91Ó°ÊÓ

The angles at which the dark fringes are observed in a single slit diffraction experiment follow a condition given below,

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{=}\mathit{m}\mathit{λ}\mathbf{}\mathbf{}\mathbf{}\mathit{m}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{3}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$(minima, i.e., dark fringes)

Where a is the slit width and λ is the wavelength of light.

The minimum intensity for 600 nm orange light is observed at 1.00 mrad. The condition for these minima is

$a\sin \left(1.00mrad\right)=m\left(600nm\right)$

For 500 nm blue-green light, the minimum intensity is observed at 1.00 mrad. The condition for these minima is

$a\sin \left(1.00mrad\right)=m\text{'}\left(500nm\right)$

It is unknown to this point at which order of the minimum intensity for both wavelengths has the same location, i.e., angle.

As the slit is common to both wavelengths, therefore equating the two condition equations

$m\left(600nm\right)=m\text{'}\left(500nm\right)$

The above equation will only be true if$\text{m}=5$ and $\text{m'}=6$. Therefore, insert any one value of the order of minima into its corresponding condition.

$$\begin{gathered} a\sin \left(1.00mrad\right)=\left(6\right)\left(500nm\right) \\ a=\frac{\left(6\right)\left(500nm\right)}{\sin \left(1.00mrad\right)} \\ a=3×{10}^{-3}m\left(\frac{1mm}{{10}^{-3}m}\right) \\ a=3mm \end{gathered}$$

Hence the minimum slit width is 3 mm.