(a) (i) What are the requisite conditions for observation of interference pattern on a screen ? (5 marks) (ii) Derive the…

(a)(i) Conditions for observing an interference pattern

The two sources must be coherent, with the same frequency and a constant phase difference. The light must be nearly monochromatic, so that fringes of different wavelengths do not wash out. The amplitudes of the two waves should be nearly equal, for good fringe contrast (visibility). The path difference must be within the coherence length of the source, and the source must be small enough (spatial coherence). The waves must have the same polarisation state; orthogonally polarised waves do not interfere. The slits must be narrow and close, and the screen far away (D ≫ d), so that the fringe width is large enough to resolve.

(a)(ii) Fringe width and intensity in Young's double-slit experiment

Let the slits S₁ and S₂ be separated by d, the screen at distance D, and let P be at distance y from the central point O, with D ≫ d and y. The path difference is Δ = S₂P − S₁P ≈ d sinθ ≈ dy/D, giving a phase difference φ = 2πΔ/λ = 2πdy/(λD). For two waves of equal amplitude a: E = a cos ωt + a cos(ωt + φ) = 2a cos(φ/2) cos(ωt + φ/2). The amplitude is 2a cos(φ/2), so the intensity is I = 4a² cos²(φ/2) = I₀ cos²(πdy/λD), with I₀ = 4a² the maximum (4 times the intensity of one slit). Bright fringes: φ = 2nπ, so Δ = nλ and yₙ = nλD/d. Dark fringes: Δ = (n + ½)λ, I = 0. Fringe width: β = yₙ₊₁ − yₙ = λD/d, the same for bright and dark fringes (valid for small angles). In a medium of index μ, β = λD/(μd).

(b)(i) Separation of colliding particles in lab and CM frames

Let r₁ and r₂ be the lab positions of the particles of masses m₁ and m₂. The centre of mass is at R = (m₁r₁ + m₂r₂)/(m₁ + m₂). In the CM frame, with origin at R, the positions are r₁′ = r₁ − R and r₂′ = r₂ − R, both at the same instant (time is absolute in Newtonian mechanics). The separation vector is r₁′ − r₂′ = (r₁ − R) − (r₂ − R) = r₁ − r₂. So the separation vector and its magnitude |r₁ − r₂| are the same in both systems. In the same way the relative velocity v₁ − v₂ = v₁′ − v₂′ is invariant, because the CM frame differs from the lab frame only by the velocity V of the CM, which cancels in a difference. This is why the two-body problem can be reduced to one body of reduced mass μ = m₁m₂/(m₁ + m₂) moving with the relative coordinate.

(b)(ii) Kinetic energy of the rotating disc

For a thin disc about the axis through its centre perpendicular to its plane, I = ½MR² = ½ × 0.5 × (0.2)² = 0.01 kg m². ω = 2πf = 2π × 100 = 628.3 rad/s. K = ½Iω² = ½ × 0.01 × (628.3)² = 1.97×10³ J (exactly 200π² J ≈ 1974 J).

(c) Damped harmonic oscillations and logarithmic decrement

Equation: m d²x/dt² + b dx/dt + kx = 0, or d²x/dt² + 2γ dx/dt + ω₀²x = 0, where 2γ = b/m and ω₀² = k/m. For light damping (γ < ω₀), the trial x = e^(λt) gives λ = −γ ± iω with ω = √(ω₀² − γ²), so x = A₀e^(−γt) cos(ωt + φ). The amplitude is A(t) = A₀e^(−γt). Amplitudes one period T = 2π/ω apart have ratio Aₙ/Aₙ₊₁ = e^(γT). The logarithmic decrement is δ = ln(Aₙ/Aₙ₊₁) = γT = 2πγ/√(ω₀² − γ²). After n oscillations, Aₙ = A₀e^(−nδ), so δ = (1/n) ln(A₀/Aₙ). Numerical: A₀ = 10 cm, A₅₀ = 2 cm, n = 50 (T = 4 s): δ = (1/50) ln(10/2) = (1/50)(1.609) = 0.0322. For 25% amplitude, Aₙ = 0.25A₀, so e^(nδ) = 4 and n = ln 4/δ = 1.386/0.0322 = 43.07, about 43 oscillations (just over 43, i.e. in the 44th), which takes about 172 s.