3605 The Diffraction Grating

An array of a large number of parallel slits, all with the same width a and spaced equal distances d between centers, is called a diffraction grating.

Intensity maxima multiple slits

dsin⁡θ=mλ(m=0,±1,±2,⋯)(36.13)d\sin\th = m \lambda \quad (m = 0, \pm 1, \pm 2, \cdots) \quad\bold{(36.13)}

dd - Distance between slits
θ\th - Angle of line from center of the slit array to mth bright region on screen
λ\lambda - Wavelength

EXAMPLE 36.4 Width of a Grating Spectrum

The wavelengths of the visible spectrum are approximately 380 nm (violet) to 750 nm (red). (a) Find the angular limits of the firstorder visible spectrum produced by a plane grating with 600 slits per millimeter when white light falls normally on the grating. (b) Do the first-order and second-order spectra overlap? What about the second-order and third-order spectra? Do your answers depend on the grating spacing?

Solution

We must find the angles spanned by the visible specttrum in the first-, second-, and third-order spectra. These correspond to m=1, 2, and 3 in Eq.(36.13).
a. The grating spacing is

d=1600 slits/mm=1.67×10−6 m\begin{aligned} d= \frac{1}{600 \text{ slits/mm}} = 1.67 \times 10^{-6} \text{ m} \end{aligned}

Solve for θ\th

θ=arcsin⁡mλd\begin{aligned} \th &= \arcsin \frac{m \lambda}{d} \end{aligned}

Then for m=1, the angle deviation θv1\th_{v1} and θr1\th_{r1} for violet and red light, respectively, are

θv1=arcsin⁡(3.8×10−71.67×10−6)=13.2°θr1=arcsin⁡(7.5×10−71.67×10−6)=26.7°\begin{aligned} \th_{v1} &= \arcsin(\frac{3.8 \times 10^{-7}}{1.67 \times 10^{-6}}) = 13.2\degree\\ \th_{r1} &= \arcsin(\frac{7.5 \times 10^{-7}}{1.67 \times 10^{-6}}) = 26.7\degree \end{aligned}

When m = 2 and 3, our equation θ=arcsin⁡(mλ/d)\th = \arcsin(m \lambda/d) for violet light and red light yields

θv2=arcsin⁡(23.8×10−71.67×10−6)=27.1°θr2=arcsin⁡(27.5×10−71.67×10−6)=63.9°θv3=arcsin⁡(33.8×10−71.67×10−6)=43.0°θr3=arcsin⁡(37.5×10−71.67×10−6)=arcsin⁡(1.35)=undefined\begin{aligned} \th_{v2} &= \arcsin(2\frac{3.8 \times 10^{-7}}{1.67 \times 10^{-6}}) = 27.1\degree\\ \th_{r2} &= \arcsin(2\frac{7.5 \times 10^{-7}}{1.67 \times 10^{-6}}) = 63.9\degree\\ \th_{v3} &= \arcsin(3\frac{3.8 \times 10^{-7}}{1.67 \times 10^{-6}}) = 43.0\degree\\ \th_{r3} &= \arcsin(3\frac{7.5 \times 10^{-7}}{1.67 \times 10^{-6}}) = \arcsin(1.35) = \text{undefined} \end{aligned}

Hence the second-order spectrum extends from 27.1°27.1\degree to 63.9°63.9\degree and the third-order spectrum extends from 43.0°43.0\degree to 90°90 \degree (the largest possible value of u). The undefined value of ur3 means that the third-order spectrum reaches u=90°=arcsin⁡(1)u = 90\degree= \arcsin(1) at a wavelength shorter than 750 nm; you should be able to show that this happens for λ=557\lambda = 557 nm. Hence the first-order spectrum (from 13.2°13.2\degree to 26.7°26.7\degree) does not overlap with the second-order spectrum, but the second- and third-order spectra do overlap. You can convince yourself that this is true for any value of the grating spacing d.

Exercises

23, 24, 26, 28, 59

36.23 When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ±17.8°\pm 17.8\degree from the central maximum.
(a) What is the line density (in lines/cm) of this grating?
(b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

Solution

a. The line density is 4830 lines/cm.

dsin⁡θ=mλθ=17.8°,m=1,λ=6.328×10−7m⇒d=mλsin⁡θ=1×6.328×10−7msin⁡17.8°=2.07×10−4cm1cmd=4830 lines/cm\begin{aligned} d\sin\th &= m \lambda\\ \th &= 17.8\degree, m = 1, \lambda = 6.328 \times 10^{-7}\text{m}\\ \To d &= \frac{m \lambda}{\sin\th}\\ &= \frac{1 \times 6.328 \times 10^{-7}\text{m}}{\sin{17.8\degree}} = 2.07\times 10^{-4} \text{cm}\\ \frac{1\text{cm}}{d} &= 4830\text{ lines/cm} \end{aligned}

b. There are two additional bright spots beyond the first one. And their angles are as follows

m=2⇒θ2=arcsin⁡2λd=37.7°m=3⇒θ3=arcsin⁡3λd=66.5°m=4⇒θ4=arcsin⁡4λd=undefined\begin{aligned} m &= 2 \To \th_2 = \arcsin \frac{2 \lambda}{d} = 37.7\degree\\ m &= 3 \To \th_3 = \arcsin \frac{3 \lambda}{d} = 66.5\degree\\ m &= 4 \To \th_4 = \arcsin \frac{4 \lambda}{d} = \text{undefined} \end{aligned}

36.24 Monochromatic light is at normal incidence on a plane transmission grating. The first-order maximum in the interference pattern is at an angle of 11.3°11.3\degree. What is the angular position of the fourth-order maximum?

Solution

θ=arcsin⁡(4λd)\th = \arcsin(\dfrac{4 \lambda}{d}) and arcsin⁡(λd)=11.3°\arcsin(\dfrac{\lambda}{d}) = 11.3 \degree. θ=51.6°\th = 51.6\degree

36.26 If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 nm) at 65.0°65.0\degree from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 nm)?

Solution

Todo

36.28 . The wavelength range of the visible spectrum is approximately 380–750 nm. White light falls at normal incidence on a diffraction grating that has 350 slits/mm. Find the angular width of the visible spectrum in (a) the first order and (b) the third order. (Note: An advantage of working in higher orders is the greater angular spread and better resolution. A disadvantage is the overlapping of different orders, as shown in Example 36.4.)

Solution

Todo

36.59 A diffraction grating has 650 slits/mm. What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately 400-700 nm.)

Solution

Second-order.

dsin⁡θ=mλθ=arcsin⁡mλdd=10−3650mλ1=4×10−7mλ2=7×10−7mm=2⇒θ1=31.3°,θ2=65.5°m=3⇒θ1=51.3°,θ2=undefined\begin{aligned} d\sin\th &= m \lambda\\ \th &= \arcsin \frac{m \lambda}{d}\\ d &= \frac{10^{-3}}{650} \text{m}\\ \lambda_1 &= 4 \times 10^{-7} \text{m}\\ \lambda_2 &= 7 \times 10^{-7} \text{m}\\ m &= 2 \To \th_1 = 31.3 \degree , \th_2 = 65.5\degree\\ m &= 3 \To \th_1 = 51.3 \degree , \th_2 = \text{undefined} \end{aligned}

Therefore, the highest order that contains the entire visible specturm is second-order.