(a) What is meant by achromatic combination of lenses? Derive the condition for achromatization of a pair of lenses separated by…

(a) Achromatic combination of two separated lenses An achromatic combination is a lens system whose focal length (power) is the same for two chosen wavelengths, usually red (C) and blue (F), so that chromatic aberration is removed to first order.

For a thin lens, P = 1/f = (μ − 1)K, where K = 1/R₁ − 1/R₂ depends only on the surfaces. Hence dP = K dμ = [dμ/(μ − 1)]P = ωP, where ω = (μ_B − μ_R)/(μ_Y − 1) is the dispersive power (μ_Y for the mean, yellow, wavelength).

Two thin lenses of powers P₁ = 1/f₁, P₂ = 1/f₂ and dispersive powers ω₁, ω₂, separated by x, have equivalent power P = P₁ + P₂ − xP₁P₂ Condition for achromatism, dP = 0: dP₁ + dP₂ − x(P₂dP₁ + P₁dP₂) = 0 ω₁P₁ + ω₂P₂ − xP₁P₂(ω₁ + ω₂) = 0 Dividing by P₁P₂: ω₁f₂ + ω₂f₁ = x(ω₁ + ω₂) x = (ω₁f₂ + ω₂f₁)/(ω₁ + ω₂)

Special cases:

• Same glass (ω₁ = ω₂): x = (f₁ + f₂)/2, as in Huygens' eyepiece.
• Lenses in contact (x = 0): ω₁/f₁ + ω₂/f₂ = 0, so f₁ and f₂ must have opposite signs (a converging crown lens with a diverging flint lens). Achromatism holds exactly only for the two chosen wavelengths; a small residual secondary spectrum remains.

(b) Circular aperture: Airy pattern and resolving power Intensity distribution: take a circular aperture of radius a (diameter D = 2a) illuminated normally by a plane wave, and a point in direction θ in the focal plane. An element ρ dρ dφ at (ρ, φ) contributes with path difference ρ sin θ cos φ, so the amplitude is E(θ) ∝ ∫₀ᵃ ∫₀²π e^−ikρ sin θ cos φ ρ dρ dφ = 2π ∫₀ᵃ J₀(kρ sin θ) ρ dρ Using ∫₀ᵘ uJ₀(u) du = uJ₁(u), with v = ka sin θ = (πD/λ) sin θ: E(θ) ∝ πa²[2J₁(v)/v] I(θ) = I₀[2J₁(v)/v]² This is the Airy pattern: a bright central disc surrounded by faint rings (the first ring has about 1.75% of I₀).

First minimum: J₁(v) = 0 first at v = 3.832, so (πD/λ) sin θ₁ = 3.832 ⇒ sin θ₁ = 1.22λ/D

Resolving power: by Rayleigh's criterion two point objects are just resolved when the central maximum of one image falls on the first minimum of the other. The minimum resolvable angle is therefore θ_min = 1.22λ/D, and the resolving power is 1/θ_min = D/(1.22λ). A larger aperture or a shorter wavelength improves resolution.

Numerical: D = 10 cm = 0.10 m, λ = 500 nm = 5.0 × 10⁻⁷ m θ_min = 1.22 × 5.0 × 10⁻⁷/0.10 = 6.1 × 10⁻⁶ rad ≈ 1.26″ (1 rad = 206265″) Resolving power = 1/θ_min ≈ 1.64 × 10⁵.

(c) Lagrangian of the heavy symmetric top Use Euler angles: θ (nutation, tilt of the symmetry axis from the vertical), φ (precession about the vertical) and ψ (spin about the symmetry axis). With the fixed tip as the common origin, let I₁ = I₂ be the principal moments about transverse axes through the tip (I₁ = I_cm + Ml²) and I₃ that about the symmetry axis; l is the distance from the tip to the centre of mass.

Body-frame angular velocity components: ω₁ = φ̇ sin θ sin ψ + θ̇ cos ψ, ω₂ = φ̇ sin θ cos ψ − θ̇ sin ψ, ω₃ = ψ̇ + φ̇ cos θ T = ½I₁(ω₁² + ω₂²) + ½I₃ω₃² = ½I₁(θ̇² + φ̇² sin²θ) + ½I₃(ψ̇ + φ̇ cos θ)² V = Mgl cos θ L = ½I₁(θ̇² + φ̇² sin²θ) + ½I₃(ψ̇ + φ̇ cos θ)² − Mgl cos θ

Conserved angular momenta: φ and ψ do not appear in L (cyclic), so their conjugate momenta are constants of motion.

1. Angular momentum about the body symmetry axis: p_ψ = ∂L/∂ψ̇ = I₃(ψ̇ + φ̇ cos θ) = I₃ω₃ = constant
2. Angular momentum about the fixed vertical axis: p_φ = ∂L/∂φ̇ = I₁φ̇ sin²θ + I₃(ψ̇ + φ̇ cos θ) cos θ = constant Physically, the gravitational torque about the tip, l × Mg, lies along the line of nodes, which is perpendicular to both the vertical and the symmetry axis, so the components of angular momentum along these two axes do not change. Since L has no explicit time dependence, the energy E = T + V is also conserved.