If we treat each of the slits as a point source of circular wavefronts. The condition for constructive interference (bright fringes) is

$d\sin\theta=m\lambda$ (m=0,1,2,..)

and for destructive interference (dark fringes)

$d\sin\theta=(m+\frac{1}{2})\lambda$ (m=0,1,2,..)

As in the case of a wave on a rope that is incident on a heavier rope and is reflected with a 180^{o} phase change when a light wave is reflected from a more optically dense media a 180^{o} phase change occurs. This effect is important when we want to consider interference effects in thin films.

For a single wavelength dark stripes will occur whenever

$2t=m\lambda$ (m=0,1,2,..)

and bright stripes will occur whenever

$2t=(m+\frac{1}{2})\lambda$ (m=0,1,2,..)

Ideally we would use a coating that produced an equal amount of reflection at both interfaces, but there is no suitable material with the required refractive index, $n=1.26$, so we use magnesium flouride MgF_{2}.

As the two reflections both occur from more optically dense media they both experience a phase change of $\pi$ on reflection which corresponds to advancing the wave by $\frac{\lambda}{2}$ .

To have the light be out of phase we need to light that goes through the coating to have advanced by $\frac{\lambda}{2}$ for destructive interference to occur.

Critically, when destructive interference occurs the light is not lost, but is instead transmitted.

As the wavelength of light in a medium is given by $\lambda=\frac{\lambda_{0}}{n}$ where $n$ is the refractive index of the medium and $\lambda_{0}$ is the wavelength of the light in free space, the thickness of the coating should be $\frac{\lambda}{4n_{2}}$.

As we now have the intensity in terms of the phase difference $\delta$

$I_{\theta}=I_{0}\cos^{2}\frac{\delta}{2}$

we should return to our diagram to find the intensity in terms of $d$ and $\theta$

We can see that

$\frac{\delta\lambda}{2\pi}=d \sin \theta$

giving

$\delta=\frac{2\pi}{\lambda}d\sin\theta$

and the intensity is

$I_{\theta}=I_{0}\cos^{2}(\frac{\pi d \sin \theta}{\lambda})$

Rather than writing our equation in terms of k

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{Dk}{2}\sin\theta)$

it is convenient to write it in terms of the wavelength, using $k=\frac{2\pi}{\lambda}$

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{D\pi}{\lambda}\sin\theta)$

This function has a central maxima and then minima at

$D\sin\theta=m\lambda$ $m=\pm 1,\pm 2,\pm 3,..$

Now we have an expression for the intensity from a slit of width $D$

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{D\pi}{\lambda}\sin\theta)$

we can use consider slits as sources instead of the point sources we considered earlier which gave

$I_{\theta}=I_{0}\cos^{2}(\frac{d \pi}{\lambda}\sin \theta)$

The combination of these gives

$I_{theta}=I_{0}\mathrm{sinc^{2}}(\frac{D\pi}{\lambda}\sin\theta)\cos^{2}(\frac{d\pi}{\lambda}\sin \theta)$

Irrespective of the number of slits $n$ the condition for maxima is the same as for the double slit

$d\sin\theta=m\lambda$ (m=0,1,2,..)

but that the larger the number of slits from which diffraction occurs the sharper the maxima will be.

n=2 $\frac{I_{\theta}}{I_{0}}=\frac{(2+2\cos\delta)}{4}$

n=3 $\frac{I_{\theta}}{I_{0}}=\frac{3+4\cos\delta+2\cos 2\delta}{9}$

n=4 $\frac{I_{\theta}}{I_{0}}=\frac{4+6\cos\delta+4\cos 2\delta+2\cos 3\delta}{16}$

For any $n$

$\frac{I_{\theta}}{I_{0}}=\frac{n+\sum\limits_{k=1}^{n-1}2(n-k)\cos(k\delta)}{n^{2}}$

The polarization of the light after it has passed through the polarizer is the direction defined by the polarizer, so the way to approach the calculation of the magnitude of the intensity that passes through the polarizer is by finding the component of the electric field which is in that direction

$E=E_{0}\cos\theta$

the intensity is then

$I=E^{2}=E_{0}^{2}\cos^{2}\theta=I_{0}\cos^{2}\theta$

This formula is known as Malus' Law.