AP Physics 2- 14.7 Diffraction- Study Notes- New Syllabus
AP Physics 2- 14.7 Diffraction – Study Notes
AP Physics 2- 14.7 Diffraction – Study Notes – per latest Syllabus.
Key Concepts:
- Diffraction of Waves
- Single-Slit Diffraction: Small-Angle Approximation
Diffraction of Waves
Diffraction is the bending or spreading of waves when they encounter an obstacle or pass through a narrow opening. It is most noticeable when the size of the opening or obstacle is comparable to the wavelength of the wave.
Behavior of a Wave through a Single Opening
When a wave passes through a slit, the wavefront spreads out on the other side, forming a pattern of alternating regions of constructive and destructive interference.
The amount of spreading (diffraction) depends on the ratio of the wavelength (\(\lambda\)) to the width of the slit (\(a\)):
- If \(\lambda \ll a\): Little diffraction occurs, wave travels almost straight.
- If \(\lambda \approx a\): Significant diffraction occurs, wave spreads widely.
- If \(\lambda > a\): Wave spreads in almost all directions.
Diffraction Pattern
For a single slit of width \(\mathrm{a}\), illuminated by monochromatic light of wavelength \(\lambda\), the diffraction pattern on a screen consists of:
- A central bright fringe (maximum intensity) that is the widest.
- Alternating dark and bright fringes on either side due to interference.
Position of Minima (Dark Fringes):
$\mathrm{a \sin \theta = m \lambda}, \quad m = 1, 2, 3, \dots $
- \(\theta\) = angle from the central maximum to the m-th minimum
- \(\mathrm{a}\) = slit width
- \(\lambda\) = wavelength of the wave
- \(m\) = order of the minimum (first, second, etc.)
Width of Central Maximum:
$ \mathrm{W = 2 L \tan \theta_1 \approx 2 L \sin \theta_1 = 2 L \frac{\lambda}{a}}$
- \(\mathrm{L}\) = distance from the slit to the screen
- \(\theta_1\) = angle to the first minimum on either side of the central maximum
Example :
Light of wavelength \(\mathrm{600 \, nm}\) passes through a slit of width \(\mathrm{0.2 \, mm}\). The screen is at \(\mathrm{2 \, m}\) from the slit. Find the width of the central maximum.
▶️ Answer/Explanation
Step 1: Width of central maximum: \(\mathrm{W = 2 L \frac{\lambda}{a}}\)
Step 2: Substitute values: \(\mathrm{W = 2 \times 2 \frac{600 \times 10^{-9}}{0.2 \times 10^{-3}}}\)
Step 3: \(\mathrm{W = 1.2 \times 10^{-2} \, m = 1.2 \, cm}\)
Final Answer: Width of central maximum = \(\mathrm{1.2 \, cm}\)
Example :
A water wave of wavelength \(\mathrm{0.05 \, m}\) passes through a narrow gap of width \(\mathrm{0.1 \, m}\). Determine whether significant diffraction occurs.
▶️ Answer/Explanation
Step 1: Compare \(\lambda\) and slit width \(a\): \(\mathrm{\lambda = 0.05 \, m, \; a = 0.1 \, m}\)
Step 2: \(\lambda / a = 0.05 / 0.1 = 0.5\), significant diffraction occurs since \(\lambda \sim a\).
Final Answer: The wave spreads significantly after passing through the gap.
Single-Slit Diffraction: Small-Angle Approximation
For a wave passing through a single slit, the angle \(\theta\) to the m-th minimum on a screen far away is usually small. In this case, the small-angle approximation can be applied:
$ \mathrm{\sin \theta \approx \tan \theta \approx \theta \quad (in \, radians)} $
Relationship Between Slit Width, Screen Distance, and Minima Position
Using geometry (see figure below), the distance \(\mathrm{y_m}\) from the center of the central maximum to the m-th minimum on the screen is:
$ \mathrm{y_m = L \tan \theta_m \approx L \sin \theta_m} $
Substitute the condition for minima in single-slit diffraction: \(\mathrm{a \sin \theta_m = m \lambda}\), we get:
$ \mathrm{y_m \approx L \frac{m \lambda}{a}} $
- \(\mathrm{a}\) = slit width
- \(\mathrm{L}\) = distance from slit to screen
- \(\mathrm{\lambda}\) = wavelength of wave
- \(\mathrm{m = 1, 2, 3, \dots}\) = order of the minimum
Width of Central Maximum
The width of the central bright fringe is the distance between the first minima on either side of the center:
$ \mathrm{W = y_1 + y_1 = 2 y_1 = 2 L \frac{\lambda}{a}} $
Example :
Light of wavelength \(\mathrm{500 \, nm}\) passes through a slit of width \(\mathrm{0.25 \, mm}\). The screen is \(\mathrm{2 \, m}\) away. Find the position of the first minimum from the central maximum using the small-angle approximation.
▶️ Answer/Explanation
Step 1: Use \(\mathrm{y_1 \approx L \frac{\lambda}{a}}\)
Step 2: Substitute: \(\mathrm{y_1 \approx 2 \frac{500 \times 10^{-9}}{0.25 \times 10^{-3}} = 4 \times 10^{-3} \, m = 4 \, mm}\)
Final Answer: Position of first minimum = \(\mathrm{4 \, mm}\) from center
Example :
A water wave of wavelength \(\mathrm{0.02 \, m}\) passes through a slit of width \(\mathrm{0.1 \, m}\). The screen is \(\mathrm{1.5 \, m}\) away. Find the distance to the second minimum.
▶️ Answer/Explanation
Step 1: Use \(\mathrm{y_m \approx L \frac{m \lambda}{a}}\)
Step 2: \(\mathrm{y_2 \approx 1.5 \frac{2 \times 0.02}{0.1} = 0.6 \, m}\)
Final Answer: Distance to second minimum = \(\mathrm{0.6 \, m}\) from center