Welcome to Superposition, Interference and Diffraction
Have you ever wondered how noise-cancelling headphones manage to wipe out background roar, or why looking at the back of a DVD reveals a rainbow of bright colors? The answers lie in the way waves interact with each other and their surroundings. In this chapter of AS 2: Waves, Photons and Astronomy, we will explore the rules that govern wave combinations, standing waves, and the patterns created when waves squeeze through tiny gaps.
Don't worry if wave behavior feels abstract at first! We will break every concept down into simple, visual steps with clear equations, real-life analogies, and exam tips to help you secure full marks in your CCEA AS Level Physics exam.
---1. The Principle of Superposition
When two particles collide, they bounce off each other. Waves, however, do something much more interesting: they pass straight through each other! While they overlap, their effects combine.
What is the Principle of Superposition?
The Principle of Superposition states that: When two or more waves meet or overlap at a point in space, the resultant displacement is equal to the vector sum of the individual displacements of the waves at that point.
Crucial Exam Note: Always use the words vector sum (or algebraic sum) and displacement. Because displacement is a vector quantity, a positive displacement (a crest) combined with a negative displacement (a trough) can cancel each other out!
• Constructive Superposition: Occurs when two crests (or two troughs) meet in phase. The resultant displacement is larger than that of either individual wave.
• Destructive Superposition: Occurs when a crest meets a trough. The displacements work against each other, reducing the overall amplitude or cancelling it out completely.
Stationary (Standing) Waves
A stationary wave (or standing wave) is a special wave pattern that stays in a fixed position—it stores vibrational energy rather than transferring energy from one place to another.
How is a stationary wave formed?
A stationary wave is formed when two progressive waves of the same frequency, wavelength, and comparable amplitude travel in opposite directions along the same medium and superpose.
A classic example is a guitar string fixed at both ends: a wave travels down the string, reflects at the fixed end, and travels back, continuously superposing with the incoming wave.
Key Anatomy of a Stationary Wave
• Nodes: Points along the medium where destructive interference is complete. The displacement and amplitude at a node are permanently zero.
• Antinodes: Points along the medium where constructive interference occurs. The displacement reaches its maximum possible amplitude.
Crucial Distances and Phase Relationships
• Distance between two adjacent nodes = \( \frac{\lambda}{2} \)
• Distance between two adjacent antinodes = \( \frac{\lambda}{2} \)
• Distance between an adjacent node and antinode = \( \frac{\lambda}{4} \)
Phase between particles:
• All particles located between two adjacent nodes vibrate in phase with each other (they reach their maximum displacements at the exact same instant, though with different amplitudes).
• Particles on opposite sides of a node oscillate in antiphase (a phase difference of \( 180^\circ \) or \( \pi\text{ rad} \)).
Energy Note: Unlike progressive waves, no net energy is transmitted along a stationary wave; energy remains trapped between the nodes.
Key Takeaway: Stationary waves need two identical waves moving in opposite directions. Nodes have zero amplitude (\( \text{distance} = \frac{\lambda}{2} \) apart), antinodes have maximum amplitude, and no net energy is transmitted.
---2. Phase Difference and Coherence
To produce clear, stable interference patterns that we can observe and measure, our wave sources must satisfy specific conditions.
Coherence
Two wave sources are defined as coherent if they maintain a constant phase difference and have the same frequency (and wavelength).
Analogy: Imagine two synchronized swimmers performing the exact same routine at the same speed. If one swimmer is always precisely half a stroke behind the other, their phase difference never changes over time. They are coherent!
Phase Difference (\( \Delta\phi \))
Phase difference measures how much one wave cycle leads or lags behind another cycle, measured in degrees (\( ^\circ \)) or radians (\( \text{rad} \)).
• In Phase: \( \Delta\phi = 0^\circ, 360^\circ, 720^\circ \dots \) or \( 0, 2\pi, 4\pi\text{ rad} \)
• In Antiphase: \( \Delta\phi = 180^\circ, 540^\circ \dots \) or \( \pi, 3\pi, 5\pi\text{ rad} \)
Key Takeaway: Coherent sources must have the same frequency and a constant phase difference over time. Without coherence, interference patterns shift so rapidly that they blur out completely.
---3. Interference and Path Difference
When coherent waves from two sources travel through space to reach a specific point, they often travel different distances. This difference in distance is called the path difference.
Conditions for Interference
1. Constructive Interference (Maxima / Bright Fringes):
Waves arrive at the point in phase (\( \Delta\phi = 0, 2\pi, 4\pi\dots \)), reinforcing each other to produce maximum amplitude.
• Condition: \( \text{Path Difference} = n\lambda \quad (n = 0, 1, 2, 3\dots) \)
2. Destructive Interference (Minima / Dark Fringes):
Waves arrive at the point in antiphase (\( \Delta\phi = \pi, 3\pi, 5\pi\dots \)), cancelling each other out to produce zero (or minimum) amplitude.
• Condition: \( \text{Path Difference} = \left(n + \frac{1}{2}\right)\lambda \quad (n = 0, 1, 2, 3\dots) \)
Memory Trick: Whole wavelengths (\( 1\lambda, 2\lambda, 3\lambda \)) create Whole (Bright) spots. Half wavelengths (\( 0.5\lambda, 1.5\lambda, 2.5\lambda \)) Halt (Dark) the light!
Key Takeaway: Path difference determines the phase match. Integer wavelengths give constructive interference (\( n\lambda \)); half-integer wavelengths give destructive interference (\( (n + \frac{1}{2})\lambda \)).
---4. Two-Source Interference: Young's Double-Slit Experiment
In 1801, Thomas Young provided definitive evidence for the wave nature of light by shining light through two narrow slits, producing an alternating pattern of bright and dark bands (fringes) on a screen.
The Double-Slit Equation
\( \lambda = \frac{a y}{D} \quad \text{or rearranged as} \quad y = \frac{\lambda D}{a} \)
Where:
• \( \lambda \) = Wavelength of the light source in metres (\( \text{m} \))
• \( a \) = Slit separation (distance between the centres of the two slits) in metres (\( \text{m} \))
• \( y \) = Fringe separation (distance between adjacent bright fringes or adjacent dark fringes) in metres (\( \text{m} \))
• \( D \) = Perpendicular distance from the double slit to the screen in metres (\( \text{m} \))
Experimental Constraints
This formula is accurate only when the screen distance is much greater than the slit spacing (\( D \gg a \)), and the fringe separation \( y \) is much smaller than \( D \).
Step-by-Step Example Problem
Problem: Laser light of wavelength \( \lambda = 6.00 \times 10^{-7}\text{ m} \) illuminates two slits separated by \( a = 0.30\text{ mm} \). An interference pattern is observed on a screen positioned at \( D = 2.50\text{ m} \). Calculate the fringe spacing \( y \).
Step 1: Convert all units to standard metres (\( \text{m} \)):
\( a = 0.30\text{ mm} = 0.30 \times 10^{-3}\text{ m} = 3.0 \times 10^{-4}\text{ m} \)
\( D = 2.50\text{ m} \)
\( \lambda = 6.00 \times 10^{-7}\text{ m} \)
Step 2: Use the rearranged equation:
\( y = \frac{\lambda D}{a} \)
Step 3: Substitute and solve:
\( y = \frac{(6.00 \times 10^{-7}\text{ m}) \times (2.50\text{ m})}{3.0 \times 10^{-4}\text{ m}} = 5.0 \times 10^{-3}\text{ m} = 5.0\text{ mm} \)
Key Takeaway: Fringe spacing \( y \) increases if you increase the screen distance \( D \) or wavelength \( \lambda \), but decreases if you spread the slits further apart (\( a \)).
---5. Diffraction and Diffraction Gratings
What is Diffraction?
Diffraction is the spreading out of a wave front as it passes through a gap or around the edge of an obstacle.
• Condition for Maximum Diffraction: Significant spreading occurs when the gap width is approximately equal to the wavelength of the wave (\( \text{gap} \approx \lambda \)).
• If the gap is much wider than \( \lambda \), the wave passes through with minimal spreading, creating sharp shadows.
The Diffraction Grating
A diffraction grating is an optical component consisting of a large number of closely spaced, parallel, equidistant slits. Because it has hundreds of slits per millimetre, the constructive interference peaks (maxima) are much sharper, brighter, and more widely spaced than those from a double slit.
The Grating Equation
\( d \sin \theta = n \lambda \)
Where:
• \( d \) = Grating spacing (distance between the centres of adjacent slits/lines) in metres (\( \text{m} \))
• \( \theta \) = Angle of diffraction for the \( n \)-th order maximum relative to the straight-through central beam (\( 0^\circ \))
• \( n \) = Order number (\( n = 0, \pm 1, \pm 2\dots \))
• \( \lambda \) = Wavelength of the incident monochromatic light in metres (\( \text{m} \))
Calculating the Grating Spacing (\( d \))
Diffraction gratings are typically labelled with the number of lines per millimetre (\( N \)). To find \( d \) in metres:
\( d = \frac{1 \times 10^{-3}\text{ m}}{N} \)
Example: If a grating has \( 500\text{ lines/mm} \):
\( d = \frac{1 \times 10^{-3}\text{ m}}{500} = 2.0 \times 10^{-6}\text{ m} \)
Finding the Maximum Observable Order (\( n_{\text{max}} \))
Because the maximum possible value of \( \sin \theta \) is \( 1 \) (corresponding to an angle of \( 90^\circ \)):
\( n \le \frac{d}{\lambda} \)
• To find \( n_{\text{max}} \), calculate \( \frac{d}{\lambda} \) and round down to the nearest whole integer.
• To find the total number of visible maxima (bright spots), remember to count the positive orders, negative orders, and the central zero order (\( n = 0 \)):
\( \text{Total Maxima} = 2n_{\text{max}} + 1 \)
Step-by-Step Grating Calculation
Problem: Light of wavelength \( \lambda = 532\text{ nm} \) is directed normally at a grating with \( 400\text{ lines/mm} \). Find the angle of the second-order maximum and the total number of maxima produced.
Step 1: Calculate \( d \) in metres:
\( d = \frac{1 \times 10^{-3}\text{ m}}{400} = 2.50 \times 10^{-6}\text{ m} \)
Step 2: Calculate the angle \( \theta \) for \( n = 2 \):
\( \sin \theta = \frac{n \lambda}{d} = \frac{2 \times (532 \times 10^{-9}\text{ m})}{2.50 \times 10^{-6}\text{ m}} = \frac{1.064 \times 10^{-6}}{2.50 \times 10^{-6}} = 0.4256 \)
\( \theta = \sin^{-1}(0.4256) \approx 25.2^\circ \)
Step 3: Determine maximum observable order:
\( n \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{532 \times 10^{-9}} \approx 4.70 \)
Rounding down gives \( n_{\text{max}} = 4 \).
\( \text{Total Maxima} = 2(4) + 1 = 9\text{ bright spots} \).
Key Takeaway: Diffraction gratings produce sharp spectral lines governed by \( d \sin \theta = n \lambda \). Always convert lines/mm to slit spacing \( d \) in metres before calculating.
---6. Examiner Pitfalls & Quick Review
Avoid these common traps to protect your marks in the AS 2 exam:
• Incomplete Superposition Definition: Never just say "the sum of waves". State that the resultant displacement is the vector sum of individual displacements.
• Incomplete Definition of Coherence: Stating only "same frequency" will lose marks. You must state: same frequency AND constant phase difference.
• Confusing \( a \) and \( d \): Use \( a \) for the slit separation in Young’s double-slit experiment (\( \lambda = \frac{ay}{D} \)), and \( d \) for the grating line spacing in the grating formula (\( d \sin \theta = n\lambda \)).
• Grating Spacing Units: A grating with \( 600\text{ lines/mm} \) does NOT have \( d = 600 \). Remember: \( d = \frac{1 \times 10^{-3}\text{ m}}{600} \).
• Nodes vs Antinodes: Remember: Nodes have No movement (zero amplitude); Antinodes have Ample movement (maximum amplitude).