Question

If, in a two-slit interference pattern, there are$\mathbf{8}$bright fringes within the first side peak of the diffraction envelope and diffraction minima coincide with two-slit interference maxima, then what is the ratio of slit separation to slit width?

Short answer

The ratio of the slit separation to the slit width is $9:1$.

Step-by-step solution

Given data

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

Diffraction and interference from a double slit

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

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }$ …(i)

Here, $\lambda$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

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{n}\mathit{\lambda }$ …(ii)

Here, $\lambda$is the wavelength of the incident light and $d$ is the separation between the two slits.

Determining the ratio of the slit separation to the slit width

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

$\begin{array}{l}d\sin \theta =9\lambda \\ a\sin \theta =\lambda \end{array}$

Divide the two equations to get

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

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