explain the meaning of the term diffraction

Diffraction

Objective

To explain the meaning of diffraction, derive the key mathematical relations for single‑slit and grating diffraction, and show how diffraction limits the resolution of optical instruments – covering Cambridge AS & A‑Level Physics (9702) syllabus sections 8.2 & 8.4.

1. Definition and Physical Origin

• Diffraction is the bending and spreading of a wave‑front when it encounters an obstacle or aperture whose dimensions are comparable to the wavelength λ.
• It is a direct consequence of Huygens’ principle: every point on a wave‑front acts as a source of secondary spherical wavelets; the superposition of these wavelets produces the observed pattern.
• The effect is most noticeable when the ratio λ / size of the aperture is not << 1. Hence longer wavelengths and smaller apertures give stronger diffraction.
• In the syllabus the term “diffraction” usually refers to the Fraunhofer (far‑field) condition, i.e. the observation screen is at a distance L ≫ ²/λ, so that the rays reaching the screen are essentially parallel.

2. Single‑Slit Diffraction – Derivation of the Minima Condition

Consider a monochromatic plane wave of wavelength λ incident normally on a slit of width . Divide the slit into 2m equally spaced narrow strips (Fig. 1). For a ray making an angle θ with the normal, the path‑difference between the top and bottom of the slit is

\$\Delta = a\sin\theta\$

Pair‑wise cancellation argument – each strip in the top half can be paired with a strip in the bottom half that is a distance  below it. The two contributions have a phase difference of

\$\delta = \frac{2\pi}{\lambda}\,\frac{a}{2m}\sin\theta = \frac{\pi}{m}\,a\sin\theta/\lambda.\$

When the total path‑difference equals an integer multiple of the wavelength, i.e.

\$a\sin\theta = m\lambda \qquad (m = \pm1,\pm2,\dots)\$

the phase difference between each pair is π, causing complete destructive interference. This is the condition for minima in the single‑slit diffraction pattern.

2.1 Angular Width of the Central Maximum and the Rayleigh Criterion

• The first minima occur for m = ±1. Using the small‑angle approximation (sin θ ≈ θ in radians), the half‑angular width of the central bright fringe is

\$\theta_{1} \approx \frac{\lambda}{a}\$

• Therefore the full angular width of the central maximum is approximately

\$\Delta\theta_{\text{central}} \approx \frac{2\lambda}{a}\$

• For a circular aperture of diameter D the diffraction pattern is an Airy disc. The first zero of the Bessel function occurs at 1.22 λ/D, giving the Rayleigh resolution limit

\$\theta_{\text{Rayleigh}} \approx 1.22\,\frac{\lambda}{D}\$

The factor 1.22 arises from the first zero of the J₁ Bessel function that describes the intensity distribution of a circular aperture.

3. Intensity Distribution (Fraunhofer Approximation)

The intensity as a function of angle θ is

\$\$I(\theta)=I_{0}\left(\frac{\sin\beta}{\beta}\right)^{2},\qquad

\beta=\frac{\pi a\sin\theta}{\lambda}\$\$

where I₀ is the maximum intensity on the central axis (θ = 0). The pattern consists of a bright central maximum flanked by weaker side fringes whose positions are given by the minima condition above.

4. Experimental Demonstration – Single‑Slit Set‑up (Fraunhofer)

1. Mount a low‑power laser (≈ 5 mW, red λ ≈ 650 nm) on an optical bench.
2. Place an adjustable single slit (micrometer‑controlled) a few centimetres from the laser so that the beam is perpendicular to the slit.
3. Position a screen (or a calibrated photodiode on a translation stage) at a distance L ≈ 1–2 m (satisfying L ≫ a²/λ).
4. Measure the lateral positions yₘ of the first‑order minima on either side of the centre.

For small angles, sin θ ≈ y/L, so the slit width can be calculated from

\$a = \frac{m\lambda L}{y_{m}}\$

Sample Paper 5 Question

A laser of wavelength 632.8 nm illuminates a single slit. The first‑order minima are observed 3.2 cm either side of the central maximum on a screen 1.50 m away. Calculate the slit width.

Solution

• y₁ = 3.2 cm = 0.032 m, L = 1.50 m, m = 1
• \$\$a = \frac{\lambda L}{y_{1}} = \frac{632.8\times10^{-9}\times1.50}{0.032}

  \approx 2.97\times10^{-5}\,\text{m}=29.7\,\mu\text{m}\$\$

5. Diffraction Grating (Syllabus 8.4)

A diffraction grating consists of many equally spaced parallel slits (or ruled lines). For Fraunhofer diffraction the condition for principal maxima is

\$d\sin\theta = n\lambda \qquad (n=0,\pm1,\pm2,\dots)\$

where d is the grating spacing (inverse of the line density). The angular separation of the maxima is much larger than for a single slit, making gratings ideal for wavelength measurement.

Worked Example

A grating with 500 lines mm⁻¹ is illuminated by white light. The first‑order (n = 1) maximum for the yellow component (λ ≈ 580 nm) appears at an angle of 20.5°. Find the line density of the grating.

First calculate the spacing:

\$\$d = \frac{n\lambda}{\sin\theta}= \frac{1\times580\times10^{-9}}{\sin20.5^{\circ}}

\approx 1.68\times10^{-6}\,\text{m}\$\$

Line density = 1/d ≈ 595 lines mm⁻¹, confirming the specification.

6. Diffraction vs Interference – Quick Comparison

7. Practical Activity Checklist (AO3)

• Equipment: laser pointer, micrometer‑adjustable single slit, diffraction grating (optional), screen or photodiode, ruler or vernier, laser‑safety goggles.
• Safety

  • Never look directly into the laser beam.
  • Wear appropriate laser‑safety goggles.
  • Secure all optics on the bench to avoid accidental displacement.

• Procedure

  1. Align the laser so the beam is perpendicular to the slit.
  2. Record the slit width set on the micrometer.
  3. Place the screen at a measured distance L; note the positions of the first‑order minima (±y₁).
  4. Calculate the experimental slit width using a = λL/y₁ and compare with the micrometer reading.
  5. Repeat with a diffraction grating to determine the wavelength of a known laser or the grating line density.

• Data recording tips

  • Use a table with columns for L, y₁ (both sides), average y, calculated a, % error.
  • Include measurement uncertainties (e.g., ±0.5 mm on y, ±1 cm on L) and propagate them to the final result.

8. Cross‑Reference Box

Diffraction concepts re‑appear throughout the syllabus:

• 7 Superposition – Huygens’ principle is an application of superposition of secondary wavelets.
• 8.3 Interference – Multi‑slit interference patterns are the product of diffraction from each slit combined with their mutual interference.
• 20 Electromagnetic Waves – Diffraction of radio waves around buildings illustrates the same principle at much longer λ.
• 21 Astronomy & Cosmology – The diffraction limit (Rayleigh criterion) determines the resolving power of telescopes and the size of the Airy disc in stellar imaging.

9. Everyday Examples

1. Faint shadows behind a hair when illuminated by a laser pointer.
2. Sound spreading around a doorway – you can still hear someone even when the source is not directly visible.
3. Bright edge patterns (Fresnel fringes) seen on the rim of a razor blade in a darkened room.

10. Why Diffraction Matters in A‑Level Physics

• Provides a quantitative test of the wave nature of light.
• Underpins the design and performance limits of optical instruments (microscopes, telescopes, spectrometers).
• Enables students to solve typical exam problems on slit width, grating spacing, and resolving power, meeting AO2–AO4 outcomes.

11. Summary

Diffraction is the bending and spreading of a wave‑front when it meets an aperture or obstacle comparable in size to its wavelength. Using Huygens’ principle and the Fraunhofer (far‑field) approximation we obtain:

• Minima condition for a single slit: a sin θ = mλ.
• Intensity distribution: I(θ)=I₀(sin β/β)² with β=πa sinθ/λ.
• Full width of the central maximum ≈ 2λ/a; for a circular aperture the Rayleigh limit is θ≈1.22 λ/D.
• Grating equation: d sin θ = nλ, giving large angular separation of orders.

Mastery of these concepts, together with practical experimental skills, equips students to meet all required outcomes of the Cambridge AS & A‑Level Physics syllabus.