Waves for AP Physics 2 — AP Physics 2
1. Wave Speed and Wavelength in Optical Media ★★☆☆☆ ⏱ 3 min
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}
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.
2. Young's Double-Slit Interference ★★★☆☆ ⏱ 4 min
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}
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.
3. Single-Slit Diffraction ★★★☆☆ ⏱ 4 min
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.
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').
4. AP-Style Concept Check ★★★☆☆ ⏱ 3 min
Common Pitfalls
Why: Students often use the wavelength given for the laser in air, forgetting that wavelength shrinks in media with higher $n$.
Why: Most problems that test both concepts provide both values, and test if students can match the variable to the formula.
Why: Students mix up which property is determined by the source vs. the medium.
Why: The single-slit minima formula looks identical to the double-slit maxima formula, so students mix up valid $m$ values.
Why: Students calculate the position of the first dark fringe on one side and stop, forgetting width spans both sides of the central maximum.