E. The Wave Nature of Light · The Wave Nature of LightDiffraction Gratings and Spectroscopy
E1.1 E2.1 E2.3 E2.4 E3.2 E3.3 A1.12 A1.13About 75 minutes
Learning goals- I can explain why a diffraction grating gives sharp, bright, widely spaced maxima.
- I can use d sin θₘ = mλ with d = 1/N to find angles, wavelengths and the highest order.
- I can explain how a grating separates white light into spectra, and compare continuous, emission and absorption spectra.
- I can explain how a spectroscope and the tracks on a CD or DVD act as diffraction gratings.
Words to know
diffraction grating
A surface with a very large number of equally spaced parallel slits (or lines), used to separate light into its wavelengths.
grating spacing (d)
The distance between neighbouring lines of a grating; d = 1/N, where N is the number of lines per metre.
transmission / reflection grating
A grating that light passes through, or one that reflects light from a ruled or grooved surface (such as a CD).
spectrum
The band of colours (wavelengths) obtained when light is separated.
spectroscope
An instrument that uses a prism or grating to separate light so its spectrum can be viewed and measured.
emission / absorption spectrum
A pattern of bright lines given off by a hot, low-density gas, or of dark lines where a cooler gas has absorbed light from a continuous spectrum.
Rainbows on a disc
Tilt an old CD or DVD under a lamp and it flashes with bands of colour. The disc has no dye or paint that colour; the colours come from interference. The disc's tightly packed tracks act like thousands of slits side by side: a diffraction grating. Gratings are among the most important tools in science: they let astronomers find out what distant stars are made of.
From two slits to thousands
A diffraction grating has a very large number of equally spaced slits or lines, often hundreds per millimetre. The light from all of them interferes. The condition for bright maxima is the same as for two slits:
d sin θₘ = mλ, m = 0, 1, 2, …
where d is the distance between neighbouring lines. Gratings are usually labelled by lines per millimetre, so d = 1/N. For example, 600 lines/mm gives d = 1 mm ÷ 600 = 1.67 × 10⁻³ mm = 1.67 × 10⁻⁶ m.
Why use thousands of slits instead of two?
- Sharper maxima: with two slits, brightness changes gradually from bright to dark. With many slits, light from all the slits adds up only at almost exactly the right angle; slightly off that angle, contributions from slits far apart cancel. The maxima become very narrow lines with wide dark regions between them.
- Brighter maxima: many slits let through much more light.
- Wider spread: d is tiny, so the angles are large (often tens of degrees). Wavelengths that differ by only a little are separated clearly, which is what makes precise wavelength measurement possible.
The angles are large, so do not use the small-angle approximation for gratings: find θ with sin θ = mλ/d, and if you need a distance on a screen, use x = L tan θ.
Worked exampleExample 1: Angles and the highest order
Light of wavelength 500 nm shines straight onto a grating with 600 lines/mm. Find the angles of the first- and second-order maxima, and the highest order that can be seen.
- d = 1 ÷ (600 000 lines/m) = 1.667 × 10⁻⁶ m.
- m = 1: sin θ₁ = 5.00 × 10⁻⁷ ÷ 1.667 × 10⁻⁶ = 0.300, so θ₁ = 17.5°.
- m = 2: sin θ₂ = 0.600, so θ₂ = 36.9°.
- Highest order: sin θ ≤ 1 means m ≤ d/λ = 3.33, so the highest order is m = 3 (at sin θ = 0.900, θ = 64.2°).
17.5° and 36.9°; orders up to m = 3 (on each side of the centre).
Worked exampleExample 2: Identifying a spectral line
A student views a hydrogen lamp through a grating with 300 lines/mm. A red line appears in first order at 11.35°. Find its wavelength.
- d = 1 ÷ 300 000 = 3.333 × 10⁻⁶ m.
- λ = d sin θ / m = (3.333 × 10⁻⁶)(sin 11.35°) ÷ 1.
- λ = 3.333 × 10⁻⁶ × 0.1968 = 6.56 × 10⁻⁷ m.
λ ≈ 656 nm, the well-known red line of hydrogen.
Spectra and spectroscopy
Shine white light through a grating and each wavelength makes its maximum at its own angle. Violet (short λ) is bent least and red (long λ) most, the opposite order from a prism. Each order (m = 1, 2, …) becomes a full spectrum; the central maximum (m = 0) stays white because every wavelength has its m = 0 maximum straight ahead.
A spectroscope uses a grating (or prism) to spread out the light from a source:
- A hot, dense object (an incandescent filament, the glowing surface of the Sun) gives a continuous spectrum: all colours, no gaps.
- A hot, low-density gas (a gas-discharge tube, a neon sign) gives an emission spectrum: only a few bright lines at wavelengths that act like a fingerprint for each element. Hydrogen's visible lines are at about 656, 486, 434 and 410 nm.
- Light from a hot source passing through a cooler gas gives an absorption spectrum: a continuous spectrum with dark lines at the same wavelengths the gas would emit. The Sun's spectrum has many such dark lines.
By measuring line wavelengths with a grating, chemists identify substances in a sample, and astronomers identify elements in stars and galaxies they can never visit. Why each element has its own lines is explained by energy levels in atoms (unit Quantum Mechanics).
Worked exampleExample 3: A CD as a reflection grating
The tracks on a CD are 1.6 µm apart. A red laser (λ = 650 nm) shines straight down onto the disc. At what angles from the incoming beam do the reflected maxima appear?
- The tracks act as a reflection grating with d = 1.6 × 10⁻⁶ m.
- m = 1: sin θ = 6.50 × 10⁻⁷ ÷ 1.6 × 10⁻⁶ = 0.406, θ = 24.0°.
- m = 2: sin θ = 0.8125, θ = 54.3°.
- m = 3: sin θ = 1.22 > 1, impossible. So only m = 0, 1 and 2 appear.
Reflected maxima straight back (m = 0) and at 24.0° and 54.3° on each side.
A laser shining straight down on a CD. The evenly spaced tracks act as a reflection grating, sending bright beams back at the angles given by d sin θ = mλ (side view, angles to scale, track spacing greatly enlarged).Discs and their tracks
Optical discs store data as tiny pits along a spiral track, read by a laser. The track spacing and the laser wavelength shrank with each generation, so each disc holds more data:
- CD: tracks about 1.6 µm apart, read with infrared light of about 780 nm.
- DVD: tracks about 0.74 µm apart, read with red light of about 650 nm.
- Blu-ray: tracks about 0.32 µm apart, read with violet light of about 405 nm (the "blue" in the name).
The reading laser must be focused to a spot small enough to read one track. Diffraction sets a limit on how small a spot can be (lesson on resolution): a shorter wavelength gives a smaller spot, which is why Blu-ray needed a violet laser.
Just for funA grating with 600 lines per millimetre goes to a job interview. "Tell me about yourself," says the interviewer. The grating replies, "I'm very well-spaced, I spread myself across many orders, and I'm good at separating things."
Hands-on activityBuild a CD spectroscope
You need: an old CD or DVD you no longer need, a cereal box or shoe box, scissors, tape, black paper or a black marker, and a grown-up for the cutting.
Grown-up note: an adult should do the cutting. If you cut a disc, it can crack into sharp pieces; this design uses the whole disc, so do not break it.
- At one end of the box, cut a narrow slit about 1 mm wide and 3 cm long (two pieces of tape or card edges can make the slit narrow).
- At the other end, cut a viewing hole and tape the disc inside the box at about 45°, shiny side facing the slit, so you can look at the disc through the hole and see light from the slit reflected off it. Tape black paper inside the box to block stray light.
- Point the slit at different light sources and look at the disc: an incandescent or "warm" bulb, a fluorescent tube or compact fluorescent bulb, an LED bulb, a computer screen, and a bright sky away from the Sun. Never point the slit at the Sun.
- Sketch each spectrum.
What to look for: an incandescent bulb gives a continuous spectrum. A fluorescent bulb shows a few bright lines (from mercury vapour inside) on top of a weaker continuous background. White LEDs give a broad, continuous-looking spectrum with a strong blue part. Order of colours: violet nearest the centre, red farthest.
Your turn
1A grating has 500 lines/mm. Find d in metres.
d = 1 ÷ 500 000 = 2.0 × 10⁻⁶ m.
2Sodium light (589 nm) falls on that grating. Find the first-order angle.
sin θ = 5.89 × 10⁻⁷ ÷ 2.0 × 10⁻⁶ = 0.2945, θ = 17.1°.
3What is the highest order you could see for the sodium light in the last question?
m ≤ d/λ = 3.40, so m = 3.
4Give two advantages of a grating over a double slit for measuring wavelength.
The maxima are sharper (so their positions can be measured precisely) and brighter, and they are more widely spread (large angles).
5In a grating spectrum, which colour is bent the most?
Red (longest wavelength).
6What kind of spectrum does a neon sign produce?
An emission (bright-line) spectrum.
7Why is the central maximum of a grating white when white light is used?
Every wavelength has its m = 0 maximum at θ = 0, so all colours overlap there.
More practice: warm-up
1Write the grating equation.
2A grating has 1000 lines/mm. What is d?
3What acts as the 'slits' on a CD?
The tracks (rows of pits).
4What does a hot, dense object such as a glowing filament produce: a continuous, emission or absorption spectrum?
5Should you use the small-angle approximation for most grating problems?
No: the angles are usually large.
More practice: on your own
1White light falls on a grating with 800 lines/mm. Find the first-order angles for 450 nm and 650 nm light.
d = 1.25 × 10⁻⁶ m. 450 nm: sin θ = 0.360, θ = 21.1°. 650 nm: sin θ = 0.520, θ = 31.3°.
2A grating with 1000 lines/mm gives a first-order line at 25.0°. Find λ and suggest the colour.
λ = d sin θ = 1.0 × 10⁻⁶ × sin 25.0° = 4.23 × 10⁻⁷ m (423 nm): violet.
3A laser at 500 nm passes through a grating with 600 lines/mm onto a screen 2.0 m away (screen parallel to the grating). How far from the central spot is the first-order spot?
sin θ = 5.00 × 10⁻⁷ × 6.00 × 10⁵ = 0.300, θ = 17.46°. x = L tan θ = 2.0 × 0.3145 = 0.63 m.
4Show that the 1st- and 2nd-order white-light spectra from a grating with 300 lines/mm do not overlap, but the 2nd and 3rd do (visible light 400 to 700 nm).
d = 3.33 × 10⁻⁶ m. 1st-order red ends at sin θ = 700/3333 = 0.210, and 2nd-order violet starts at 2 × 400/3333 = 0.240, so no overlap. 2nd-order red ends at 2 × 700/3333 = 0.420, but 3rd-order violet starts at 3 × 400/3333 = 0.360 < 0.420, so they overlap.
5The tracks of a DVD are 0.74 µm apart. Find the first-order angle for 650 nm light at normal incidence. Is there a second order?
sin θ = 0.650/0.74 = 0.878, θ = 61.4°. Second order would need sin θ = 1.76 > 1: no.
Challenge
1Sodium's yellow light is really two lines, 589.0 nm and 589.6 nm. With a grating of 600 lines/mm, find the angle between the two lines in first order and in second order. Which order separates them better?
First order: θ = 20.695° and 20.717°, difference ≈ 0.022°. Second order: θ = 44.975° and 45.033°, difference ≈ 0.058°. The second order separates them better (higher orders spread wavelengths more).
2Blu-ray tracks are 0.32 µm apart. Show that violet light of 405 nm shining straight onto a Blu-ray disc cannot produce a first-order maximum. What does this tell you about the visible light diffracted from such a disc at normal incidence?
sin θ = 0.405/0.32 = 1.27 > 1: impossible. Every visible wavelength (400 to 700 nm) is longer than 0.32 µm, so at normal incidence there is no first-order maximum for any visible colour; only the m = 0 reflection. Coloured bands can appear only for light arriving at a steep angle.
3A grating produces a second-order maximum for 600 nm light at 30.0°. Find the number of lines per millimetre, and the largest wavelength that could appear in third order.
d = mλ/sin θ = 2 × 6.00 × 10⁻⁷ ÷ 0.500 = 2.40 × 10⁻⁶ m, so N = 1/d = 417 lines/mm. Third order needs 3λ ≤ d: λ ≤ 2.40 × 10⁻⁶ ÷ 3 = 800 nm (all visible wavelengths can appear in third order).
Quiz
1. What is the spacing d of a grating with 400 lines/mm?
2. Compared with a double slit of the same d, a grating produces maxima that are…
3. Light of 600 nm hits a grating with d = 2.0 × 10⁻⁶ m. The first-order angle is…
4. For λ = 500 nm and d = 1.2 × 10⁻⁶ m, the highest order is…
5. A spectrum of bright lines on a dark background comes from…
6. In a grating spectrum, which colour is closest to the central maximum?
Big idea: gratings (E2.3, E2.4, E3.2, E3.3) and one technology based on them, the spectroscope (E1.1). The grating equation is the double-slit bright-fringe equation; the new skills are converting lines/mm to d and finding the highest order.
How to help: the CD spectroscope is cheap and fun; please do the cutting yourself. Ask your teen to compare an incandescent bulb, a fluorescent tube and an LED bulb, and to explain the differences.
Safety: never point a spectroscope or grating at the Sun.
Common difficulties:
- Using N (lines per mm) instead of d = 1/N, or forgetting to convert mm to m.
- Using the small-angle approximation for large grating angles; use x = L tan θ.
- Rounding the highest order up instead of down.
- Forgetting that a grating bends red the most, the reverse of a prism.