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d sin θ = , for m = 0, 1, –1, 2, –2, (constructive), size 12{d"sin"θ=mλ,~m="0,"`"1,"`"2,"` dotslow } {}

where d size 12{d} {} is the distance between slits in the grating, λ size 12{λ} {} is the wavelength of light, and m size 12{m} {} is the order of the maximum. Note that this is exactly the same equation as for double slits separated by d size 12{d} {} . However, the slits are usually closer in diffraction gratings than in double slits, producing fewer maxima at larger angles.

The figure shows a schematic of a diffraction grating, which is represented by a vertical black line into which are cut five small gaps. The gaps are evenly spaced a distance d apart. From the left five rays arrive, with one ray arriving at each gap. To the right of the line with the gaps the rays all point down and to the right at an angle theta below the horizontal. At each gap a triangle is formed where the hypotenuse is length d, one angle is theta, and the side opposite theta is labeled delta l. At the top is written delta l equals d sine theta.
Diffraction grating showing light rays from each slit traveling in the same direction. Each ray travels a different distance to reach a common point on a screen (not shown). Each ray travels a distance d sin θ size 12{d`"sin"θ} {} different from that of its neighbor.

Where are diffraction gratings used? 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 frequency 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 specific wavelengths for such use.

Take-home experiment: rainbows on a cd

The spacing d size 12{d} {} of the grooves in a CD or DVD can be well determined by using a laser and the equation d sin θ = , for m = 0, 1, –1, 2, –2, size 12{d"sin"θ=mλ,`m="0,"`"1,"`"2,"` dotslow } {} . However, we can still make a good estimate of this spacing by using white light and the rainbow of colors that comes from the interference. Reflect sunlight from a CD onto a wall and use your best judgment of the location of a strongly diffracted color to find the separation d size 12{d} {} .

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). (b) What is the distance between the ends of the rainbow of visible light produced on the screen for first-order interference? (See [link] .)

The image shows a vertical black bar at the left labeled grating. From the midpoint of this bar four lines fan out to the right, with two lines angled above the horizontal centerline and two lines angled symmetrically below the horizontal centerline. These four lines hit a vertical black line to the right that is labeled screen. On the screen between the two upper lines is a rainbow region, with violet nearer the centerline and red farther from the centerline. The same is true for the two lower lines, except that they are below the centerline instead of above. The distance from the centerline to the upper violet zone is labeled y sub v equals question mark and the distance from the centerline to the upper red zone is labeled y sub r equals question mark. The angle between the centerline and the line leading to the upper violet zone is labeled theta V equals question mark and the angle between the line leading to the upper red zone is labeled theta R equals question mark. The distance between the grating and the screen is labeled x equals two point zero zero meters.
The diffraction grating considered in this example produces a rainbow of colors on a screen a distance x = 2 . 00 m size 12{x=2 "." "00"`m} {} from the grating. The distances along the screen are measured perpendicular to the x size 12{x} {} -direction. In other words, the rainbow pattern extends out of the page.


The angles can be found using the equation

d sin θ = (for m = 0, 1, –1, 2, –2, …) size 12{d"sin"θ=mλ,`m="0,"`"1,"`"2,"` dotslow } {}

once a value for the slit spacing d size 12{d} {} has been determined. Since there are 10,000 lines per centimeter, each line is separated by 1/10,000 of a centimeter. Once the angles are found, the distances along the screen can be found using simple trigonometry.

Solution for (a)

The distance between slits is d = ( 1 cm ) / 10 , 000 = 1 . 00 × 10 4 cm size 12{d= \( 1`"cm" \) /"10","000"=1 "." "00" times "10" rSup { size 8{ - 4} } `"cm"} {} or 1 . 00 × 10 6 m size 12{1 "." "00" times "10" rSup { size 8{ - 6} } `m} {} . Let us call the two angles θ V size 12{θ rSub { size 8{V} } } {} for violet (380 nm) and θ R size 12{θ rSub { size 8{R} } } {} for red (760 nm). Solving the equation d sin θ V = size 12{d"sin"θ rSub { size 8{V} } =mλ} {} for sin θ V size 12{"sin"θ rSub { size 8{V} } } {} ,

sin θ V = V d , size 12{"sin"θ rSub { size 8{V} } = { {mλ rSub { size 8{V} } } over {d} } ,} {}

where m = 1 size 12{m=1} {} for first order and λ V = 380 nm = 3 . 80 × 10 7 m size 12{λ rSub { size 8{V} } ="380"`"nm"=3 "." "80" times "10" rSup { size 8{ - 7} } `m} {} . Substituting these values gives

Questions & Answers

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derivation of resistance
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A diverging lens
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