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 27.20.)

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, 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.

Figure 27.20 The diffraction grating considered in this example produces a rainbow of colors on a screen a distance

x = 2 . 00 m

from the grating. The distances along the screen are measured perpendicular to the

x

-direction. In other words, the rainbow pattern extends out of the page.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The angles can be found using the equation
$d sin θ = mλ (for m = 0, 1, –1, 2, –2, \dots)$
27.12

once a value for the slit spacing $d$ 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 $d = (1 cm) / 10, 000 = 1 . 00 \times 10^{- 4} cm$ or $1 . 00 \times 10^{- 6} m$ . Let us call the two angles $θ_{V}$ for violet (380 nm) and $θ_{R}$ for red (760 nm). Solving the equation $d sin θ_{V} = mλ$ for $sin θ_{V}$ ,
$sin θ_{V} = \frac{{mλ}_{V}}{d},$
27.13

where $m = 1$ for first order and $λ_{V} = 380 .80 \times 10^{- 7}$ . Substituting these values gives
$sin θ_{V} = \frac{3 . 80 \times 10^{- 7} m}{1 . 00 \times 10^{- 6} m} = 0 . 380.$
27.14

Thus the angle $θ_{V}$ is

$θ_{V} = {sin}^{- 1} 0 . 380 = 22 . 33º.$
27.15

Similarly,

$sin θ_{R} = \frac{7 . 60 \times 10^{- 7} m}{1.00 \times 10^{- 6} m} .$
27.16

Thus the angle $θ_{R}$ is

$θ_{R} = {sin}^{- 1} 0 . 760 = 49.46º.$
27.17

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 $y_{V}$ and $y_{R}$ in Figure 27.20. Noting that $tan θ = y / x$ , we can solve for $y_{V}$ and $y_{R}$ . That is,
$y_{V} = x tan θ_{V} = (2.00 m) (tan 22.33º) = 0.815 m$
27.18

and

$y_{R} = x tan θ_{R} = (2.00 m) (tan 49.46º) = 2.338 m.$
27.19

The distance between them is therefore

$y_{R} - y_{V} = 1.52 m.$
27.20
Discussion
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.

0 of 4 steps open

How did it go?

Strategy

Solution for (a)

Solution for (b)

Discussion