91Ó°ÊÓ

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

$\mathbf{Ï•}\mathbf{=}\left(2m+1\right)\mathbf{Ï€}\mathbf{,}\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

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

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

$$\begin{gathered} \frac{7\mathrm{λ}}{2}=1750nm \\ =1.75\mathrm{μ}m \end{gathered}$$

Hence, the path length difference to the fourth dark band is $1.75\mathrm{μ}m$.

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

Therefore, it gives:

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

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