# C.6 Interference (AHL)

> IB Physics Higher Level · IB Physics AHL Theme C: Wave Behaviour
> Source: https://www.owlsprep.com/study/ib-physics-hl-u3-c-6-interference/

This sub-topic covers wave interference from superposition, including double-slit two-source interference, diffraction grating patterns, and thin-film interference effects. You will learn to apply key equations to solve common exam problems.

**Prerequisites:** [Superposition principle](https://www.owlsprep.com/study/ib-physics-hl-u3-c-1-superposition/); [Single-slit diffraction](https://www.owlsprep.com/study/ib-physics-hl-u3-c-5-diffraction/)

## Learning objectives

- Explain the conditions required for stable observable interference
- Calculate fringe spacing for double-slit interference setups
- Solve problems using the diffraction grating equation
- Distinguish between constructive and destructive interference
- Account for phase change on reflection in thin-film interference

## Conditions for Observable Interference

Interference occurs when two or more waves superpose to form a resultant wave of greater, lower, or equal amplitude. For a stable, visible interference pattern to form, two core conditions must be satisfied.

**Coherent Sources** — Sources that produce waves with the same frequency and a constant phase difference over time. Incoherent sources (such as two separate incandescent bulbs) cannot produce stable patterns because phase difference changes randomly.

The type of interference at any point depends on the path difference between the two waves reaching that point:

- Constructive interference (maximum intensity): path difference $\Delta = n\lambda$, where $n = 0, 1, 2, ...$
- Destructive interference (minimum intensity): path difference $\Delta = (n + \frac{1}{2})\lambda$, where $n = 0, 1, 2, ...$

**Worked example:** Two coherent microwave sources emit 3.0 cm wavelength waves. What path differences give points of minimum intensity?

1. Minimum intensity corresponds to destructive interference, which requires:
2. $$\Delta = (n + \frac{1}{2})\lambda, \quad n = 0, 1, 2, ...$$
3. Substitute $\lambda = 3.0$ cm:
4. $$\Delta = 1.5 \text{ cm}, 4.5 \text{ cm}, 7.5 \text{ cm}, ...$$

> **Exam tip:** Always confirm if the question asks for maximum or minimum intensity before applying the path difference condition.

## Double-Slit Interference

Young's double-slit experiment was the first definitive proof that light behaves as a wave. When monochromatic coherent light passes through two narrow slits, it produces a pattern of equally spaced bright and dark fringes on a distant screen.

**Derivation:** Derive the fringe spacing equation for double-slit interference

*Starting from:* Condition for constructive interference: $d \sin\theta = n\lambda$, where $d$ is slit separation

1. For small angles (when screen distance $D \gg d$):
2. $$\sin\theta \approx \tan\theta = \frac{x_n}{D}$$
3. Where $x_n$ is the distance from the central maximum to the nth maximum. Equate the two expressions:
4. $$d \cdot \frac{x_n}{D} = n\lambda \implies x_n = \frac{n\lambda D}{d}$$
5. Fringe spacing $\Delta x$ is the distance between adjacent maxima: $\Delta x = x_{n+1} - x_n$

*Conclusion:* The fringe spacing equation is:

$$\Delta x = \frac{\lambda D}{d}$$

**Worked example:** A double slit with slit separation 0.25 mm is placed 1.2 m from a screen. Fringe spacing is measured as 2.8 mm. Calculate the wavelength of the light.

1. Convert all units to SI (metres):
2. $$d = 0.25 \times 10^{-3} \text{ m}, \quad D = 1.2 \text{ m}, \quad \Delta x = 2.8 \times 10^{-3} \text{ m}$$
3. Rearrange for $\lambda$:
4. $$\lambda = \frac{\Delta x \cdot d}{D}$$
5. Substitute values:
6. $$\lambda = \frac{(2.8 \times 10^{-3})(0.25 \times 10^{-3})}{1.2} = 5.83 \times 10^{-7} \text{ m} = 580 \text{ nm}$$

> **Exam tip:** Always convert all units to SI before calculation to avoid order of magnitude errors.

## Diffraction Gratings

A diffraction grating consists of hundreds or thousands of equally spaced parallel slits etched into a transparent substrate. It produces sharp, widely spaced bright maxima, making it ideal for accurate wavelength measurement.

**Grating Spacing** — The distance between the centres of two adjacent slits, calculated as $d = \frac{1}{N}$ where $N$ is the number of slits per unit length.

*Notation:* $d$

The condition for constructive interference (bright maximum) from adjacent slits is:

$$d \sin\theta = n \lambda$$

The maximum possible order of maximum is limited by $\sin\theta \leq 1$, so $n_{max} = \lfloor \frac{d}{\lambda} \rfloor$.

**Worked example:** A diffraction grating has 500 lines per mm. Calculate the angle of the second order maximum for 500 nm green light.

1. Calculate grating spacing $d$ in metres:
2. $$d = \frac{1 \times 10^{-3} \text{ m}}{500} = 2.0 \times 10^{-6} \text{ m}$$
3. Rearrange the grating equation for $\sin\theta$:
4. $$\sin\theta = \frac{n\lambda}{d}$$
5. Substitute $n=2$, $\lambda = 500 \times 10^{-9}$ m:
6. $$\sin\theta = \frac{2 \times 500 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.5$$
7. Solve for $\theta$:
8. $$\theta = \sin^{-1}(0.5) = 30^\circ$$

> **Exam tip:** If you calculate $\sin\theta > 1$, the order you are calculating does not exist.

## Thin-Film Interference

Thin-film interference occurs when light reflected from the top and bottom surfaces of a thin transparent film superposes to produce an interference pattern. This is the effect that creates rainbow colours on soap bubbles and oil slicks.

> **info**
>
> When light reflects off a medium with a higher refractive index, it undergoes a phase change of $\pi$, which is equivalent to an extra path difference of $\frac{\lambda}{2}$. This flips the conditions for constructive and destructive interference.

**Check your understanding**

Test your understanding of phase change:

1. Light travels from air ($n=1.0$) to glass ($n=1.5$) and reflects off the boundary. What phase change occurs?

   - No phase change
   - Phase change of $\frac{\pi}{2}$
   - Phase change of $\pi$
   - Phase change of $2\pi$

   *Answer:* Phase change of $\pi$

   *Why:* Correct. Reflection off a higher refractive index medium always produces a $\pi$ phase change.

## Common pitfalls

- **Wrong:** Forgetting to convert all units to SI before calculation
  - Why it fails: Grating spacing is often given in lines per mm and wavelength in nm, mismatched units give wrong order of magnitude results
  - Correct: Convert all lengths to metres before substituting into any interference equation
- **Wrong:** Mixing up constructive and destructive interference path difference conditions
  - Why it fails: It is easy to confuse the conditions for maxima and minima leading to wrong answers
  - Correct: Remember: Constructive = $n\lambda$, Destructive = $(n + \frac{1}{2})\lambda$
- **Wrong:** Ignoring phase change on reflection in thin-film questions
  - Why it fails: Phase change adds an extra $\frac{\lambda}{2}$ path difference that flips the interference conditions
  - Correct: Always check for reflections off higher refractive index boundaries and account for phase change
- **Wrong:** Calculating grating spacing as $N$ (lines per unit length) instead of $1/N$
  - Why it fails: Grating spacing is the distance between slits, not the number of slits per unit length
  - Correct: If there are $N$ lines per mm, $d = \frac{1}{N}$ mm, convert to metres for calculation
- **Wrong:** Using small angle approximation for diffraction grating angles
  - Why it fails: Diffraction grating maxima are often at large angles where $\sin\theta \neq \tan\theta$
  - Correct: Always use $d\sin\theta = n\lambda$ directly for gratings, do not use the double-slit fringe formula

## Cheatsheet

| Concept | Equation | Key Notes |
| --- | --- | --- |
| Constructive interference | $\Delta = n\lambda$, $n=0,1,2...$ | Phase difference = $2n\pi$ |
| Destructive interference | $\Delta = (n+\frac{1}{2})\lambda$, $n=0,1,2...$ | Phase difference = $(2n+1)\pi$ |
| Double-slit fringe spacing | $\Delta x = \frac{\lambda D}{d}$ | Small angles only, $D$ = screen distance |
| Diffraction grating equation | $d\sin\theta = n\lambda$ | $d = 1/N$, $N$ = lines per unit length |
| Phase change on reflection | $\Delta \phi = \pi$, $\Delta x = \lambda/2$ | Only when reflecting off higher $n$ medium |

## What's next

Interference is a core wave phenomenon that underpins many topics in IB Physics, from standing waves to X-ray crystallography and quantum mechanics. The calculation skills you learned here for path difference and interference conditions are frequently tested in both Paper 1 multiple choice and Paper 2 extended response questions. Interference also demonstrates wave behaviour that is central to understanding wave-particle duality in quantum physics. Next, you can build on this knowledge by exploring related topics that rely on the superposition and interference principles you have mastered.

- [C.7 Diffraction and resolution (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u3-c-7-diffraction-and-resolution/)
- [Theme D: Fields](https://www.owlsprep.com/study/ib-physics-hl-u4-overview/)

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