Two parallel slits are illuminated with monochromatic light of wavelength 500 nm. An interference pattern is formed on a screen some distance from the slits, and the fourth dark band is located 1.68 cm from the central bright band on the screen. (a) What is the path length difference corresponding to the fourth dark band? (b) What is the distance on the screen between the central bright band and the first bright band on either side of the central band? (hint: The angle to the fourth dark band and the angle to the first band are small enough that $\mathbf{tan\theta }\mathbf{\approx }\mathbf{sin\theta }$)

When the interference between two waves is completely destructive, their phase difference is given by:

$\mathbf{\varphi }\mathbf{=}\left(2m+1\right)\mathbf{\pi }\mathbf{,}\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

The equivalent condition is that their path length difference is an odd multiple of$\frac{\mathit{\lambda }}{\mathit{2}}$, where $\mathbf{\lambda }$ is the wavelength of the light.

The first dark band is produced by a path length difference of$\frac{\lambda }{2}$, the second dark band by a path length difference of$\frac{3\lambda }{2}$, and so on. The fourth dark band therefore represents a path length difference of,

$$\begin{gathered} \frac{7\lambda }{2}=1750nm \\ =1.75\mu m \end{gathered}$$

Hence, the path length difference to the fourth dark band is $1.75\mu m$.

The fringes in the small angle approximation are evenly spaced, therefore if$∆y$ represents the distance between one maxima, then $16.8\text{mm}=3.5\Delta y$is the distance from the pattern's centre to the fourth dark band.

Therefore, it gives:

$$\begin{gathered} \Delta y=\frac{16.8\text{mm}}{3.5} \\ =4.8\text{mm} \end{gathered}$$

Hence, the distance between the central bright band and first bright band is$4.8mm$ .