91Ó°ÊÓ

In a two-slit interference pattern, there are $8$ bright fringes within the first side peak of the diffraction envelope.

The angular distance of the $m_{th}$ order diffraction minima is given by

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{=}\mathit{m}\mathit{λ}$ …(¾±)

Here, $\mathrm{λ}$is the wavelength of the incident light and $a$ is the slit width.

The angular distance of the $n_{th}$order interference maxima is given by

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{=}\mathit{n}\mathit{λ}$ …(¾±¾±)

Here, $\mathrm{λ}$is the wavelength of the incident light and $d$ is the separation between the two slits.

The $9_{th}$ interference maxima coincide with the $1_{st}$diffraction minima. From equations (i) and (ii) the angular distance of the $1_{st}$diffraction minima and the $9_{th}$interference maxima is

$\begin{array}{l}d\sin \mathrm{θ}=9\mathrm{λ}\\ a\sin \mathrm{θ}=\mathrm{λ}\end{array}$

Divide the two equations to get

$\frac{d}{a}=\frac{9}{1}$

Thus, the required ratio is$9:1$ .