E. The Wave Nature of Light · The Wave Nature of LightSingle-Slit Diffraction
E2.1 E2.3 E2.4 E3.2 A1.8 A1.13About 70 minutes
Learning goals- I can describe the single-slit diffraction pattern: a wide, bright central maximum with dimmer side fringes.
- I can explain the dark fringes by dividing the slit into pairs of wavelets that cancel.
- I can use w sin θₙ = nλ and the central-maximum width 2Lλ/w to solve problems.
- I can compare single-slit and double-slit patterns and explain the dimming of double-slit fringes.
Words to know
single-slit diffraction
The pattern of light and dark fringes made when waves pass through one narrow opening.
central maximum
The wide, bright band in the middle of a single-slit pattern, between the first dark fringes on either side.
slit width (w)
The width of the single opening the light passes through.
secondary maxima
The dimmer bright fringes on each side of the central maximum.
diffraction pattern
The arrangement of bright and dark fringes made by waves spreading through or around an object.
One slit, many fringes
Shine a laser through one narrow slit and you might expect a single bright line on the wall, maybe a little blurry. Instead you get a striking pattern: a broad, bright central band with dark gaps and fainter bands on either side. With only one opening, what is interfering with what?
The answer uses Huygens' principle: every point across the slit is a source of wavelets, and wavelets from different parts of the same slit interfere with each other.
Where are the dark fringes?
Picture the slit as a row of wavelet sources from the top edge A to the bottom edge C, with B in the middle. Look at light leaving at an angle θ toward a far-away point on the screen.
- Straight ahead (θ = 0) all wavelets travel the same distance and arrive in phase: the centre is bright.
- Now tilt θ until the bottom edge C is exactly one wavelength farther from the point than the top edge A. Then the middle point B is half a wavelength farther than A.
- Pair each wavelet in the top half with the one exactly w/2 below it (A with B, the next point down with the point just below B, and so on). Every pair is λ/2 out of step, so every pair cancels. The whole slit gives darkness.
The path difference from edge to edge is w sin θ, so the first dark fringe is where w sin θ = λ. Dividing the slit into 4, 6, 8… strips gives more dark fringes at 2λ, 3λ, …:
Dark fringes: w sin θₙ = nλ, n = 1, 2, 3, …
Be careful: this looks like the double-slit bright fringe equation, but here it gives dark fringes, and w is the width of one slit, not the distance between two slits.
Width of the pattern
For small angles, sin θ ≈ x/L, so the dark fringes are at xₙ = nLλ/w.
- The side fringes are each Lλ/w wide.
- The central maximum stretches from the first dark fringe on one side to the first dark fringe on the other: its width is 2Lλ/w, twice the width of the others.
- The side (secondary) maxima are much dimmer: the first is less than 5% as bright as the centre.
The key idea: the narrower the slit, the wider the pattern. Squeeze the slit and the light spreads more, exactly as the ripple tank showed. If w is smaller than λ, even the first dark fringe would need sin θ > 1, so there are no dark fringes at all: the light spreads out in every forward direction.
Worked exampleExample 1: Angles and widths
Light of wavelength 600 nm passes through a slit 0.10 mm wide onto a screen 2.0 m away. Find the angle to the first dark fringe and the width of the central maximum.
- sin θ₁ = λ/w = 6.00 × 10⁻⁷ ÷ 1.0 × 10⁻⁴ = 6.0 × 10⁻³, so θ₁ = 0.34°.
- Central maximum width = 2Lλ/w = 2(2.0)(6.00 × 10⁻⁷) ÷ 1.0 × 10⁻⁴.
- Width = 0.024 m.
θ₁ = 0.34°; the central maximum is 2.4 cm wide.
Worked exampleExample 2: Measuring a slit
A helium-neon laser (λ = 633 nm) shines through an adjustable slit. On a wall 1.5 m away, the central maximum is 1.6 cm wide. How wide is the slit?
- Rearrange width = 2Lλ/w: w = 2Lλ/width.
- w = 2(1.5)(633 × 10⁻⁹) ÷ 0.016.
- w = 1.19 × 10⁻⁴ m.
Worked exampleExample 3: Measuring a hair
A hair placed in a laser beam (λ = 650 nm) makes a diffraction pattern just like a slit of the same width. On a wall 2.0 m away, the first dark fringes on either side of the centre are 3.0 cm apart. Estimate the hair's thickness.
- The distance between the first dark fringes is the central-maximum width: 2Lλ/w = 0.030 m.
- w = 2Lλ/0.030 = 2(2.0)(6.50 × 10⁻⁷) ÷ 0.030 = 8.7 × 10⁻⁵ m.
The hair is about 0.087 mm (87 µm) thick. (An obstacle and a slit of the same width give the same diffraction pattern apart from the very centre; this is called Babinet's principle.)
Single slit meets double slit
Real double slits each have a width w. Each slit makes its own single-slit pattern, and the two slits' light interferes inside it. So a real double-slit pattern is a set of evenly spaced fringes (spacing Lλ/d) whose brightness follows the single-slit envelope (width 2Lλ/w). Because d is always larger than w, many double-slit fringes fit inside the central maximum. That is why the outer fringes in Young's experiment fade away.
How to tell them apart:
- Double slit: evenly spaced, similar bright fringes; the central fringe is the same width as the others.
- Single slit: a central maximum twice as wide as the others and much brighter than the rest.
Just for funA laser pointer walks up to a single slit and asks, "Can I squeeze through?" The slit says, "Sure, but the narrower I am, the more you'll spread yourself thin."
Hands-on activityMeasure the thickness of a hair
You need: a low-power red laser pointer, a hair (from a hairbrush), tape, a cardboard frame (an index card with a hole about 1 cm wide), a metre stick or tape measure, a sheet of white paper taped to a wall, a pencil, and a grown-up.
Laser safety (grown-up needed): use only a low-power laser pointer labelled Class 2 (or Class II) with output under 1 mW, ideally a red one. Never look into the beam or point it at anyone, at a pet, out a window or toward aircraft or vehicles. Keep the beam below eye level, aim it only at a wall or a sheet of paper, and keep shiny objects (mirrors, glass, watches, jewellery) out of its path so it cannot reflect into someone's eye. Do not use a green or blue pointer of unknown power, and never look at the beam through binoculars or a lens. Switch it off as soon as you finish each measurement.
- Tape the hair straight and tight across the hole in the card.
- Set the card about 2 m from the wall. Measure L, the distance from the hair to the paper.
- With the grown-up, rest the laser on a stack of books so it points at the hair (below eye level) and the beam hits the paper. Switch it on briefly.
- Mark the centres of the first dark gaps on each side of the bright central band. Switch the laser off and measure the distance between the marks (the central-maximum width).
- Calculate w = 2Lλ/width, using the wavelength printed on the laser (often 650 nm).
- Repeat with a hair from another person, or a thin strand of thread.
What to look for: a bright central band with dimmer bands on each side, at right angles to the hair. Many human hairs give results of a few hundredths to about a tenth of a millimetre. Thicker hair gives a narrower pattern.
Your turn
1Write the condition for dark fringes in single-slit diffraction.
w sin θₙ = nλ, n = 1, 2, 3, …
2Light of λ = 500 nm passes through a 0.050 mm slit. Find the angles to the first and second dark fringes.
sin θ₁ = 5.00 × 10⁻⁷ ÷ 5.0 × 10⁻⁵ = 0.010, θ₁ = 0.57°; sin θ₂ = 0.020, θ₂ = 1.15°.
3For the same slit with the screen 1.5 m away, how wide is the central maximum?
2Lλ/w = 2 × 1.5 × 5.00 × 10⁻⁷ ÷ 5.0 × 10⁻⁵ = 0.030 m = 3.0 cm.
4A slit is made narrower. What happens to the central maximum?
It gets wider (width = 2Lλ/w).
5How does the width of the central maximum compare with the width of the other bright fringes?
6Why is the centre of a single-slit pattern bright?
Straight ahead, wavelets from every part of the slit travel the same distance and arrive in phase.
7Red or blue light: which gives a wider single-slit pattern through the same slit?
More practice: warm-up
1In single-slit diffraction, what does w stand for?
2What is the path difference from one edge of the slit to the other at the first dark fringe?
3Is the central maximum bright or dark?
Bright (the brightest part of the pattern).
4Convert 0.020 mm to metres.
5What happens to the pattern if the screen is moved farther away?
It gets wider (all widths are proportional to L).
More practice: on your own
1A laser (λ = 633 nm) makes a central maximum 2.0 cm wide on a screen 2.5 m from a slit. Find the slit width.
w = 2Lλ/width = 2 × 2.5 × 633 × 10⁻⁹ ÷ 0.020 = 1.6 × 10⁻⁴ m (0.16 mm).
2Light of λ = 450 nm passes through a slit 2.0 µm wide. Find the angle to the first dark fringe. Why can't the small-angle formula be used here?
sin θ = 4.5 × 10⁻⁷ ÷ 2.0 × 10⁻⁶ = 0.225, θ = 13.0°. The angle is too large for sin θ ≈ tan θ to be accurate.
3Explain why there are no dark fringes at all when a slit is narrower than the wavelength.
The first dark fringe needs sin θ = λ/w. If w < λ this would be greater than 1, which is impossible, so the light spreads out with no dark fringes.
4A double slit has slits 0.040 mm wide and 0.20 mm apart. How many double-slit bright fringes fit inside the single-slit central maximum (count the central fringe)?
Single-slit first dark: sin θ = λ/w. Double-slit bright: sin θ = mλ/d. They coincide at m = d/w = 0.20/0.040 = 5 (so the 5th order is missing). Inside: m = 0, ±1, ±2, ±3, ±4: 9 fringes.
5How can you tell from a photo whether a pattern came from a single slit or a double slit?
Single slit: a central band twice as wide as the rest and much brighter than the side bands. Double slit: evenly spaced fringes of similar width (fading slowly at the edges).
Challenge
1A slit is illuminated first with 600 nm light and then with an unknown light. The 3rd dark fringe of the unknown light is at the same place as the 2nd dark fringe of the 600 nm light. Find the unknown wavelength.
Same angle: 3λ = 2(600 nm), so λ = 400 nm.
2Light of λ = 600 nm passes through a slit 0.040 mm wide onto a screen 1.2 m away. Find the distance between the 3rd dark fringes on opposite sides of the centre.
x₃ = 3Lλ/w = 3 × 1.2 × 6.0 × 10⁻⁷ ÷ 4.0 × 10⁻⁵ = 0.054 m; between opposite sides: 2 × 0.054 = 0.108 m ≈ 11 cm.
3Explain, using wavelets, why the second dark fringe occurs where w sin θ = 2λ.
Divide the slit into four equal strips. When the edge-to-edge path difference is 2λ, neighbouring strips differ by λ/2. Strip 1 cancels strip 2 and strip 3 cancels strip 4, wavelet by wavelet, so the direction is dark.
Quiz
1. The equation w sin θ = nλ gives the positions of…
2. Compared with the side bright fringes, the central maximum is…
3. A slit is made narrower. The diffraction pattern…
4. Light of λ = 500 nm passes through a 0.10 mm slit onto a screen 2.0 m away. The central maximum is…
5. Why do the outer fringes of a real double-slit pattern fade?
6. A slit is 300 nm wide and the light has λ = 500 nm. What happens?
Big idea: single-slit diffraction (E2.3, E2.4, E3.2). The trap is the equation: w sin θ = nλ gives dark fringes, the opposite of the double-slit equation, which gives bright fringes with mλ.
How to help: the hair-measuring activity is memorable and gives a real measurement, but it needs a grown-up for laser safety. Read the safety box together first. If you have no suitable laser, PhET Wave Interference has a single-slit option.
Common difficulties:
- Confusing w (slit width) with d (slit separation).
- Using Lλ/w instead of 2Lλ/w for the central maximum.
- Expecting a narrower slit to give a narrower pattern. It is the reverse.