Unit 14 · Topic 14.7

14.7 Diffraction

Diffraction means a wave fans out as it passes through an opening or around an obstacle, and it's strongest when the opening is about as wide as the wavelength. Light through a single narrow slit makes a wide bright central band with dimmer bands on each side, because waves from different parts of the slit interfere. Dark bands sit where the path length difference across the slit is a whole number of wavelengths: $\Delta D = a\sin\theta = m\lambda$ . Before 2024–25 this sat in one optics unit with mirrors and lenses; now it's grouped with the other wave topics.

Key terms

diffraction
single slit
path length difference
central maximum
small-angle approximation

What diffraction is

Ocean waves passing through a gap in a harbor wall don't stay in a narrow beam; they fan out on the other side. That spreading is diffraction. It happens to every kind of wave.

Diffraction is strongest when the opening is roughly the same size as the wavelength. Sound has wavelengths around a meter, about the size of a doorway, so it spreads around corners and you can hear someone in the next room. Visible light has wavelengths under a micrometer, so a doorway is enormous by comparison and light barely spreads. You can't see around the corner.

This is why the ray model of Unit 13 works for big openings but fails for tiny ones.

The single-slit pattern

Shine monochromatic light (light of a single wavelength λ), such as a laser, through a slit of width a onto a screen a distance L away. You see a bright central band, called the central maximum, with dimmer bright bands on each side, separated by dark bands.

The central maximum is twice as wide as the other bright bands and much brighter. The pattern comes from interference: every point across the slit acts as a source of wavelets, and those wavelets travel different distances to each point on the screen.

Where the dark bands are

Look at light leaving the slit at an angle θ. The wavelets from the top and bottom edges of the slit travel different distances; the path length difference is $\Delta D = a\sin\theta$ .

When that difference is exactly one wavelength, pair each wavelet in the top half of the slit with one in the bottom half, half a slit width lower. Each pair differs in path by half a wavelength, so they cancel. Every wavelet has a partner, so the screen is dark there.

The same argument works for 2λ, 3λ and so on, so the dark bands are at $a\sin\theta = m\lambda$ , with m = 1, 2, 3, …. Note that m = 0 is the bright center, not a dark band.

For small angles (less than about 10°), $\sin\theta \approx \tan\theta = \frac{y}{L}$ . The equation sheet gives $a\left(\frac{y_{\min}}{L}\right) \approx m\lambda$ , where $y_{\min}$ is the distance from the center of the pattern to the mth dark band.

How the pattern changes

Solve for the position: $y_{\min} = \frac{m\lambda L}{a}$ . So the pattern spreads out when the wavelength is longer, when the screen is farther away, or when the slit is narrower. A narrower slit spreads light more, which surprises many students.

Red light (long λ) makes a wider pattern than blue light (short λ) through the same slit.

The shape of the opening matters too. A circular hole makes a bright central disk surrounded by rings, and a long slit makes bands spread out perpendicular to the slit's length.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Locating the dark bands
Red laser light with λ = 633 nm passes through a slit 0.10 mm wide onto a screen 2.0 m away. How far from the center is the first dark band, and how wide is the central maximum?
Show the solutionHide the solution
1. Step 1: Use the small-angle form with m = 1: $y_{\min} = \dfrac{m\lambda L}{a} = \dfrac{(1)(633 \times 10^{-9})(2.0)}{0.10 \times 10^{-3}} = 0.0127$ m, about 1.3 cm.
2. Step 2: The central maximum stretches from the first dark band on one side to the first dark band on the other, so its width is 2 × 1.27 cm ≈ 2.5 cm.
3. Step 3: Check the small-angle approximation: $\frac{y}{L} = \frac{0.0127}{2.0} \approx 0.006$ , a tiny angle, so it's fine.
Answer: First dark band 1.3 cm from center; central maximum about 2.5 cm wide.
Example 2
Factor of change
A single-slit pattern has its first dark bands 1.2 cm from the center. Predict the new distance if (a) the light's frequency is doubled, or (b) the slit width is cut in half (each change made separately).
Show the solutionHide the solution
1. Step 1: (a) Doubling the frequency halves the wavelength (λ = c/f). Since $y_{\min} \propto \lambda$ , the distance halves to 0.60 cm.
2. Step 2: (b) Since $y_{\min} \propto \frac{1}{a}$ , halving the slit width doubles the distance, to 2.4 cm.
3. Step 3: A common slip is to think a narrower slit squeezes the pattern. It spreads it.
Answer: (a) 0.60 cm, (b) 2.4 cm.
Example 3Calculator allowed
Finding a slit width
Green light (λ = 532 nm) passes through a slit onto a screen 1.5 m away. The central bright band is 1.8 cm wide. How wide is the slit?
Show the solutionHide the solution
1. Step 1: The first dark band is half the central width from the center: $y_{\min}$ = 0.90 cm = 0.0090 m.
2. Step 2: Rearrange $a\left(\dfrac{y_{\min}}{L}\right) = m\lambda$ with m = 1: $a = \dfrac{\lambda L}{y_{\min}} = \dfrac{(532 \times 10^{-9})(1.5)}{0.0090} = 8.9 \times 10^{-5}$ m.
3. Step 3: That's about 0.089 mm, a typical slit width.
Answer: a ≈ 8.9 × 10⁻⁵ m (about 0.09 mm)

Common mistakes

Using the full width of the central maximum as $y_{\min}$ . $y_{\min}$ is measured from the center, so it's half the central width.
Thinking a narrower slit gives a narrower pattern. Narrower slits spread light more.
Using $a\sin\theta = m\lambda$ for bright bands. For a single slit, that equation locates the dark bands.
Forgetting to convert nm and mm to meters before calculating.

On the exam

Expect proportional reasoning: predict how the pattern changes if λ, a or L changes. State the relationship ( $y_{\min} \propto \frac{\lambda L}{a}$ ) as your justification.
You may be asked why you can hear around a corner but not see around it. Compare each wavelength with the size of the opening.

Connected topics

Videos

AP Physics 2: Single Slit Diffraction (Unit 14) - Step-by-Step Derivations & Examples
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Wave Diffraction
Bozeman ScienceWatch on YouTube (opens in a new tab)
Diffraction and interference of light | Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Single Slit Diffraction - Physics Problems
The Organic Chemistry TutorWatch on YouTube (opens in a new tab)
Interference, Reflection, and Diffraction
Professor Dave ExplainsWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 14.7 Diffraction. Pick an answer to see if you got it, and why.

Green laser light with a wavelength of 500 nm passes through a single slit 2.0 × 10⁻⁵ m wide. The diffraction pattern falls on a screen 2.0 m away. Use the small-angle approximation.

Described experiment

Question 1 of 4Calculator allowed

How far from the center of the pattern is the first dark band on the screen?

Question 2 of 4Calculator allowed

The slit is replaced with one half as wide, with everything else the same. How does the central bright band change?

Question 3 of 4Calculator allowed

The green laser is replaced with a red laser (longer wavelength), using the original slit. How does the pattern change?

Question 4 of 4Calculator allowed

A person in a hallway can hear people talking in a room through an open doorway, but can't see them. Which explanation is best?

0 of 4 answered