# 4.4 Diffraction gratings  (Page 2/4)

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## Applications of diffraction gratings

Where are diffraction gratings used in applications? Diffraction gratings are commonly used for spectroscopic dispersion and analysis of light. What makes them particularly useful is the fact that they form a sharper pattern than double slits do. That is, their bright fringes are narrower and brighter while their dark regions are darker. Diffraction gratings are key components of monochromators used, for example, in optical imaging of particular wavelengths from biological or medical samples. A diffraction grating can be chosen to specifically analyze a wavelength emitted by molecules in diseased cells in a biopsy sample or to help excite strategic molecules in the sample with a selected wavelength of light. Another vital use is in optical fiber technologies where fibers are designed to provide optimum performance at specific wavelengths. A range of diffraction gratings are available for selecting wavelengths for such use.

## Calculating typical diffraction grating effects

Diffraction gratings with 10,000 lines per centimeter are readily available. Suppose you have one, and you send a beam of white light through it to a screen 2.00 m away. (a) Find the angles for the first-order diffraction of the shortest and longest wavelengths of visible light (380 and 760 nm, respectively). (b) What is the distance between the ends of the rainbow of visible light produced on the screen for first-order interference? (See [link] .)

## Strategy

Once a value for the diffraction grating’s slit spacing d has been determined, the angles for the sharp lines can be found using the equation

$d\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =m\lambda \phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}m=0,\phantom{\rule{0.2em}{0ex}}±1,\phantom{\rule{0.2em}{0ex}}±2,\phantom{\rule{0.2em}{0ex}}...\phantom{\rule{0.2em}{0ex}}.$

Since there are 10,000 lines per centimeter, each line is separated by 1/10,000 of a centimeter. Once we know the angles, we an find the distances along the screen by using simple trigonometry.

## Solution

1. The distance between slits is $d=\left(1\phantom{\rule{0.2em}{0ex}}\text{cm}\right)\text{/}10,000=1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{cm or}\phantom{\rule{0.2em}{0ex}}1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{m}.$ Let us call the two angles ${\theta }_{\text{V}}$ for violet (380 nm) and ${\theta }_{\text{R}}$ for red (760 nm). Solving the equation $d\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{\text{V}}=m\lambda \phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{\text{V}},$
$\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{\text{V}}=\frac{m{\lambda }_{\text{V}}}{d},$

where $m=1$ for the first-order and ${\lambda }_{\text{V}}=380\phantom{\rule{0.2em}{0ex}}\text{nm}=3.80\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{m}.$ Substituting these values gives
$\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{\text{V}}=\frac{3.80\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{m}}{1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{m}}=0.380.$

Thus the angle ${\theta }_{\text{V}}$ is
${\theta }_{\text{V}}={\text{sin}}^{-1}\phantom{\rule{0.2em}{0ex}}0.380=22.33\text{°}.$

Similarly,
$\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{\text{R}}=\frac{7.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{m}}{1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{m}}=0.760.$

Thus the angle ${\theta }_{\text{R}}$ is
${\theta }_{\text{R}}={\text{sin}}^{-1}\phantom{\rule{0.2em}{0ex}}0.760=49.46\text{°}.$

Notice that in both equations, we reported the results of these intermediate calculations to four significant figures to use with the calculation in part (b).
2. The distances on the secreen are labeled ${y}_{\text{V}}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{y}_{\text{R}}$ in [link] . Notice that $\text{tan}\phantom{\rule{0.2em}{0ex}}\theta =y\text{/}x.$ We can solve for ${y}_{\text{V}}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{y}_{\text{R}}.$ That is,
${y}_{\text{V}}=x\phantom{\rule{0.2em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}{\theta }_{V}=\left(2.00\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(\text{tan}\phantom{\rule{0.2em}{0ex}}22.33\text{°}\right)=0.815\phantom{\rule{0.2em}{0ex}}\text{m}$

and
${y}_{\text{R}}=x\phantom{\rule{0.2em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}{\theta }_{R}=\left(2.00\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(\text{tan}\phantom{\rule{0.2em}{0ex}}49.46\text{°}\right)=2.338\phantom{\rule{0.2em}{0ex}}\text{m}.$

The distance between them is therefore
${y}_{\text{R}}-{y}_{\text{V}}=1.523\phantom{\rule{0.2em}{0ex}}\text{m}.$

in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
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