1. Overview
Diffraction is a fundamental wave phenomenon that occurs when a wave encounters an obstacle or an aperture. It is defined by the spreading of a wave into the "geometrical shadow" region behind an object. This effect is a direct consequence of the wave nature of energy transfer and is observed in all types of waves, including mechanical waves (water, sound) and electromagnetic waves (light, radio).
The extent of diffraction is governed by the relationship between the wavelength ($\lambda$) of the incident wave and the physical dimensions of the gap or obstacle ($b$). Significant diffraction only occurs when these two values are of a similar order of magnitude. This principle explains why we can hear sound around a corner but cannot see light around the same corner, and it provides the primary evidence that light travels as a wave rather than purely as a stream of particles.
Key Definitions
- Diffraction: The spreading of a wave as it passes through a gap (aperture) or past the edge of an obstacle.
- Wavelength ($\lambda$): The distance between two consecutive points in a wave that are in phase, such as from one crest to the next.
- Aperture: A narrow opening or slit through which a wave passes.
- Wavefront: A line or surface representing the points of a wave that are all at the same stage of their cycle (in phase).
- Huygens' Principle: A geometric construction stating that every point on a wavefront may be considered a source of secondary spherical wavelets which spread out in the forward direction at the speed of the wave. The new wavefront is the tangential surface to all of these secondary wavelets.
Content
3.1 The Mechanism of Diffraction (Huygens' Principle)
To understand why waves spread, we apply Huygens' Principle. When a plane (straight) wavefront reaches a gap:
- Most of the wavefront is blocked by the barrier.
- Only the portion of the wavefront that meets the gap is allowed to pass through.
- According to Huygens, the points on the wavefront within the gap act as sources of secondary wavelets.
- At the edges of the gap, there are no neighboring wavelets to the side to interfere and keep the wave moving in a straight line. Consequently, these wavelets spread out into the regions that were previously "shadowed" by the barrier.
Physical Constants during Diffraction: When a wave undergoes diffraction, the following properties do not change:
- Speed ($v$): The wave remains in the same medium.
- Frequency ($f$): The source of the wave is unchanged.
- Wavelength ($\lambda$): Since $v$ and $f$ are constant, $\lambda$ must remain constant ($v = f\lambda$).
The property that does change is:
- Amplitude ($A$): As the wave spreads out, the energy is distributed over a larger area (or wavefront length), leading to a decrease in intensity and amplitude.
3.2 Qualitative Effect of Gap Width ($b$)
The degree of diffraction is determined by the ratio $\lambda / b$.
| Condition | Visual Description | Degree of Spreading |
|---|---|---|
| $b \gg \lambda$ (Gap much wider than $\lambda$) | The wave passes through the gap almost unaffected. The wavefronts remain straight, with only very slight curving at the extreme edges. | Negligible |
| $b > \lambda$ (Gap wider than $\lambda$) | The wave begins to spread. The central part of the wavefront remains straight, but the edges curve significantly into the shadow region. | Moderate |
| $b \approx \lambda$ (Gap similar to $\lambda$) | The gap acts almost like a point source. The wavefronts emerging from the gap are semi-circular or circular. | Maximum |
| $b < \lambda$ (Gap smaller than $\lambda$) | The waves spread out as circular wavefronts. However, if the gap is extremely small, the amount of energy (intensity) passing through becomes very low. | Significant (but low intensity) |
3.3 Experiments Demonstrating Diffraction
A. Water Waves in a Ripple Tank A ripple tank is the standard laboratory method for visualizing diffraction qualitatively.
- Apparatus: A shallow glass-bottomed tank filled with water. A vibrating motor-driven bar creates plane waves. Metal barriers are placed in the water to create gaps of varying widths.
- Observation of Gap Width: If the gap is wide, the waves pass through with straight centers. As the metal barriers are moved closer together (decreasing $b$), the wavefronts become increasingly curved.
- Observation of Wavelength: By increasing the frequency of the motor, the wavelength $\lambda$ decreases (since $v$ is constant for a fixed depth). For a fixed gap $b$, decreasing $\lambda$ results in less diffraction (straighter waves).
B. Diffraction of Light Light diffraction is difficult to observe in daily life because visible light has an extremely small wavelength ($400 \text{ nm}$ to $700 \text{ nm}$).
- Single Slit Experiment: A laser (monochromatic light) is shone through a very narrow slit (approx. $0.1 \text{ mm}$ wide). Instead of seeing a single sharp line of light on a distant screen, we see a central bright fringe that has spread out, flanked by smaller, dimmer fringes.
- Red vs. Blue Light: Red light ($\lambda \approx 7.0 \times 10^{-7} \text{ m}$) has a longer wavelength than blue light ($\lambda \approx 4.0 \times 10^{-7} \text{ m}$). Therefore, for the same slit width, red light diffracts more than blue light.
C. Diffraction of Sound Sound waves have wavelengths ranging from roughly $1.7 \text{ cm}$ (at $20 \text{ kHz}$) to $17 \text{ m}$ (at $20 \text{ Hz}$).
- Because these wavelengths are comparable to the size of everyday objects like doors, windows, and walls, sound diffracts significantly. This is why you can hear someone speaking in a hallway even if you are inside a room around the corner. The sound waves spread out as they pass through the doorway.
3.4 Diffraction around an Obstacle
Diffraction also occurs when a wave passes the edge of an obstacle.
- As the wavefront passes the edge, the secondary wavelets at the edge spread into the region behind the obstacle.
- If the obstacle is small compared to the wavelength (e.g., a radio mast for long-wave radio signals), the wave can diffract so much that it "fills in" the shadow behind the obstacle entirely.
4. Worked Examples
Worked example 1 — Calculating Wavelength and Predicting Diffraction
A sound wave travels through the air at a speed of $340 \text{ m s}^{-1}$. The wave is produced by a speaker with a frequency of $850 \text{ Hz}$ and approaches a doorway that is $0.80 \text{ m}$ wide. (a) Calculate the wavelength of the sound wave. (b) Determine whether significant diffraction will occur as the sound passes through the doorway.
Step 1: Identify the known variables.
- $v = 340 \text{ m s}^{-1}$
- $f = 850 \text{ Hz}$
- $b = 0.80 \text{ m}$
Step 2: Use the wave equation to find wavelength. $$v = f\lambda$$ $$\lambda = \frac{v}{f}$$ $$\lambda = \frac{340}{850}$$ $$\lambda = 0.40 \text{ m}$$
Step 3: Compare $\lambda$ to the gap width $b$.
- $\lambda = 0.40 \text{ m}$
- $b = 0.80 \text{ m}$
Answer: (a) The wavelength is $0.40 \text{ m}$. (b) Since the wavelength ($0.40 \text{ m}$) is of the same order of magnitude as the gap width ($0.80 \text{ m}$), significant diffraction will occur. The sound waves will spread out into the room after passing through the doorway.
Worked example 2 — Comparing Light and Radio Waves
A mountain stands between a transmitter and a receiver. The transmitter sends out both a visible light signal ($\lambda = 5.0 \times 10^{-7} \text{ m}$) and a radio wave signal ($\lambda = 300 \text{ m}$). Explain, in terms of diffraction, why the receiver can detect the radio signal but not the light signal.
Answer:
- State the condition for diffraction: Significant diffraction occurs when the wavelength of the wave is comparable to the size of the obstacle.
- Analyze the light signal: The wavelength of visible light ($5.0 \times 10^{-7} \text{ m}$) is many orders of magnitude smaller than the size of the mountain (typically $10^2$ to $10^3 \text{ m}$). Therefore, the light undergoes negligible diffraction and travels in straight lines, leaving the receiver in a "shadow zone."
- Analyze the radio signal: The wavelength of the radio wave ($300 \text{ m}$) is comparable to the dimensions of the mountain.
- Conclusion: The radio waves diffract significantly over the peak and around the edges of the mountain, spreading into the shadow region where the receiver is located.
Key Equations
| Equation | Description | Status |
|---|---|---|
| $v = f\lambda$ | Wave Equation: Relates speed, frequency, and wavelength. | Memorise (Data sheet) |
| $b \approx \lambda$ | Condition for Maximum Diffraction: Gap width approximately equals wavelength. | Conceptual |
| $\text{Diffraction} \propto \frac{\lambda}{b}$ | Proportionality: Diffraction increases as $\lambda$ increases or $b$ decreases. | Conceptual |
Common Mistakes to Avoid
- ❌ Wrong: Thinking that the wavelength or frequency changes when a wave diffracts.
- ✓ Right: $\lambda$, $f$, and $v$ are constant. Only the direction of travel and the amplitude change.
- ❌ Wrong: Using the word "bending" to describe diffraction.
- ✓ Right: Use the term spreading. In physics exams, "bending" is often reserved for refraction (change in medium).
- ❌ Wrong: Stating that diffraction only happens with light.
- ✓ Right: Diffraction is a property of all waves. It is often easier to demonstrate with water or sound because their wavelengths are larger.
- ❌ Wrong: Confusing the effects of gap width.
- ✓ Right: Remember: Smaller gap = More spreading. Larger wavelength = More spreading.
- ❌ Wrong: Forgetting to convert units to SI (e.g., nm to m).
- ✓ Right: Always convert: $1 \text{ nm} = 1 \times 10^{-9} \text{ m}$ and $1 \text{ mm} = 1 \times 10^{-3} \text{ m}$.
Exam Tips
- Drawing Wavefronts: This is a common exam task.
- Use a ruler to ensure the distance between wavefronts ($\lambda$) is identical before and after the gap.
- If $b \approx \lambda$, draw the waves as semi-circles centered at the center of the gap.
- Ensure the "ends" of the diffracted wavefronts extend into the region behind the barriers.
- The "Order of Magnitude" Argument: When explaining why diffraction is not seen for light in a specific scenario, always compare the wavelength ($10^{-7} \text{ m}$) to the object size ($10^0 \text{ m}$) to show they are not comparable.
- Keyword Checklist: For a 2 or 3-mark definition/explanation, ensure you have used: spreading, gap/obstacle, and wavelength.
- Ripple Tank Variables: If asked how to increase diffraction in a ripple tank, you have two options:
- Decrease the gap width.
- Decrease the motor frequency (which increases the wavelength).