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Q101P

Expert-verifiedFound in: Page 1115

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Show that the dispersion of a grating is ${\mathit{D}}{\mathbf{=}}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\theta}}{\mathbf{\lambda}}$**

It is proved that the dispersion of a grating is $\frac{\mathrm{tan}\theta}{\lambda}$

**An optical element with a periodic structure that divides light into a number of beams that move in diverse directions is known as a diffraction grating. **

The angular distance $\theta $of the ${m}_{th}$order diffraction pattern produced from a grating having line separation is

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

Here, $\lambda $ is the wavelength of the incident light.

The dispersion is given by

$D=\frac{d\theta}{d\lambda}$

Differentiate equation (i) with respect to $\lambda $ to get

$d\mathrm{cos}\theta \frac{d\theta}{d\lambda}=m\phantom{\rule{0ex}{0ex}}\frac{d\theta}{d\lambda}=\frac{m}{d\mathrm{cos}\theta}$

Substitute form of $m$ from equation (i) to get

$\frac{d\theta}{d\lambda}=\frac{\frac{d\mathrm{sin}\theta}{\lambda}}{d\mathrm{cos}\theta}\phantom{\rule{0ex}{0ex}}\frac{d\theta}{d\lambda}=\frac{\mathrm{tan}\theta}{\lambda}$

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