Learning Objectives

Learning Objectives

By the end of this section, you will be able to do the following:

  • Discuss the pattern obtained from diffraction grating
  • Explain diffraction grating effects

The information presented in this section supports the following AP® learning objectives and science practices:

  • 6.C.3.1 The student is able to qualitatively apply the wave model to quantities that describe the generation of interference patterns to make predictions about interference patterns that form when waves pass through a set of openings whose spacing and widths are small, but larger than the wavelength. (S.P. 1.4, 6.4)

An interesting thing happens if you pass light through a large number of evenly spaced parallel slits, called a diffraction grating. An interference pattern is created that is very similar to the one formed by a double slit (see Figure 10.16). A diffraction grating can be manufactured by scratching glass with a sharp tool in a number of precisely positioned parallel lines, with the untouched regions acting like slits. These can be photographically mass produced rather cheaply. Diffraction gratings work both for transmission of light, as in Figure 10.16, and for reflection of light, as on butterfly wings and the Australian opal in Figure 10.17 or the CD pictured in the opening photograph of this chapter, Figure 10.1. In addition to their use as novelty items, 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 regions are narrower and brighter, while their dark regions are darker. Figure 10.18 shows idealized graphs demonstrating the sharper pattern. Natural diffraction gratings occur in the feathers of certain birds. Tiny, finger-like structures in regular patterns act as reflection gratings, producing constructive interference that gives the feathers colors not solely due to their pigmentation. This is called iridescence.

On the left side of the figure is a diffraction grating represented by a vertical bar with five horizontal slits cut through it. A single horizontal arrow, representing white light, points at the center slit from the left side. On the right side, five arrows spread symmetrically above and below the horizontal centerline. The arrow that is on the horizontal centerline points at a white block labeled central white. The first arrows above and below the centerline point to rainbow-colored blocks labeled first
Figure 10.16 A diffraction grating is a large number of evenly spaced parallel slits. (a) Light passing through is diffracted in a pattern similar to a double slit, with bright regions at various angles. (b) The pattern obtained for white light incident on a grating. The central maximum is white, and the higher-order maxima disperse white light into a rainbow of colors.
Colorful photos of an Australian opal and a butterfly. The opal is full of fiery reds and yellows and deep blues and purples. The butterfly has its yellow wings spread and you can see its characteristic red, blue, and black spots and fringing.
Figure 10.17 (a) This Australian opal and (b) the butterfly wings have rows of reflectors that act like reflection gratings, reflecting different colors at different angles. ((a) Opals-On-Black.com, via Flickr (b) whologwhy, Flickr)
The upper graph, which is labeled double slit, shows a smooth curve similar to a sine curve that is shifted up so that its minimum value is zero. Three peaks are shown: the middle peak is labeled m equals zero and the left and right peaks are labeled m equals one. The lower graph, which is labeled grating, is aligned under the upper graph and also shows three peaks, with each peak aligned directly underneath the peaks in the upper graph. These three peaks are also labeled m equals zero or one, as in the u
Figure 10.18 Idealized graphs of the intensity of light passing through a double slit (a) and a diffraction grating (b) for monochromatic light. Maxima can be produced at the same angles, but those for the diffraction grating are narrower and hence sharper. The maxima become narrower and the regions between darker as the number of slits is increased.

The analysis of a diffraction grating is very similar to that for a double slit (see Figure 10.19). As we know from our discussion of double slits in Young's Double Slit Experiment, light is diffracted by each slit and spreads out after passing through. Rays traveling in the same direction (at an angle θθ size 12{θ} {} relative to the incident direction) are shown in the figure. Each of these rays travels a different distance to a common point on a screen far away. The rays start in phase, and they can be in or out of phase when they reach a screen, depending on the difference in the path lengths traveled. As seen in the figure, each ray travels a distance dsinθdsinθ size 12{d`"sin"θ} {} different from that of its neighbor, where dd size 12{d} {} is the distance between slits. If this distance equals an integral number of wavelengths, the rays all arrive in phase, and constructive interference (a maximum) is obtained. Thus, the condition necessary to obtain constructive interference for a diffraction grating is

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

where dd size 12{d} {} is the distance between slits in the grating, λλ size 12{λ} {} is the wavelength of light, and mm size 12{m} {} is the order of the maximum. Note that this is exactly the same equation as for double slits separated by d.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 t
Figure 10.19 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 dsinθdsinθ 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 dd size 12{d} {} of the grooves in a CD or DVD can be well determined by using a laser and the equation dsinθ=,form=0,1,–1,2,–2,.dsinθ=,form=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.d. size 12{d} {}

Example 10.3 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 Figure 10.20)?

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, exc
Figure 10.20 The diffraction grating considered in this example produces a rainbow of colors on a screen a distance x=2.00mx=2.00m size 12{x=2 "." "00"`m} {} from the grating. The distances along the screen are measured perpendicular to the x-x- size 12{x} {}direction. In other words, the rainbow pattern extends out of the page.


The angles can be found using the equation

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

once a value for the slit spacing dd size 12{d} {} has been determined. Since there are 10,000 lines per centimeter, each line is separated by 1/10,0001/10,000 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=1.00×104cmd=(1 cm)/10,000=1.00×104cm size 12{d= \( 1`"cm" \) /"10","000"=1 "." "00" times "10" rSup { size 8{ - 4} } `"cm"} {} or 1.00×106m.1.00×106m. size 12{1 "." "00" times "10" rSup { size 8{ - 6} } `m} {} Let us call the two angles θVθV size 12{θ rSub { size 8{V} } } {} for violet (380 nm) and θRθR size 12{θ rSub { size 8{R} } } {} for red (760 nm). Solving the equation dsinθV=dsinθV= size 12{d"sin"θ rSub { size 8{V} } =mλ} {} for sinθVsinθV size 12{"sin"θ rSub { size 8{V} } } {}

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

where m=1m=1 size 12{m=1} {} for first order and λV=380nm=3.80×107m.λV=380nm=3.80×107m. size 12{λ rSub { size 8{V} } ="380"`"nm"=3 "." "80" times "10" rSup { size 8{ - 7} } `m} {} Substituting these values gives

10.16 sinθV=3.80×107m1.00×106m=0.380.sinθV=3.80×107m1.00×106m=0.380. size 12{"sin"θ rSub { size 8{V} } = { {3 "." "80" times "10" rSup { size 8{ - 7} } `m} over {1 "." "00" times "10" rSup { size 8{ - 6} } `m} } =0 "." "380"} {}

Thus the angle θVθV size 12{θ rSub { size 8{V} } } {} is

10.17 θV=sin10.380=22.33º.θV=sin10.380=22.33º. size 12{θ rSub { size 8{V} } ="sin" rSup { size 8{ - 1} } 0 "." "380"="22" "." 3°} {}


10.18 sinθR=7.60×107m1.00×106m.sinθR=7.60×107m1.00×106m. size 12{"sin"θ rSub { size 8{R} } = { {7 "." "60" times "10" rSup { size 8{ - 7} } `m} over {1 "." "00" times "10" rSup { size 8{ - 6} } `m} } } {}

Thus the angle θRθR size 12{θ rSub { size 8{R} } } {} is

10.19 θR=sin10.760=49.46º.θR=sin10.760=49.46º. size 12{θ rSub { size 8{R} } ="sin" rSup { size 8{ - 1} } 0 "." "760"="49" "." 5°} {}

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).

Solution for (b)

The distances on the screen are labeled yVyV size 12{y rSub { size 8{V} } } {} and yRyR size 12{y rSub { size 8{R} } } {} in Figure 10.20. Noting that tanθ=y/x,tanθ=y/x, size 12{"tan"θ=y/x} {} we can solve for yVyV size 12{y rSub { size 8{V} } } {} and yR.yR. size 12{y rSub { size 8{R} } } {} That is

10.20 y V = x tan θ V = ( 2.00 m ) ( tan 22.33º ) = 0.815 m y V = x tan θ V = ( 2.00 m ) ( tan 22.33º ) = 0.815 m size 12{y rSub { size 8{V} } =x"tan"θ rSub { size 8{V} } = \( 2 "." "00"`m \) \( "tan""22" "." 3° \) =0 "." "822"`m} {}


10.21 yR=xtanθR=(2.00 m)(tan 49.46º)=2.338 m.yR=xtanθR=(2.00 m)(tan 49.46º)=2.338 m. size 12{y rSub { size 8{R} } =x"tan"θ rSub { size 8{R} } = \( 2 "." "00"`m \) \( "tan""49" "." 5° \) =2 "." "339"`m} {}

The distance between them is therefore

10.22 yRyV=1.52 m.yRyV=1.52 m. size 12{y rSub { size 8{R} } - y rSub { size 8{V} } =1 "." 52`m} {}


The large distance between the red and violet ends of the rainbow produced from the white light indicates the potential this diffraction grating has as a spectroscopic tool. The more it can spread out the wavelengths—greater dispersion—the more detail can be seen in a spectrum. This depends on the quality of the diffraction grating—it must be very precisely made in addition to having closely spaced lines.