(a) (i) Explain the phenomenon of double refraction. What are positive and negative crystals? Give their examples. (5…

(a)(i) Double Refraction

When a ray of light enters certain anisotropic crystals (e.g., calcite, quartz), it splits into two refracted rays: the ordinary ray (O-ray) and the extraordinary ray (E-ray). This phenomenon is double refraction. The O-ray obeys Snell's law and has refractive index n₀ independent of direction; the E-ray does not obey Snell's law and its refractive index nₑ varies with direction. The cause is the anisotropic arrangement of atoms, which makes the velocity of light direction-dependent. The two rays are plane-polarised in mutually perpendicular planes, and along the optic axis there is no double refraction. In Huygens' picture the O-wavelet is a sphere and the E-wavelet an ellipsoid of revolution.

Crystals are classified by comparing nₑ and n₀. If nₑ > n₀, the crystal is positive (e.g., quartz). If nₑ < n₀, it is negative (e.g., calcite).

(a)(ii) Optical Activity

Optical activity is the rotation of the plane of polarization of linearly polarized light as it passes through certain substances (e.g., quartz, sugar solution). The rotation arises because right-circularly and left-circularly polarized components travel with different velocities (different refractive indices), so a phase difference develops between them, rotating the plane of polarization.

For thickness d, rotation θ = (πd/λ)(n_L − n_R).

Given d = 0.2 cm = 2×10⁻³ m, λ = 0.5 μm = 5×10⁻⁷ m, n_L − n_R = 7×10⁻⁵:

θ = (π × 2×10⁻³ / 5×10⁻⁷) × 7×10⁻⁵ = π × (2×10⁻³ × 7×10⁻⁵)/(5×10⁻⁷) = π × 0.28 = 0.88 rad ≈ 50.4°.

(Using θ = πd(n_L−n_R)/λ directly gives 0.88 rad ≈ 50.4°.)

(b)(i) Attenuation in Optical Fibers

Attenuation is the loss of optical power as light propagates through a fiber, measured in dB/km. Factors: (1) Absorption — intrinsic (UV and IR absorption of silica) and extrinsic (OH⁻ and metal ion impurities); (2) Scattering — Rayleigh scattering from density fluctuations (∝ λ⁻⁴) and Mie scattering from imperfections; (3) Bending losses — micro-bending and macro-bending causing mode leakage.

(b)(ii) Power Calculation

P_out = P_in × 10^(−αL/10). Here α = 0.5 dB/km, L = 50 km, so αL = 25 dB.

P_out = 10 mW × 10^(−2.5) = 10 × 3.16×10⁻³ = 0.0316 mW.

(c)(i) Hooke's Law

Hooke's law states that within the elastic limit, stress is directly proportional to strain: σ = Eε, where E is Young's modulus. The stress-strain diagram for a wire shows: (1) proportional limit — linear region obeying Hooke's law; (2) elastic limit — beyond which permanent deformation begins; (3) yield point — marked increase in strain without stress increase; (4) ultimate stress — maximum stress the wire can bear; (5) breaking stress — point of fracture. The plastic region lies between yield and fracture.

(c)(ii) Poiseuille's Equation

Assume steady, laminar (streamline) flow of an incompressible liquid with zero velocity at the wall. Consider a coaxial cylinder of radius x inside a tube of radius r and length l. The pressure force Pπx² balances the viscous drag −η(2πxl)(dv/dx), so dv/dx = −Px/(2ηl). Integrating with v = 0 at x = r gives the parabolic profile v(x) = P(r² − x²)/(4ηl). The volume flow rate is

Q = ∫₀^r v(x) 2πx dx = πPr⁴/(8ηl).

For two tubes in series, the same Q flows through both. Pressure drops: P₁ = 8ηl₁Q/(πr₁⁴), P₂ = 8ηl₂Q/(πr₂⁴). Total P = P₁ + P₂ = (8ηQ/π)(l₁/r₁⁴ + l₂/r₂⁴).

Thus Q = (πP/8η)(l₁/r₁⁴ + l₂/r₂⁴)⁻¹.

This is analogous to electrical resistances in series, where the equivalent resistance is the sum of individual resistances.