# Young's Double Slit Experiment

If a light source is placed above a horizontal mirror in front of a vertical screen, why is there a dark spot formed at the junction of the screen and mirror? Shouldn't it work the same way as a a double slit apparatus and form a bright spot?

Just guessing here: Does the reflection cause a phase difference of π which causes destructive interference?

The reflection on the mirror causes a phase difference of $\pi$ which causes destructive interference. A mirror is often made of a thin layer of metal, which, in first approximation, doesn't let the electromagnetic field to pass through. Thus, we can say that
$$\mathbf{E}_{inc} + \mathbf{E}_{ref} = \frac{\sigma}{\epsilon_0}\mathbf{u}$$
where $\mathbf{E}_{inc}$ (resp. $\mathbf{E}_{ref}$) is the incident (resp. reflected) field, $\sigma$ the surface charge density and $\mathbf{u}$ a unit vector perpendicular to the mirror, directed toward the exterior of the mirror. Now, since here the electric field is locally flat, you can show that there is a phase difference of $\pi$ between $\mathbf{E}_{inc}$ and $\mathbf{E}_{ref}$, ie. the relfection cause a phase difference of $\pi$.