Checkpoint: Classical Waves Recap

This checkpoint pulls together the key ideas from Module 4 — oscillations, wave quantities, superposition, and interference — and asks you to apply them in a single calculation.

Wave quantities at a glance

A monochromatic plane wave of wavelength $\lambda$ and period $T$ is described by

$y(x, t) = A\cos(kx - \omega t + \varphi),$

where the wave number $k = 2\pi/\lambda$, the angular frequency $\omega = 2\pi/T = 2\pi f$, and the wave speed $v = \lambda f = \omega / k$. Every wave you encounter in this module — mechanical or electromagnetic — is characterised by exactly these four quantities, plus an amplitude $A$ and an initial phase $\varphi$.

Path-length difference and the interference condition

When two coherent, in-phase waves travel different distances to the same point, the extra distance one wave travels is the path-length difference $\Delta\ell$. The phase difference that results is

$\delta = \frac{2\pi}{\lambda}\,\Delta\ell.$

Bright fringes (constructive interference) occur when $\delta = 2\pi m$, i.e. when

$\Delta\ell = m\lambda, \quad m = 0, \pm 1, \pm 2, \ldots$

Dark fringes (destructive interference) occur when $\delta = (2m+1)\pi$, i.e. when

$\Delta\ell = \left(m + \tfrac{1}{2}\right)\lambda.$

For a double-slit geometry with slit separation $d$, the path-length difference to a point at angle $\theta$ from the central axis is $\Delta\ell = d\sin\theta$ (exact for plane waves; derived by dropping a perpendicular from the farther slit to the wave-front reaching the closer slit).

Fringe spacing on a distant screen

In the small-angle limit ($\theta \ll 1$, equivalently $y \ll L$), $\sin\theta \approx \tan\theta = y/L$, where $y$ is the transverse displacement on the screen. Substituting into the constructive-interference condition $d\sin\theta_m = m\lambda$ gives the position of the $m$-th bright fringe:

$y_m = \frac{m\lambda L}{d}.$

Consecutive bright fringes are therefore uniformly spaced by

$\Delta y = y_{m+1} - y_m = \frac{\lambda L}{d}.$

This single formula encodes the entire module: a larger wavelength or a longer throw $L$ spreads the fringes further apart; a wider slit separation $d$ crowds them together.

Numerical check

As a concrete example, take $\lambda = 500\,\text{nm}$, $d = 0.25\,\text{mm}$, $L = 1.0\,\text{m}$:

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