(a) (i) Write down the system matrix for a combination of two thin lenses in paraxial approximation. Hence obtain the focal…

(a)(i) In paraxial approximation, sin θ ≈ tan θ ≈ θ. Take a ray vector (y, θ), where y is height and θ is slope. The translation matrix through distance d is T(d) = [[1, d], [0, 1]] and the thin-lens matrix is L(f) = [[1, 0], [-1/f, 1]]. For two thin lenses f₁ and f₂ separated by d, the system matrix from just before L₁ to just after L₂ is M = L(f₂) T(d) L(f₁) = [[1 - d/f₁, d], [-(1/f₁ + 1/f₂ - d/(f₁ f₂)), 1 - d/f₂]]. For any system matrix M = [[A, B], [C, D]], the effective focal length satisfies C = -1/f. Hence 1/f = 1/f₁ + 1/f₂ - d/(f₁ f₂). For unit (principal) planes, h₁ = (D - 1)/C and h₂ = (1 - A)/C, measured from the first and second lenses respectively, positive to the right. Substituting A, C, D gives h₁ = f d/f₂, h₂ = - f d/f₁. Thus the first unit plane lies f d/f₂ to the right of L₁, and the second unit plane lies f d/f₁ to the left of L₂.

(a)(ii) Given f₁ = +10 cm, f₂ = +20 cm, d = 25 cm. 1/f = 1/10 + 1/20 - 25/(10×20) = 0.10 + 0.05 - 0.125 = 0.025 cm⁻¹. Therefore f = 1/0.025 = 40 cm. The first unit plane is h₁ = f d/f₂ = (40×25)/20 = 50 cm to the right of L₁. Since L₂ is 25 cm to the right of L₁, this is 25 cm to the right of L₂. The second unit plane is h₂ = - f d/f₁ = -(40×25)/10 = -100 cm relative to L₂, i.e. 100 cm to the left of L₂, or 75 cm to the left of L₁. A ray trace confirms this: a ray parallel to the axis at unit height leaves L₂ at height −1.5 with slope −0.025, so the back focal point is 60 cm to the left of L₂ (virtual), which is 40 cm from H′. Final: f = +40 cm; H at +50 cm from L₁; H′ at -75 cm from L₁.

(b) The central-zone diameter is 2.4 mm, so its radius is r₁ = 1.2 mm = 1.2×10⁻³ m. Wavelength λ = 600 nm = 6.0×10⁻⁷ m. For a source at distance u = 5.0 m and first image at distance v, the path difference between the axial ray and the ray through the first-zone edge is approximately r₁²/2 (1/u + 1/v). For the first image, this equals λ/2, so r₁²(1/u + 1/v) = λ. Thus 1/u + 1/v = λ/r₁² = 1/f, where f = r₁²/λ. Now f = (1.2×10⁻³)²/(6.0×10⁻⁷) = 1.44×10⁻⁶/6.0×10⁻⁷ = 2.4 m. Then 1/v = 1/f - 1/u = 1/2.4 - 1/5.0 = 5/12 - 1/5 = 13/60 m⁻¹. Hence v = 60/13 m = 4.615... m ≈ 4.62 m. Final: the first image is real and lies on the opposite side of the zone plate from the source at 60/13 m ≈ 4.62 m from the plate.

(c)(i) Let β = V/c and γ = 1/√(1 - β²). The Lorentz transformation from O to O′ is x′ = γ(x - Vt), y′ = y, z′ = z, t′ = γ(t - Vx/c²). Let β′ = V′/c and γ′ = 1/√(1 - β′²). The transformation from O′ to O″ is x″ = γ′(x′ - V′t′), y″ = y′, z″ = z′, t″ = γ′(t′ - V′x′/c²). Substituting the first pair into the second pair: x″ = γ′[γ(x - Vt) - V′γ(t - Vx/c²)] = γγ′[(1 + VV′/c²)x - (V+V′)t]. Similarly, t″ = γ′[γ(t - Vx/c²) - V′γ(x - Vt)/c²] = γγ′[(1 + VV′/c²)t - (V+V′)x/c²]. Define u = (V+V′)/(1 + VV′/c²), Γ = γγ′(1 + VV′/c²) = 1/√(1 - u²/c²). Then x″ = Γ(x - ut), y″ = y, z″ = z, t″ = Γ(t - ux/c²). This is again a Lorentz transformation with velocity u along the x-axis. Hence the velocity addition law is u = (V+V′)/(1 + VV′/c²).

(c)(ii) For a receding source, the relativistic Doppler formula is λ_obs = λ_emit √((1+β)/(1-β)), where β = v/c. Here β = 15000/(3.00×10⁵) = 0.05. Thus λ_obs = 550 nm × √(1.05/0.95) = 550 nm × √(21/19) ≈ 550 nm × 1.05132 = 578.22 nm. Final: λ_obs ≈ 578.2 nm.