(a) (i) In Newton's ring experiment, the space between planoconvex lens and glass plate is filled with a liquid of refractive…

(a) (i) Newton's rings with a liquid film With a liquid of refractive index μ in the gap, the optical path of the film becomes μ times its geometric path, so the pattern changes compared with air. At thickness t, the reflected rays have path difference 2μt + λ/2 (normal incidence; the λ/2 comes from the phase change of π at one of the two reflections, as long as μ lies below or above both glass indices consistently). Dark ring (destructive interference): 2μt + λ/2 = (2n + 1)λ/2 ⇒ 2μt = nλ, n = 0, 1, 2, … Geometry: for a lens of radius of curvature R, r² = 2Rt − t² ≈ 2Rt, so t = rₙ²/2R. 2μ(rₙ²/2R) = nλ ⇒ rₙ = √(nλR/μ), i.e. Dₙ² = 4nλR/μ.

Change compared with air (rₙ = √(nλR)): every ring shrinks by the factor 1/√μ, the rings come closer together and more rings are seen in the same field; the central spot stays dark (n = 0). If μ increases: the rings contract further (rₙ ∝ 1/√μ) and get more crowded. As μ approaches the refractive index of the glass, the reflections at the film surfaces weaken and the fringe contrast falls; at μ = μ_glass the rings disappear. The relation D²ₙ₊ₚ − D²ₙ = 4pλR/μ is used to measure μ.

(a) (ii) Fabry–Pérot: wavelength separation of successive maxima Transmitted maxima at normal incidence (air gap): 2d = mλ. Order: m = 2d/λ = (2 × 0.8 × 10⁻³)/(500 × 10⁻⁹) = 3200. Successive maxima satisfy 2d = mλ = (m − 1)(λ + Δλ) ⇒ Δλ = λ/(m − 1) ≈ λ/m = λ²/2d Δλ = (500 × 10⁻⁹)²/(2 × 0.8 × 10⁻³) = 1.56 × 10⁻¹⁰ m = 0.156 nm (1.56 Å) This is the wavelength interval between neighbouring transmitted orders (the free spectral range) set by the given data.

Why it depends on d: the phase difference between successive beams is δ = 4πd/λ, so dδ/dλ = −4πd/λ². A larger d makes the phase change faster with wavelength, so a smaller change of λ is needed to move from one maximum to the next (Δλ = λ²/2d) and the interference order m = 2d/λ is higher. The resolving power λ/δλ = mF (F = finesse, fixed by the mirror reflectivity) therefore rises with d, while the free spectral range shrinks.

(b) Single-slit Fraunhofer diffraction Minima: a sin θ = mλ, m = ±1, ±2, … (m = 0 is the central maximum).

(i) λ = 600 nm = 6.0 × 10⁻⁷ m, a = 0.3 mm = 3.0 × 10⁻⁴ m, f = 1 m. First minimum: θ₁ = λ/a = 2.0 × 10⁻³ rad Angular width of central maximum = 2λ/a = 4.0 × 10⁻³ rad Linear width on the screen in the focal plane = 2fλ/a = 2 × 1 × 2.0 × 10⁻³ = 4.0 × 10⁻³ m = 4.0 mm

(ii) Slit halved (a′ = 0.15 mm): angular width = 2λ/a′ = 8.0 × 10⁻³ rad and linear width = 8.0 mm, both doubled. All minima move out twice as far. The light through the slit is halved and the central peak intensity (I₀ ∝ a²) falls to one quarter.

(iii) Minima coincide if m₁λ₁ = m₂λ₂, i.e. m₁/m₂ = λ₂/λ₁ = 450/600 = 3/4. So the 3rd minimum of 600 nm coincides with the 4th minimum of 450 nm (3 × 600 = 4 × 450 = 1800 nm): sin θ = 1800 × 10⁻⁹/3.0 × 10⁻⁴ = 6.0 × 10⁻³, y = fθ = 6.0 mm from the centre (and again at 12 mm, 18 mm, …). The first minima (2.0 mm and 1.5 mm) do not coincide. So yes, some minima coincide: every (3k)th minimum of 600 nm with the (4k)th of 450 nm.

(c) Particle under the central force F = −kr (i) Angular momentum about the force centre: J = r × p = m(r × ṙ), of magnitude J = mr²θ̇, perpendicular to the plane of motion. dJ/dt = ṙ × mṙ + r × F = 0 + r × (−kr) = 0 Since the force is along r, its torque about the centre vanishes and J is conserved.

(ii) In polar coordinates, a = (r̈ − rθ̇²) r̂ + (rθ̈ + 2ṙθ̇) θ̂ and F = −kr r̂: Radial: m(r̈ − rθ̇²) = −kr Transverse: m(rθ̈ + 2ṙθ̇) = (m/r) d(r²θ̇)/dt = 0 ⇒ mr²θ̇ = J = constant Eliminate θ̇ = J/(mr²): mr̈ = −kr + J²/(mr³), a one-dimensional equation for r. Writing mr̈ = −dU_eff/dr with U(r) = ½kr²: U_eff(r) = ½kr² + J²/(2mr²) Energy: E = ½mṙ² + U_eff(r) is conserved. The J²/(2mr²) term is the centrifugal barrier; U_eff has a minimum (circular orbit) at r₀ = (J²/mk)¹/4.