# Waves for AP Physics 2

> AP Physics 2 · Unit 6: Geometric and Physical Optics
> Source: https://www.owlsprep.com/study/ap-physics-2-u6-waves/

This subtopic covers core wave properties of light for AP Physics 2 Unit 6, including wave speed/wavelength in media, Young's double-slit interference, and single-slit diffraction, with worked examples and exam problem-solving guidance.

**Prerequisites:** Basic wave properties (period, frequency, wavelength) from AP Physics 1; Electromagnetic spectrum and energy-wavelength relationships; Snell's law of refraction for light

## Learning objectives

- Calculate wave speed and wavelength of light in optical media
- Solve problems involving Young's double-slit interference
- Calculate properties of single-slit diffraction patterns
- Identify common misconceptions about wave behavior and interference

## Wave Speed and Wavelength in Optical Media

All electromagnetic waves travel at speed $c = 3.0 \times 10^8 \text{ m/s}$ in vacuum, but slow down when entering a transparent medium with index of refraction $n > 1$. When a wave crosses a boundary between two media, the frequency $f$ does not change, because frequency is a property of the source, not the medium.

$$n = \frac{c}{v}$$

Since the fundamental wave relationship $v = f\lambda$ always holds, wavelength changes proportionally with speed. The wavelength of light in a medium $\lambda_n$ is related to its vacuum wavelength $\lambda_0$ by:

$$\lambda_n = \frac{\lambda_0}{n}$$

**Index of Refraction** — A dimensionless property of a transparent medium equal to the ratio of the speed of light in vacuum to the speed of light in the medium.

*Notation:* n

*Example:* $n_{water} \approx 1.33$, $n_{air} \approx 1.00$

**Worked example:** A blue laser has a wavelength of 420 nm in air (assume $n_{air} = 1$). The laser is shone through a layer of water with $n_{water} = 1.33$ toward a diffraction grating. What is the frequency and wavelength of the laser light inside the water?

1. Calculate frequency in air, which equals frequency in water because frequency is source-dependent:

   $$f = \frac{c}{\lambda_0} = \frac{3.0 \times 10^8 \text{ m/s}}{420 \times 10^{-9} \text{ m}} \approx 7.14 \times 10^{14} \text{ Hz}$$
2. Frequency does not change across the boundary, so frequency inside water is also $7.14 \times 10^{14} \text{ Hz}$.
3. Calculate wavelength in water:

   $$\lambda_n = \frac{\lambda_0}{n} = \frac{420 \text{ nm}}{1.33} \approx 316 \text{ nm}$$

> **Exam tip:** Always check which medium the light is traveling through when calculating interference. If interference occurs inside a medium, use the wavelength in that medium, not the vacuum wavelength given in the problem.

*Calculator:* allowed

## Young's Double-Slit Interference

Young's double-slit experiment was the first definitive confirmation that light behaves as a wave, producing an interference pattern from superposition of waves from two narrow slits separated by distance $d$. Constructive interference (bright fringes) occurs when the path difference between the two waves is an integer multiple of the wavelength:

$$d \sin\theta = m\lambda \quad (m = 0, \pm 1, \pm 2, ...)$$

Where $m$ is the order of the bright fringe, and $\theta$ is the angle from the central maximum ($m=0$, the brightest fringe at the center of the screen). Destructive interference (dark fringes) occurs when the path difference is a half-integer multiple of wavelength: $d \sin\theta = (m + 1/2)\lambda$. For small angles (when $L \gg y$, where $y$ is the fringe position on the screen and $L$ is the distance from slits to screen), $\sin\theta \approx \tan\theta = \frac{y}{L}$, so the formula simplifies to:

$$y_m = \frac{m \lambda L}{d}$$

**Worked example:** In a double-slit experiment, slit separation is 0.30 mm, the screen is 2.5 m from the slits, and the distance between the central bright fringe and the second-order bright fringe is 8.2 mm. What is the wavelength of the light used, in nanometers?

1. Convert all units to SI:

   $$d = 0.30 \times 10^{-3} \text{ m}, \quad L = 2.5 \text{ m}, \quad y_2 = 8.2 \times 10^{-3} \text{ m}, \quad m=2$$
2. Rearrange the small-angle formula to solve for $\lambda$:

   $$\lambda = \frac{y_m d}{m L}$$
3. Substitute values:

   $$\lambda = \frac{(8.2 \times 10^{-3} \text{ m})(0.30 \times 10^{-3} \text{ m})}{(2)(2.5 \text{ m})} = 4.92 \times 10^{-7} \text{ m}$$
4. Convert to nanometers:

   $$4.92 \times 10^{-7} \text{ m} = 492 \text{ nm}$$

> **Exam tip:** If asked for the distance between two non-central fringes (e.g., between $m=1$ and $m=-2$), calculate the absolute difference of their $y$-positions, don't just multiply the fringe separation by an integer.

*Calculator:* allowed

## Single-Slit Diffraction

Diffraction is the bending of light around the edges of an obstacle, a fundamental property of all waves. When monochromatic light passes through a single narrow slit of width $a$, it produces a diffraction pattern on a distant screen with a wide, intense central maximum, and smaller, dimmer secondary maxima on either side. Dark fringes (minima) in the single-slit pattern occur at angles:

$$a \sin\theta = m \lambda \quad (m = \pm 1, \pm 2, \pm 3, ...)$$

A key note: $m=0$ is not a minimum, it is the center of the bright central maximum. For small angles, the position of the $m$-th dark fringe is $y_m = \frac{m \lambda L}{a}$, and the total width of the central maximum (between the $m=-1$ and $m=1$ dark fringes) is:

$$w = \frac{2 \lambda L}{a}$$

A key conceptual result: narrowing the slit increases the width of the central maximum, a purely wave effect that is frequently tested on the AP exam.

**Worked example:** A 620 nm laser is incident on a single slit of width 0.060 mm, and the pattern is projected onto a screen 2.0 m away. What is the width of the central maximum on the screen?

1. Convert all units to SI:

   $$\lambda = 620 \times 10^{-9} \text{ m}, \quad a = 0.060 \times 10^{-3} \text{ m}, \quad L = 2.0 \text{ m}$$
2. Use the central maximum width formula:

   $$w = \frac{2 \lambda L}{a}$$
3. Substitute values:

   $$w = \frac{2(620 \times 10^{-9})(2.0)}{0.060 \times 10^{-3}} \approx 0.041 \text{ m} = 4.1 \text{ cm}$$

> **Exam tip:** For problems that combine double-slit interference with finite-width slits, remember that the bright double-slit fringes are cut off by the single-slit diffraction envelope: any double-slit bright fringe that falls on a single-slit dark fringe will disappear (a 'missing order').

*Calculator:* allowed

## AP-Style Concept Check

**Check your understanding**

Test your understanding of core relationships with this AP-style multiple choice question:

1. A student measures the fringe separation (distance between adjacent bright fringes) in a double-slit experiment as $\Delta y$. The student then replaces the double-slit with a new set that has three times the original slit separation, and uses a light source with one-third the original wavelength. The distance from the slits to the screen is unchanged. What is the new fringe separation?

   - $\Delta y / 9$
   - $\Delta y / 3$
   - $\Delta y$
   - $9 \Delta y$

   *Answer:* $\Delta y / 9$

   *Why:* Fringe separation between adjacent bright fringes is $\Delta y = \lambda L / d$. Substituting $\lambda' = \lambda/3$ and $d' = 3d$ gives $\Delta y' = (\lambda/3 \cdot L)/(3d) = \lambda L/(9d) = \Delta y/9$.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using the vacuum wavelength of light for interference that occurs inside a different medium.
  - Why it fails: Students often use the wavelength given for the laser in air, forgetting that wavelength shrinks in media with higher $n$.
  - Correct: Always note the medium where interference occurs, and divide the vacuum wavelength by the medium's index of refraction before doing any calculations.
- **Wrong:** Confusing $d$ (distance between slits in double-slit) with $a$ (width of a single slit) in formulas.
  - Why it fails: Most problems that test both concepts provide both values, and test if students can match the variable to the formula.
  - Correct: Label every given variable on your paper before plugging in: $d$ = distance between two slits, $a$ = width of one slit.
- **Wrong:** Treating frequency as the variable that changes when light enters a new medium, keeping wavelength constant.
  - Why it fails: Students mix up which property is determined by the source vs. the medium.
  - Correct: Memorize: frequency = source property, never changes across a boundary; wavelength and speed = medium properties, change with $n$.
- **Wrong:** Including $m=0$ as a dark fringe for single-slit diffraction, or using $m=0$ as anything other than the central bright maximum.
  - Why it fails: The single-slit minima formula looks identical to the double-slit maxima formula, so students mix up valid $m$ values.
  - Correct: Always remember: $m=0$ is the central bright maximum for both experiments; single-slit minima start at $m=\pm 1$.
- **Wrong:** Calculating the width of the single-slit central maximum as $y_1 = \lambda L /a$ instead of $2\lambda L /a$.
  - Why it fails: Students calculate the position of the first dark fringe on one side and stop, forgetting width spans both sides of the central maximum.
  - Correct: Always multiply the position of the first dark fringe by 2 to get the total width of the central maximum.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Speed of light in medium | $n = \frac{c}{v}$ | $n$ = index of refraction, $c$ = speed in vacuum |
| Wavelength in medium | $\lambda_n = \frac{\lambda_0}{n}$ | $\lambda_0$ = vacuum wavelength; frequency does not change across boundaries |
| Double-slit bright fringes | $d \sin\theta = m\lambda$; $y_m = \frac{m \lambda L}{d}$ | $m = 0, \pm1, \pm2...$; small-angle approximation for $y_m$ |
| Double-slit dark fringes | $d \sin\theta = (m + 1/2)\lambda$ | $m = 0, \pm1, \pm2...$; destructive interference |
| Single-slit dark fringes | $a \sin\theta = m\lambda$ | $m = \pm1, \pm2...$; $m=0$ is central bright maximum |
| Single-slit central maximum width | $w = \frac{2 \lambda L}{a}$ | Small-angle approximation; $a$ = single slit width |
| Double-slit adjacent fringe separation | $\Delta y = \frac{\lambda L}{d}$ | Distance between consecutive bright fringes for small angles |

## What's next

This subtopic lays the foundation for all further wave optics topics in AP Physics 2 Unit 6. The interference and diffraction principles you learned here apply directly to thin film interference, diffraction gratings, and resolution problems for optical instruments, all of which are frequently tested on both multiple-choice and free-response sections of the AP exam. Understanding how wavelength changes in different media is especially critical for thin film interference problems, where you must calculate path differences inside coated lenses or thin material layers. Diffraction is also the core concept behind resolving power of telescopes and microscopes, connecting wave behavior to real-world applications.

- [Unit 6 Overview](https://www.owlsprep.com/study/ap-physics-2-u6-overview/)
- [Electromagnetic Waves](https://www.owlsprep.com/study/ap-physics-2-u6-electromagnetic-waves/)
- [Geometric Optics: Refraction and Reflection](https://www.owlsprep.com/study/ap-physics-2-u6-geometric-optics-refraction-and-reflection/)

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