Double slit experiment

If we zoom out from the two slits we can observe the interference pattern in the far field. There are certain directions where constructive interference is observed and other directions where destructive interference is observed.

The interference can be calculated at a distance $L$ from the slits and a height $y$ from the center of the slits. The distance that the waves travel from the upper slit to this point is $r_1$ and the distance traveled from the lower slit is $r_2$. The difference between these two distances is the small red section $r_2-r_1=d\sin\theta$, where $d$ is the distance between the slits and $\theta \approx \frac{y}{L}$.

There will be constructive interference if $r_2-r_1$ is an integer number of wavelengths and there will be destructive interference if $r_2-r_1$ is an integer number of wavelengths plus one half wavelength.

In the Far field: $L >> d,\,\lambda$

Constructive interference $r_2-r_2=d\sin\theta \approx\frac{dy}{L}=n\lambda$ $n\in$ integers

Destructive interference $r_2-r_2=d\sin\theta \approx\frac{dy}{L}=\left(n+\frac{1}{2}\right)\lambda$ $n\in$ integers

Simulation details

A wave train fifteen wavelengths long strikes a barrier and reflects off of it. There are two narrow slits in the barrier each with a width $a$ that is much smaller than the wavelength $\lambda$ of the waves. Circular waves radiate from the slits. To the left of the barrier, the reflected left moving waves interfere with the incoming right moving waves and with the reflection from the slits.

The interference pattern on the left of the barrier is decribed by the superposition of a right moving wave, a left moving wave, and two circular waves from the slits.

$$z = -H(\xi_R+30\pi)\sin(\xi_R)H(-\xi_R) -H(\xi_L+30\pi)\sin(\xi_L)H(-\xi_L) +H(\xi_{c1}+30\pi)\frac{\sin(\xi_{c1})}{\sqrt{r_1}}H(-\xi_{c1})+H(\xi_{c2}+30\pi)\frac{\sin(\xi_{c2})}{\sqrt{r_2}}H(-\xi_{c2}),$$

where $k$ is the wavenumber, $\omega$ is the angular frequency, $r_1 = \sqrt{(x-L/2)^2+(y-0.45L)^2}$ is the distance from slit 1, $r_2=\sqrt{(x-L/2)^2+(y-0.55L)^2}$ is the distance from slit 2, $\xi_R = kx -\omega t$, $\xi_L = 30\pi-kx -\omega t$, $L=15\lambda=30\pi/k$ is the length of the wavetrain, $\xi_{c1} = kr_1-\omega\left(t-\frac{L}{2}\frac{k}{\omega}\right)$, $\xi_{c2} = kr_2-\omega\left(t-\frac{L}{2}\frac{k}{\omega}\right)$, and $H(x)$ is the Heaviside function.

The circular waves on the right side of the barrier are described by,

$$z = -H(\xi_{c1}+30\pi)\frac{\sin(\xi_{c1})}{\sqrt{r_1}}H(-\xi_{c1})-H(\xi_{c2}+30\pi)\frac{\sin(\xi_{c2})}{\sqrt{r_2}}H(-\xi_{c2}).$$






	Physik für Geodäsie 511.018 / Physik M 513.805