12.3 Wave Optics

12.3 Wave Optics

Quick Answer: Wave-optics items on the PAF Aeronautical Engineering paper usually stay at FSc recognition level: Young’s double slit, fringe width β = λD/d, single-slit diffraction conditions, and polarisation (Malus / Brewster). You rarely need university-level Fourier optics—know the conditions and the headline formulas cold.

Ray optics assumes light travels in straight lines and forms sharp shadows. When obstacles or openings are comparable to the wavelength, interference and diffraction appear. Those phenomena confirm that light behaves as a wave.

Coherence — Why You Need Two “Related” Waves

Stable interference requires coherent sources: constant phase difference and the same frequency. Ordinary two separate bulbs are incoherent; Young’s experiment creates two coherent sources from one wavefront by splitting light at two slits.

Young’s Double-Slit Experiment (YDSE)

Light of wavelength λ illuminates two slits separated by d. A screen sits at distance D (D ≫ d). Path difference to a point at distance y from the centre is approximately:

$$\delta = \frac{yd}{D}$$

Central fringe (m = 0) is bright for sources that start in phase.

Fringe Width

Distance between adjacent bright (or adjacent dark) fringes:

$$\beta = \frac{\lambda D}{d}$$

So fringe spacing increases if λ increases or D increases, and decreases if slit separation d increases.

Worked example — fringe width

λ = 600 nm = 6.0 × 10⁻⁷ m, D = 1.0 m, d = 0.30 mm = 3.0 × 10⁻⁴ m.

$\beta = \frac{(6.0\times 10^{-7})(1.0)}{3.0\times 10^{-4}} = 2.0\times 10^{-3}\ \text{m} = 2.0\ \text{mm}$

Worked example — which fringe?

If δ = 1.5λ at a point, that is destructive interference (dark), because 1.5 = m + ½ with m = 1.

Position of Bright Fringes

$$y_m = \frac{m\lambda D}{d}$$

Third bright fringe (m = 3) sits at y₃ = 3λD/d from the centre (not counting the central fringe as “first” in some wordings—read the question: “third bright” usually means m = 3).

Diffraction

Diffraction is the spreading of waves when they pass an aperture or edge. A wider aperture relative to λ shows less noticeable spreading; a narrow slit shows a broader central maximum.

For a single slit of width a, dark fringes (minima) satisfy:

$$a\sin\theta = m\lambda \quad (m = \pm 1, \pm 2, \ldots)$$

The central bright region is wider than the secondary maxima. Qualitatively: smaller a → broader diffraction pattern.

Diffraction limits angular resolution of optical instruments (Rayleigh criterion appears in richer syllabi). For PAF AE initial MCQs, recognising “bending around obstacles / spreading at slits” and the single-slit minima formula is usually enough.

Polarisation

Longitudinal waves (sound in air) cannot be polarised. Polarisation of light proves light is transverse.

A polariser transmits the electric-field component along its transmission axis. If unpolarised light of intensity I₀ passes one ideal polariser, transmitted intensity is I₀/2. A second polariser (analyser) at angle θ to the first obeys Malus’s law:

$$I = I_0\cos^2\theta$$

(here I₀ is the intensity after the first polariser if that is how the problem defines it—read carefully).

Worked example — Malus

Intensity after polariser–analyser with θ = 60°: I = I₀ cos²60° = I₀(0.5)² = I₀/4.

Brewster’s Law

At the polarising angle i_p, reflected light is completely plane-polarised (electric field perpendicular to the plane of incidence):

$$\mu = \tan i_p$$

For glass with μ = √3, i_p = 60°.

Thin-Film Colours (Recognition)

Soap bubbles and oil films show colours because of interference between reflections from front and back surfaces, with a possible phase change of π on reflection from a denser medium. You may only need the qualitative idea: path difference depends on film thickness and wavelength, so some colours constructively reinforce while others cancel. White-light illumination therefore produces coloured bands rather than simple dark/bright monochromatic fringes.

White Light vs Monochromatic Light in YDSE

With a single wavelength (laser or filtered light), fringes are sharp and equally spaced. With white light, the central fringe is white (all wavelengths constructively overlap near δ = 0), while coloured fringes appear outward and eventually wash out because different wavelengths have different β = λD/d. That comparison is a frequent recognition MCQ: central fringe white → white light; highly coherent laser → high-contrast mono fringes.

How Wave Optics Connects to Aeronautics (Motivation)

Antenna patterns, radar lobes, and optical sensors all live in a wave world. The academic paper will ask FSc-style questions—fringe width, path difference, Malus—not full radar design. Still, the same physics of interference and diffraction underpins why a larger aperture can resolve finer detail and why coherent sources matter.

Rapid Revision Checklist

1. Coherent sources → stable fringes.
2. Bright: δ = mλ; dark: δ = (m + ½)λ.
3. β = λD/d — know how each symbol scales the pattern.
4. Diffraction: waves spread at edges; single-slit minima a sin θ = mλ.
5. Polarisation: transverse nature; I ∝ cos²θ; μ = tan i_p.

If you can apply those five points under time pressure, wave-optics MCQs become recognition-plus-arithmetic rather than surprise theory.