Intensity of the double slit experiment

Two slits emit circular waves on the left edge waves of the simulation. These slits are a distance $d$ apart. The intensity is measured on a screen on the right edge of the simulation which is a distance $L$ from the slits. The intensity is plotted as $I/I_0$ where $I-0$ is the maximum intensity which occurs at $y=0$. The wavenumber $k$ is related to the wavelength $\lambda$, $k=\frac{2\pi}{\lambda}$.

There is a red point on the intensity pattern that displays the intensity in a region of 6 × 6 cm. The coordinates of this point are $P_x$ and $P_y$. To the left of the intensity pattern is a representation in the complex plane of the harmonic oscillations at the red point. The blue and green phasors represent the harmonic motion caused by the waves traveling from the two slits. The slit 1 is indicated by the green point and slit 2 is indicated by a blue point. The red phasor is the sum of the blue and green phasors. The real part of the red phasor is the motion observed at the red point.

The distance from the slit 1 to a point at height $y$ on the screen is,

$r_1 = \sqrt{L^2+(y-d/2)^2},$

and the distance from the slit 2 to point $y$ is,

$r_2 = \sqrt{L^2+(y+d/2)^2}.$

The waves leave the slits with the same amplitude $A$ and the same phase (which we choose to be zero). The waves interfere at the screen. The amplitude of the waves at the screen is,

$\frac{A}{\sqrt{r_1}} \exp (ikr_1)+\frac{A}{\sqrt{r_2}} \exp (ikr_2).$

The intensity is the square of the amplitude,

$ I \propto \left(\frac{A}{\sqrt{r_1}} \cos(kr_1)+\frac{A}{\sqrt{r_2}} \cos(kr_2)\right)^2+\left(\frac{A}{\sqrt{r_1}} \sin(kr_1)+\frac{A}{\sqrt{r_2}} \sin(kr_2 )\right)^2.$

This is the exact result for the interference pattern. It is plotted as the solid black line above. However, by making an approximation, it can be expressed in a simpler form.

Far-field approximation   $n\lambda << d << L$

For $y < < L$, $A/\sqrt{r_1} \approx A/\sqrt{r_2} \approx A/\sqrt{L}$. Using the trigonometric identities, $\sin a+\sin b=2\sin ((a+b)/2)\cos ((a-b)/2)$ and $\cos a +\cos b=2\cos ((a+b)/2)\cos ((a-b)/2)$

$ I \propto \frac{4A^2}{L} \left(\sin^2((kr_1+kr_2)/2)+\cos^2((kr_1+kr_2)/2)\right)\cos^2((kr_1-kr_2)/2).$

$ I \propto \frac{4A^2}{L} \cos^2\left( \frac{kr_1-kr_2}{2}\right).$

$ I \propto \frac{4A^2}{L} \cos^2\left( \frac{kdy}{2L}\right).$

The factor $\cos^2\left( \frac{kdy}{2L}\right)$ has a maximum when $n\pi = \frac{kdy}{2L}$ so constructive interference will take place when $n\lambda=yd/L$ where $n$ is an integer. Destructive interference will take place when $(n+1/2)\lambda=yd/L$. The initial assumption for this approximation was that $y < < L$. The result for $y$ that was then calculated was $y = n\lambda L/d$. This implies that a condition that the far-field approximation to be valid is $n\lambda << d$. The approximate solution for the far field is plotted in light gray in the plot in the upper right.