C3-2. Diffraction and Superposition

1. Diffraction Basics

Diffraction through Slits & Around Obstacles

  • Diffraction is the spreading of a wave after passing through a gap or around an obstacle.
  • The effect is strongest when the gap or obstacle size is comparable to the wavelength ($b \approx \lambda$).
  • If the obstacle is much larger than the wavelength, a significant "shadow" region forms behind it.
Gap ≈ λ Strong spreading
Gap ≫ λ Little spreading
Shadow Object ≫ λ Significant shadow
Object ≪ λ Insignificant effect (Waves reform)

2. Interference & Superposition

When two coherent waves meet at a point, their phase difference ($\Delta \phi$) determines the resulting interference. This phase difference is directly related to the path difference ($\Delta L$), which is the difference in distance traveled by the two waves from their respective sources to the meeting point. Since a full wavelength ($\lambda$) corresponds to a phase change of $2\pi$ radians, the general relationship is:

$$\Delta \phi = \frac{2\pi}{\lambda} \Delta L$$

Constructive Interference

Constructive interference occurs when the waves arrive completely in phase. For this to happen, their phase difference must be an even multiple of $\pi$ (i.e., $0, 2\pi, 4\pi, \dots$), where $m$ is an integer ($0, 1, 2, \dots$).

$$\Delta \phi = 2m\pi \implies 2m\pi = \frac{2\pi}{\lambda} \Delta L \implies \Delta L = m\lambda$$
  • Physical Meaning: The waves are perfectly synchronized so that crests meet crests and troughs meet troughs. According to the principle of superposition, their individual displacements add together algebraically, resulting in a new wave with a maximized, larger amplitude.

Destructive Interference

Destructive interference happens when the waves arrive completely out of phase. Their phase difference must be an odd multiple of $\pi$ (i.e., $\pi, 3\pi, 5\pi, \dots$).

$$\Delta \phi = (2m+1)\pi \implies (2m+1)\pi = \frac{2\pi}{\lambda} \Delta L \implies \Delta L = \left(m+\dfrac{1}{2}\right)\lambda$$
  • Physical Meaning: The waves are exactly half a cycle out of sync. The crests of one wave perfectly align with the troughs of the other. Their opposite displacements counteract each other, and if the two waves have equal amplitudes, they will completely cancel each other out, leaving zero net amplitude.
S₁ S₂ P L₁ L₂ ΔL = |L₁ - L₂|

How to plot the resulting wave?

Wave 1 + Wave 2 = Resultant Wave

1. Constructive Interference

distance displacement

2. Destructive Interference

distance displacement

3. Partial Interference

distance displacement

Example 1: Wave Pulses Superposition

Problem: Two wave pulses (a rectangle and a triangle) move towards each other. What is the combination of the two pulses when they overlap?

Wave Pulse Superposition Scenarios Initial State: Two Pulses Approaching (1 grid unit = 1 amplitude unit) +2 -3 Which diagram correctly shows the resultant wave when the pulses completely overlap at the center? A B C D

Solution: According to the Principle of Superposition, the resultant displacement at any point is the algebraic sum of the individual wave displacements. When the two rectangular pulses completely overlap at their centers, we must consider the different regions:

  • Outer edges of the overlap: The wider wave has an amplitude of +2, while the narrower wave does not exist in this space (amplitude 0). The resultant amplitude is +2 + 0 = +2.
  • Center of the overlap: Both waves occupy this space. The positive wave has an amplitude of +2 and the negative wave has an amplitude of -3. The resultant amplitude is +2 + (-3) = -1.

This superposition creates a stepped wave shape with positive outer "shoulders" at +2 and a central rectangular dip down to -1, which corresponds to Option A.

Example 2: Lloyd's Mirror

Problem: A microwave transmitter, $T$, is positioned 12.0 cm above a perfectly reflecting flat metal surface. A receiver, $R$, is placed at a horizontal distance of 80.0 cm from $T$ and can be adjusted vertically. The transmitter emits coherent microwaves with a wavelength of $\lambda$ = 3.0 cm.

When microwaves reflect off the metal surface, they undergo a phase change of $\pi$ radians. Assuming $D \gg h, y$, calculate the lowest height $y$ above the surface where the receiver will detect a maximum signal (constructive interference).

Lloyd's Mirror Interference Setup Reflecting Metal Surface h = 12.0 cm h = 12.0 cm y D = 80.0 cm Path L₁ Path L₂ T T' (Virtual) R π phase shift at reflection

Solution:

To solve this, we must account for both the path difference ($\Delta L$) and the phase shift caused by the reflection. This is known as a Lloyd's Mirror setup.

  1. Establish the Phase Condition:
    Normally, constructive interference requires a path difference of $m\lambda$. However, because the reflected wave undergoes a $\pi$ radian shift upon bouncing off the metal, the condition for constructive interference flips to the standard destructive equation:
    $$\Delta L = \left(m + \dfrac{1}{2}\right)\lambda$$
    We are looking for the lowest height where a maximum occurs, meaning this is the first fringe. We set $m = 0$:
    $$\Delta L = \dfrac{\lambda}{2} = \dfrac{3.0}{2} = 1.5 \text{ cm}$$
  2. Define the Path Difference Geometrically:
    The direct wave travels path $L_1$. The reflected wave travels path $L_2$. Using the method of images, the reflected wave acts as though it originates from a virtual transmitter $T'$ located 12.0 cm below the surface. Using the Pythagorean theorem, the exact path lengths from the two sources (separated by $d$) to the receiver at height $y$ are: $$ \begin{aligned} L_1^2 &= D^2 + \left(y - \dfrac{d}{2}\right)^2 \\[0.2cm] L_2^2 &= D^2 + \left(y + \dfrac{d}{2}\right)^2 \end{aligned} $$ Subtracting the first equation from the second gives $$ L_2^2 - L_1^2 = \left(y + \dfrac{d}{2}\right)^2 - \left(y - \dfrac{d}{2}\right)^2 = 2yd $$ Factoring the left side as a difference of squares gives $(L_2 - L_1)(L_2 + L_1) = 2yd$. Furthermore, because $D \gg y$ and $D \gg d$, the rays are nearly horizontal, meaning $L_1 \approx D$ and $L_2 \approx D$. Therefore, $(L_2 + L_1) \approx 2D$. Substituting these into our factored equation:
    $$\Delta L (2D) \approx 2yd \implies \Delta L \approx \dfrac{d \cdot y}{D} = \dfrac{2hy}{D}$$
    The last equality follows from that the distance between the two "sources" (real and virtual) is $d = 2h$.
  3. Solve for $y$:
    Equating our required phase condition to our geometric path difference:
    $$\dfrac{2hy}{D} = 1.5$$
    Substitute the given values ($h$ = 12.0 cm, $D$ = 80.0 cm):
    $$\dfrac{2(12.0)y}{80.0} = 1.5$$
    $$\dfrac{24.0y}{80.0} = 1.5 \implies y = \dfrac{1.5 \times 80.0}{24.0} = 5.0$$

Conclusion: The receiver must be placed 5.0 cm above the metal surface to detect the first maximum constructive signal.