# C.7 Diffraction and resolution (AHL)

> IB Physics HL · IB DP Physics 2025
> Source: https://www.owlsprep.com/study/ib-physics-hl-u3-c-7-diffraction-and-resolution/

This subtopic covers single-slit diffraction patterns, the position of diffraction minima, and the Rayleigh criterion for image resolution. You will learn to solve problems about resolving power of common optical instruments like telescopes.

**Prerequisites:** [Wave properties and Huygens' principle](https://www.owlsprep.com/study/ib-physics-hl-u3-wave-basics/); [Interference of waves](https://www.owlsprep.com/study/ib-physics-hl-u3-c-6-interference/)

## Learning objectives

- Describe single-slit diffraction patterns and calculate the width of the central maximum
- Recall and apply the single-slit diffraction formula for minima
- State and explain the Rayleigh criterion for resolvability of two sources
- Solve problems involving resolution of common optical instruments

## Single-slit diffraction

**Single-slit diffraction** — The spreading of monochromatic light passing through a narrow slit produces a characteristic pattern: a wide bright central maximum, with smaller dimmer secondary maxima separated by dark minima.

*Example:* Light passing through a 0.1 mm vertical slit produces a horizontal diffraction pattern on a distant screen.

Per Huygens' principle, every point across the width of the slit acts as a secondary source of wavelets. Destructive interference between these wavelets produces dark minima at angles given by:

$$a \sin\theta = n\lambda \quad n = 1, 2, 3, ...$$

Where $a$ = slit width, $\theta$ = angular position of the $n$-th minimum, $\lambda$ = wavelength. For small angles, $\sin\theta \approx \theta \approx \frac{y}{D}$, where $y$ = distance from central maximum to the minimum on the screen, and $D$ = distance from slit to screen. The width of the central maximum is twice the distance to the first ($n=1$) minimum.

**Worked example:** A slit of width 0.10 mm is illuminated by 450 nm blue light. Find the width of the central maximum on a screen 2.0 m from the slit.

1. Convert all values to SI units:
2. $$a = 0.10 \, \text{mm} = 1.0 \times 10^{-4} \, \text{m}, \quad \lambda = 450 \, \text{nm} = 4.5 \times 10^{-7} \, \text{m}, \quad D = 2.0 \, \text{m}$$
3. Find $\sin\theta$ for the first minimum ($n=1$):
4. $$\sin\theta = \frac{n\lambda}{a} = \frac{1 \times 4.5 \times 10^{-7}}{1.0 \times 10^{-4}} = 4.5 \times 10^{-3}$$
5. Use small angle approximation to find $y$, distance from center to first minimum:
6. $$y = D \sin\theta = 2.0 \times 4.5 \times 10^{-3} = 9.0 \times 10^{-3} \, \text{m}$$
7. Central maximum width is twice $y$:
8. $$\text{Width} = 2y = 1.8 \times 10^{-2} \, \text{m} = 18 \, \text{mm}$$

> **Exam tip:** Narrower slits produce wider central maxima: remember the inverse relationship between slit width and diffraction pattern size.

## The Rayleigh criterion

**Rayleigh criterion** — Two point sources are just resolvable by an aperture when the central maximum of the diffraction pattern of one source coincides with the first minimum of the diffraction pattern of the other.

For a circular aperture (the most common case for lenses, mirrors, and pupils), the minimum angular separation between two just resolvable sources is given by:

$$\theta = 1.22 \frac{\lambda}{D}$$

Where $D$ is the diameter of the circular aperture. The factor 1.22 comes from the mathematical solution for diffraction from a circular shape, and does not apply to rectangular single slits.

> **tip**
>
> A smaller $\theta$ means better resolution: resolution improves with shorter wavelengths and larger apertures.

**Worked example:** Two point sources 1.5 m apart are 1.0 km from an observer. Find the minimum pupil diameter required to just resolve the two sources, for 550 nm light.

1. Calculate the angular separation of the sources with small angle approximation:
2. $$\theta \approx \frac{\text{separation}}{\text{distance}} = \frac{1.5}{1000} = 1.5 \times 10^{-3} \, \text{radians}$$
3. Rearrange Rayleigh criterion to solve for $D$:
4. $$D = 1.22 \frac{\lambda}{\theta}$$
5. Substitute values:
6. $$D = 1.22 \times \frac{550 \times 10^{-9}}{1.5 \times 10^{-3}} \approx 4.5 \times 10^{-4} \, \text{m} = 0.45 \, \text{mm}$$

## Resolution of optical instruments

Resolution is a key performance metric for all imaging devices. Common practical examples include:

- **Astronomical telescopes**: Large primary mirrors improve resolution by increasing $D$
- **Radio telescopes**: Require very large apertures (or arrays of telescopes) because radio wavelengths are thousands of times longer than light wavelengths
- **Electron microscopes**: Use electrons with very short de Broglie wavelengths to get much higher resolution than light microscopes

**Worked example:** Compare the angular resolution of a 76 m diameter radio telescope observing 1420 MHz radio waves, and a 1 m diameter optical telescope observing 550 nm light.

1. Calculate radio wavelength from $c = f\lambda$:
2. $$\lambda_{\text{radio}} = \frac{3.0 \times 10^8}{1420 \times 10^6} \approx 0.21 \, \text{m}$$
3. Calculate radio telescope resolution:
4. $$\theta_{\text{radio}} = 1.22 \frac{0.21}{76} \approx 3.4 \times 10^{-3} \, \text{rad} = 0.19^\circ$$
5. Calculate optical telescope resolution:
6. $$\theta_{\text{optical}} = 1.22 \frac{550 \times 10^{-9}}{1} \approx 6.7 \times 10^{-7} \, \text{rad} = 0.000038^\circ$$
7. Even with a 76x larger aperture, the radio telescope has ~5000x worse resolution than the optical telescope, due to the much longer wavelength of radio waves.

**Check your understanding**

Test your understanding:

1. Which of the following increases the resolving power of a telescope?

   - Increase the wavelength of observed light
   - Increase the diameter of the primary mirror
   - Decrease the diameter of the primary mirror
   - Increase the eyepiece focal length

   *Answer:* Increase the diameter of the primary mirror

   *Why:* Resolving power improves when minimum resolvable $\theta$ is smaller. From $\theta = 1.22 \lambda/D$, increasing $D$ decreases $\theta$, so resolving power increases.

## Common pitfalls

- **Wrong:** Forgetting the 1.22 factor for circular apertures, using the single-slit formula $\theta = \lambda/a$ instead
  - Why it fails: Most exam problems involve circular apertures (lenses, pupils, mirrors), which require the 1.22 correction from circular diffraction mathematics
  - Correct: Always check aperture shape: use $\theta = 1.22 \lambda/D$ for circular apertures, no 1.22 for rectangular single slits
- **Wrong:** Claiming narrower slits produce narrower diffraction patterns
  - Why it fails: The angular size of the diffraction pattern is proportional to $\lambda/a$, so smaller $a$ (narrower slit) gives a larger pattern
  - Correct: Remember: smaller aperture = more spreading = wider diffraction = worse resolution
- **Wrong:** Using degrees instead of radians for small angle approximation calculations
  - Why it fails: The approximation $\sin\theta \approx \theta$ only holds when $\theta$ is measured in radians
  - Correct: Always convert angular values to radians for diffraction and resolution problems
- **Wrong:** Using $n=0$ for the first minimum in single-slit diffraction
  - Why it fails: $n=0$ corresponds to the center of the pattern, which is the central maximum, not a minimum
  - Correct: The first dark minimum is at $n=1$, so central maximum width is twice the position of the $n=1$ minimum

## Cheatsheet

| Concept | Formula | Key Note |
| --- | --- | --- |
| Single-slit minima | $a \sin\theta = n\lambda$ $n=1,2,3...$ | Gives position of dark fringes |
| Central maximum width | $2y = 2D\lambda/a$ | Small angle approximation |
| Rayleigh (circular aperture) | $\theta = 1.22 \lambda/D$ | Minimum resolvable angular separation |
| Rayleigh (rectangular slit) | $\theta = \lambda/a$ | No 1.22 factor |
| Small angle approx | $\theta \approx d/L$ | $\theta$ must be in radians |

## What's next

Diffraction and resolution underpin all modern imaging technology, from the largest astronomical telescopes to the smallest medical microscopes and smartphone cameras. The wave diffraction principles you learned here are also the basis for X-ray crystallography (used to map molecular structures) and very long baseline interferometry for astronomy, which produces images of black holes. This topic completes the AHL wave behaviour component of the IB Physics syllabus, after which you will move on to apply wave concepts to other topics like astronomy and quantum mechanics.

- [C.6 Interference](https://www.owlsprep.com/study/ib-physics-hl-u3-c-6-interference/)
- [Theme D: Fields](https://www.owlsprep.com/study/ib-physics-hl-u4-overview/)

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