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Found in: Page 1115

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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# 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?

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

See the step by step solution

## Step 1: Given data

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

## Step 2: Diffraction and interference from a double slit

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

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

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

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

## Step 3: 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

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

Divide the two equations to get

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

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

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