this is the distance to a bright fringe (maximum). If the destructive interference condition is used, this is the distance to a dark fringe (minimum).

## Screen distance

Unit
The distance of the double slit to the detector screen. Here the approximation is used that the screen is far away from the double slit, compared to the fringe distance, that is: $$a$$ must be much larger than $$x$$ for this formula to be used.