Diffraction is the bending of waves around obstacles or through openings. 🎯 In physics, we often study how light or sound waves spread when they pass through a narrow gap.
When a wave encounters a gap of width \$d\$, the waves emerging from the gap spread out. The amount of spreading depends on the wavelength \$\lambda\$ and the size of the gap.
Imagine a pond with a small stone creating ripples. If you put a narrow fence (gap) in the water, the ripples will bend around the fence. The larger the ripples (longer wavelength) or the narrower the fence (smaller gap), the more pronounced the bending. 🌊
The first minimum in a single‑slit diffraction pattern occurs when:
\$d \sin \theta = m \lambda \quad \text{with } m = 1\$
For small angles, \$\sin \theta \approx \theta\$ (in radians), giving the simple approximation:
\$\theta \approx \frac{\lambda}{d}\$
So, if you double the wavelength or halve the gap, the diffraction angle roughly doubles.
| Gap Size \$d\$ (mm) | Wavelength \$\lambda\$ (nm) | Diffraction Angle \$\theta\$ (degrees) |
|---|---|---|
| 0.5 | 500 | 0.058° |
| 1.0 | 500 | 0.029° |
| 2.0 | 500 | 0.015° |
| 1.0 | 1000 | 0.058° |
🔍 Remember: The key formula for the first minimum is \$d \sin \theta = \lambda\$. For small angles, use \$\theta \approx \lambda/d\$ (in radians).
💡 Units matter: Convert all lengths to the same unit (usually metres) before plugging into the formula.
📝 Typical exam question: “If a light wave of wavelength 600 nm passes through a 0.2 mm slit, calculate the angle to the first minimum.” Use the small‑angle approximation.
📚 Practice: Vary \$\lambda\$ and \$d\$ in the table to see how the angle changes.
Diffraction is stronger when the wavelength is large compared to the gap size. The simple rule of thumb: \$\theta \propto \lambda/d\$. This relationship helps predict how waves will spread after passing through a narrow opening, a concept that appears in many physics exam questions.