College Physics

Figure 27.20The diffraction grating considered in this example produces a rainbow of colors on a screen a distancex= 2.00 m from the grating. The distances

along the screen are measured perpendicular to thex-direction. In other words, the rainbow pattern extends out of the page.

Strategy

The angles can be found using the equation

d sinθ=mλ(form= 0, 1, –1, 2, –2, ...) (27.12)

once a value for the slit spacingdhas been determined. Since there are 10,000 lines per centimeter, each line is separated by1/10,000of 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 isd= (1 cm) / 10,000 = 1.00×10

−4

cmor1.00×10

−6

m. Let us call the two anglesθVfor violet (380 nm)

andθRfor red (760 nm). Solving the equation dsinθV=mλforsinθV,

(27.13)

sinθV=

mλV

d

,

wherem= 1for first order andλV= 380 nm = 3.80×10−7m. Substituting these values gives

(27.14)

sinθV=3.80×10

−7m

1.00× 10 −6m

= 0.380.

Thus the angleθVis

θ (27.15)

V= sin

−10.380 = 22.33º.

Similarly,

(27.16)

sinθR=7.60×10

−7m

1.00× 10 −6m

.

Thus the angleθRis

θ (27.17)

R= sin

−10.760 = 49.46º.

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 labeledyVandyRinFigure 27.20. Noting thattanθ=y/x, we can solve foryVandyR. That is,

yV=xtanθV= (2.00 m)(tan 22.33º) = 0.815 m (27.18)

966 CHAPTER 27 | WAVE OPTICS

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