Two particles cannot occupy the same space – the particles will collide with each other. Two waves, however, can occupy the same space. The resulting excitation will be the sum of the excitation of the two waves. This is the superposition principle.

• $d$: the distance between the two sources.
• $D$: the distance between the sources and the screen.
• $y$: the vertical position of the screen.
• $\ell_1$: the distance between source one and a point on the screen $P$.
• $\ell_2$: the distance between source two and a point on the screen $P$.

As $D >> d$,

\[ \ell_2 - \ell_1 \approx \lambda d \sin \theta \]

\[ \sin \theta \approx \tan \theta = y/D. \]

For constructive interference, the waves coming from the two sources must come in with

\[ \ell_2 - \ell_1 = \lambda n, n \in \mathbb{Z} = \{ \ldots, -2, -1, 0, 1, 2, \ldots \}. \]

\[ d y/D = \lambda n \]

The spots on the screen will be at positions \[ y = \frac{ n \lambda D}{ d } \qquad \textrm{ for } n = 1, 2, 3, \ldots \]

For destructive interference, \[ y = \frac{ (2n - 1) \lambda D}{ 2d } \qquad \textrm{ for } n = 1, 2, 3, \ldots \]

A red laser beam with wavelength 650 nm is shined on a pair of slits separated by 2mm. The distance between the slits and screen is 1.2m. Find the position of the third bright fringe on the screen from the central maximum.

Sol: We can use the formula for constructive interference $y = \frac{n Dλ}{d}$ with $n=3$. The position will be $y_3 = \frac{ 3 \times 1.2 \times 650 x 10^{-9} }{ 0.002 } = 0.12$[cm].

The same scenario

electrons also work…

http://en.wikipedia.org/wiki/Double-slit_experiment