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Q105P

Expert-verifiedFound in: Page 1115

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Show that a grating made up of alternately transparent and opaque strips of equal width eliminates all the even orders of maxima (except${\mathit{m}}{\mathbf{=}}{\mathbf{0}}$ ).**

It is proved that a grating made up of alternately transparent andopaque strips of equal width eliminates all the even orders of maximaexcept* $m=0$*.

There is a grating made up of alternately transparent andopaque strips of equal width.

**The angular distance ${\mathit{\theta}}$** **of the ** ${{\mathit{m}}}_{\mathbf{t}\mathbf{h}}$** order diffraction maxima produced from a grating having line separation ${\mathit{d}}$** ** is**

** ${\mathit{d}}{\mathit{s}}{\mathit{i}}{\mathit{n}}{\mathit{\theta}}{\mathbf{=}}{\mathit{m}}{\mathit{\lambda}}$**** .....(1)**

**Here, ** **is the wavelength of the incident light.**

**The angular distance ${\mathit{\theta}}$** ** of the ${{\mathit{k}}}_{\mathbf{t}\mathbf{h}}$****order single slit diffraction minima for slit width ** ${\mathit{a}}$** is**

** **** ${\mathit{a}}{\mathit{s}}{\mathit{i}}{\mathit{n}}{\mathit{\theta}}{\mathbf{=}}{\mathit{k}}{\mathit{\lambda}}$ .....(2)**

Slit separation is the distance between the mid points of two slits. Hence slit separation is equal to twice the slit width, that is

$d=2a$

Thus, equation (1) becomes

$2a\mathrm{sin}\theta =m\lambda $ ….. (3)

Subtract twice of equation (2) from equation (3) to get

$\begin{array}{l}2asin\theta -2asin\theta =m\lambda -2k\lambda \\ \left(m-2k\right)\lambda =0\\ \mathrm{m}=2k\end{array}$$\begin{array}{l}2asin\theta -2asin\theta =m\lambda -2k\lambda \\ \left(m-2k\lambda \right)=0\\ \mathrm{m}=2k\end{array}$

But

$\mathrm{K}=\pm 1\pm 2\pm 3\dots $

Hence

$m=\pm 2,\pm 4,\pm 6$

Thus the even order grating diffraction maxima’s (except$m=0$ ) overlap with single slit diffraction minima’s and disappear.

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