Question

Intensity distribution for double slit experiment.

I am trying to remember how we get the intensity distribution for slits separated by a distance d, on a screen L distance away, and wavelength \(\lambda\)

Answer 1

so Intensity (or I guess it's called irradiance now) is the time average of the Poynting vector right? \[I=\left\langle S\right\rangle _T\]so for linearly polarized light I would have \[I=\epsilon_0 c\left\langle E^2\right\rangle_T\]

Answer 2

ok now i'm gonna have to deal with the superposition of two waves right? \[E=E_1+E_2\]

Answer 3

If you want to derive it, you have to deal with interference - the phase difference between the electric fields from the two slits. You add them together to get the total, and then use your correct formula for intensity above.

Answer 4

\[E^2=\langle E_1\rangle^2+\langle E_2\rangle^2+2E_1\cdot E_2\]

Answer 5

\[\mathbf{E_1}=\mathbf{E_0}\cos(\mathbf{k}\cdot\mathbf{r_1}-\omega t)\]\[\mathbf{E_2}=\mathbf{E_0}\cos(\mathbf{k}\cdot\mathbf{r_2}-\omega t+\phi)\]

Answer 6

Express it in terms of +/- phi/2, and add them together before squaring. Remember you can take time averages.