Wave Optics – Interference, Diffraction & Polarization
(KVPY – Physics – Historic Problem Set)
Goal – Understand how light behaves as a wave, predict patterns, and model the action of polarizing devices.
1. Simple Explanation
| Concept | What it is | How it shows up |
|---|---|---|
| Interference | Superposition of two or more coherent light waves. | Bright & dark fringes in Young’s double‑slit, Newton’s rings, etc. |
| Diffraction | Bending of waves when encountering obstacles or apertures. | Central bright fringe in single‑slit, Airy pattern for circular apertures. |
| Polarization | Restriction of electric‑field oscillation to a plane or ellipse. | Light becomes linear when passed through a polarizer, circular when two orthogonal polarizers are crossed at 45°. |
2. Key Points
2.1 Interference
| Quantity | Meaning | Key Equation |
|---|---|---|
| Coherence length | Distance over which two waves remain phase‑locked | (L_c = \dfrac{\lambda^2}{\Delta \lambda}) |
| Path difference | (\Delta = d \sin \theta) (Young’s double slit) | |
| Phase difference | (\Delta \phi = \dfrac{2\pi}{\lambda},\Delta) | |
| Constructive interference | (\Delta = m\lambda) | |
| Destructive interference | (\Delta = (m+\tfrac12)\lambda) |
2.2 Diffraction
| Type | Condition | Resulting Pattern |
|---|---|---|
| Fraunhofer | Light parallel to axis, far field | (\theta_n = n\lambda / a) (single slit) |
| Fresnel | Near field, finite distance | Complex integrals (not covered in depth) |
| Multiple‑slit | (N) slits, spacing (d) | Sharp principal maxima when (d\sin\theta = m\lambda) |
| Circular aperture | Airy pattern | (\theta_\text{min} \approx 1.22 \lambda / D) |
2.3 Polarization
| Item | Formula / Law | Notes |
|---|---|---|
| Malus’ Law | (I = I_0 \cos^2 \theta) | (\theta) = angle between light’s polarization and analyzer |
| Jones vector | (\begin{pmatrix} E_x \ E_y \end{pmatrix}) | Useful for linear & circular |
| Circular polarization | (E_x = E_0 \cos\omega t,; E_y = E_0 \sin\omega t) | Right‑handed / left‑handed |
| Polarizer | Transmits only one component | Ideal polarizer: (I = I_{\text{in}}\cos^2\theta) |
3. Worked Examples
3.1 Young’s Double‑Slit
Problem – Two slits, (d = 0.2,\text{mm}), wavelength (\lambda = 500,\text{nm}). Screen is (L = 2.0,\text{m}) away. Find fringe spacing (\Delta y).
Solution
- Small‑angle approximation: (\sin \theta \approx \tan \theta \approx \theta = \Delta y / L).
- Constructive condition: (d \sin \theta = m\lambda) → (d,\Delta y / L = m\lambda).
- Fringe spacing for (m=1): [ \Delta y = \frac{m\lambda L}{d} = \frac{(1)(500\times10^{-9},\text{m})(2.0,\text{m})}{2.0\times10^{-4},\text{m}} = 0.005 \text{ m} = 5 ,\text{mm}. ] (Higher orders: (\Delta y = m \times 5,\text{mm}).)
Check – 5 mm is comfortably resolvable.
3.2 Single‑Slit Diffraction Minima
Problem – A single slit of width (a = 0.1,\text{mm}) illuminated by (\lambda = 650,\text{nm}). Find angle to first minimum.
Solution
Fraunhofer minimum condition: (a \sin \theta = m\lambda) with (m = \pm 1).
[ \sin \theta = \frac{\lambda}{a} = \frac{650\times10^{-9}}{1.0\times10^{-4}} = 0.0065 ] [ \theta \approx 0.373^\circ ] (First minimum ≈ 0.37°, next at (m=2) ≈ 0.75°.)
3.3 Malus’ Law
Problem – Light of intensity (I_0 = 10,\text{mW}) passes through a polarizer, then an analyzer rotated by (30^\circ). Find transmitted intensity.
Solution
[ I = I_0 \cos^2 30^\circ = 10 \times \left(\frac{\sqrt{3}}{2}\right)^2 = 10 \times \frac