wave equation propagation interference and diffraction

The wave equation links curvature along space to acceleration in time: $\frac{\partial^2\psi}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2\psi}{\partial t^2}$ or, in three dimensions, $\nabla^2\psi=\frac{1}{v^2}\frac{\partial^2\psi}{\partial t^2}$. It arises when each point of a medium behaves as a harmonic oscillator elastically coupled to its neighbours, while the electromagnetic field in vacuum satisfies the same equation with $v=c$. Its solutions are traveling waves $\psi(x,t)=f(x-vt)+g(x+vt)$; for a harmonic wave $\psi=A\cos(kx-\omega t)$ the speed is $v=\omega/k$. Because the equation is linear, waves add: superposition holds at every point, and the outcome depends on the phase difference $\Delta\varphi=k\Delta x+\Delta\varphi_0$. Interference is the stable amplitude pattern from overlapping coherent waves: for $\Delta x=m\lambda$ the waves arrive in phase (maximum), and for $\Delta x=(m+\tfrac12)\lambda$ they arrive in antiphase (minimum). Diffraction is described by Huygens' principle: every point on a wavefront is a source of secondary waves; for one slit the minima occur at $a\sin\theta=m\lambda$, $m=\pm1,\pm2,\ldots$.

By substitution, verify that $\psi(x,t)=A\sin(kx-\omega t)$ satisfies the wave equation $\frac{\partial^2\psi}{\partial t^2}=v^2\frac{\partial^2\psi}{\partial x^2}$. Find $v$ in terms of $\omega$ and $k$.

Two coherent waves of equal amplitude $A$ arrive at a point with path difference $\Delta x=\lambda/3$; their initial phases are the same. Find the resultant amplitude and its phase relative to the first wave.

In a Young's double-slit experiment, the slit separation is $d=0{.}20$ mm, the width of each slit is $a=0{.}040$ mm, and the wavelength is $\lambda=600$ nm. Determine the order $n$ of the first interference maximum that disappears because of the first single-slit diffraction minimum.

Self-study with a simulation (no laser). Open PhET “Wave Interference” (or a similar wave-optics simulation). 1. Use the water mode with two point sources; vary the source separation and frequency, and observe the lines of maxima and minima. 2. Switch to light: for two slits, measure the distance between adjacent maxima on the screen and compare it with $\Delta y=\lambda L/d$. 3. Reduce the width to one slit and find the angle of the first minimum, checking $a\sin\theta=\lambda$. Record measured and predicted values; do not use real lasers and do not look into direct light sources.

Show how the path difference between two coherent waves becomes a phase difference and how superposition creates an interference pattern, while Huygens secondary waves produce a diffraction minimum.

Diagram: two sources $S_1$ and $S_2$ separated by $d$, a point $P$ on the screen, segments $S_1P$ and $S_2P$ with labeled path difference $\Delta x$; on the right, a graph of resultant amplitude $A_{\text{res}}=2A|\cos(\Delta\varphi/2)|$ versus $\Delta\varphi$ (radians). Below, single-slit geometry of width $a$: wavefronts of secondary waves, angle $\theta$, and the minimum condition $a\sin\theta=m\lambda$.

In a gym, a student attaches a 10 m rope to a wall and stretches it with a tension of 200 N. The rope's linear density is 0.5 kg/m. She sharply shakes one end, sending a transverse pulse. The pulse travels 10 m in 0.5 s, i.e. at a speed of 20 m/s.

This scenario uses the one-dimensional wave equation $\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$ with speed $v = \sqrt{T/\mu}$. The pulse's motion obeys this equation, and its speed is determined by tension and linear density.

At a school party, two identical speakers are placed 2 m apart on a stage. They emit a pure tone of 686 Hz. A student stands 3.0 m from the left speaker and 3.5 m from the right speaker. The speed of sound is 343 m/s. The wavelength is $\lambda = v/f = 343/686 = 0.5$ m. The path difference is $3.5 - 3.0 = 0.5$ m $= \lambda$, so the student hears a loud sound due to constructive interference.

Superposition principle: the total displacement is the sum of the two waves; when the path difference equals an integer number of wavelengths, crests align (constructive interference). The linearity of the wave equation makes this possible.

A phone plays a 1000 Hz tone. The phone is on a table, and a student listens from the next room through a doorway 0.9 m wide. The speed of sound is 343 m/s, so the wavelength is $\lambda = 343/1000 = 0.343$ m. Because $\lambda$ is comparable to the doorway width, the sound bends around the edges and fills the room, with the first diffraction minimum at an angle $\theta \approx \arcsin(\lambda/a) = \arcsin(0.343/0.9) \approx 22^\circ$.

Diffraction occurs when a wave encounters an obstacle or opening comparable to its wavelength; Huygens' principle and the wave equation predict this bending. The phone's sound wave obeys the wave equation in three dimensions.