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Q107P

Expert-verifiedFound in: Page 1115

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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 $9:1$ .

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

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

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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