Wave Optics: Huygens' Principle & Diffraction
Wave Optics: Huygens\' Principle & Diffraction
Wave optics treats light as an electromagnetic wave, revealing phenomena geometric optics cannot explain. Huygens\' principle states that every point on a wavefront acts as a source of secondary spherical wavelets; the new wavefront is their envelope. This predicts reflection, refraction, and — crucially — diffraction.
A single slit of width $a$ illuminated by a plane wave produces intensity $I(\theta)=I_0\,\text{sinc}^2(\beta)$ where $\beta=\pi a\sin\theta/\lambda$, with dark fringes at $a\sin\theta=m\lambda$, $m=\pm1,\pm2,\ldots$
Huygens-Fresnel Principle
Field at $P$: $U(P)=\frac{-i}{\lambda}\iint_\Sigma U(Q)\frac{e^{ikr}}{r}\cos\theta\,dS.$
Single-Slit Intensity
$$I(\theta)=I_0\!\left(\frac{\sin\beta}{\beta}\right)^2,\quad \beta=\frac{\pi a\sin\theta}{\lambda}.$$ Minima at $a\sin\theta=m\lambda$, $m\neq0$.
Example 1
Slit $a=0.1\,\text{mm}$, $\lambda=500\,\text{nm}$. Find the first-minimum angle.
Solution: $\sin\theta=\lambda/a=5\times10^{-3}$, so $\theta\approx0.29°$.
Example 2
How does halving slit width $a$ affect the central maximum?
Solution: First minimum moves to $\sin\theta=2\lambda/a$, so the central lobe doubles in angular width.
Practice
- State Huygens\' principle in one sentence.
- Slit $a=0.2\,\text{mm}$, $\lambda=600\,\text{nm}$: find the first dark-fringe angle.
- What is the $\text{sinc}$ function and why does it appear in single-slit diffraction?
- How does increasing $\lambda$ affect the diffraction pattern width?
Show Answer Key
1. Every point on a wavefront acts as a source of secondary spherical wavelets; the new wavefront is the envelope of these wavelets.
2. $a\sin\theta = \lambda$ → $\sin\theta = 600\times10^{-9}/(0.2\times10^{-3}) = 3\times10^{-3}$ → $\theta \approx 0.17°$.
3. $\text{sinc}(x) = \sin(\pi x)/(\pi x)$. It arises because the far-field diffraction pattern is the Fourier transform of the aperture function. A rectangular slit has a rectangular (top-hat) transmission function whose FT is a sinc. Zeros of sinc give the dark fringes at $a\sin\theta = m\lambda$.
4. Increasing $\lambda$ increases the angular width of the central maximum ($\theta_1 = \lambda/a$), spreading the pattern out. All fringe spacings scale linearly with $\lambda$. Longer wavelengths diffract more for a given aperture.