(a) Consider the diagram below with a water flow rate Q. Derive the expression for Q in terms of the difference in the manometer…

(a) Let water be incompressible, steady, non-viscous and irrotational. At the wider section 1, let pressure be p₁, speed v₁ and area A₁. At the throat section 2, let pressure be p₂, speed v₂ and area A₂. Since the duct is horizontal, the Bernoulli equation between sections 1 and 2 is p₁ + 1/2 ρv₁² = p₂ + 1/2 ρv₂². Hence p₁ − p₂ = 1/2 ρ(v₂² − v₁²). By the continuity equation for water, Q = A₁v₁ = A₂v₂, so v₁ = Q/A₁ and v₂ = Q/A₂. Substituting, p₁ − p₂ = ρQ²/2 (1/A₂² − 1/A₁²) = ρQ²/2 ((A₁² − A₂²)/(A₁²A₂²)).

The two vertical open manometer tubes measure the pressure-head difference. Since the pipe is horizontal and the same liquid water is used, p₁ − p₂ = ρgh. Equating the two expressions for p₁ − p₂, ρgh = ρQ²/2 ((A₁² − A₂²)/(A₁²A₂²)). Cancelling ρ and solving for Q, Q² = 2gh A₁²A₂²/(A₁² − A₂²). Therefore Q = A₁A₂ √(2gh/(A₁² − A₂²)). This is valid for a horizontal venturimeter, steady incompressible flow, no viscous losses, and the same liquid in the manometer as in the pipe. If a manometer liquid of density ρₘ is used, h is replaced by h(ρₘ/ρ − 1).

(b) In Fraunhofer diffraction at a single slit, the source and screen are effectively at infinity, so a plane wavefront falls normally on a slit of width a. According to Huygens’ principle, every point of the slit acts as a source of secondary wavelets. These wavelets interfere on the screen. For diffraction angle θ, the path difference between wavelets from the two edges of the slit is a sinθ.

Divide the slit into infinitesimal strips of width dx. Let x be the distance of a strip from the centre of the slit. Its phase lag relative to the centre is δ = (2π/λ)x sinθ. The contribution to the resultant amplitude is dE = (E₀/a) exp[−i(2π/λ)x sinθ] dx. Integrating from x = −a/2 to x = +a/2, E = (E₀/a) ∫ exp[−i(2π/λ)x sinθ] dx = (E₀/a) [exp[−i(2π/λ)x sinθ] / (−i(2π/λ) sinθ)] from −a/2 to a/2 = (E₀/a) (exp[−i(πa sinθ/λ)] − exp[i(πa sinθ/λ)]) / (−i(2π/λ) sinθ) = (E₀/a) (−2i sin(πa sinθ/λ)) / (−i(2π/λ) sinθ) = E₀ sin(πa sinθ/λ) / (πa sinθ/λ). Let β = (πa sinθ)/λ. Then E = E₀ sinβ/β. Therefore the intensity is I = I₀ (sinβ/β)², where I₀ is the intensity at θ = 0.

Minima occur when sinβ = 0 but β ≠ 0, i.e. β = mπ, m = ±1, ±2, ±3, … or a sinθ = mλ. The central maximum is at β = 0. Secondary maxima are obtained from dI/dβ = 0: d/dβ (sin²β/β²) = 0 ⇒ (2 sinβ cosβ β² − sin²β 2β)/β⁴ = 0 ⇒ 2 sinβ (β cosβ − sinβ)/β³ = 0. For nonzero maxima, tanβ = β. The exact roots of tanβ = β are close to βₙ ≈ (2n+1)π/2 for n = 1, 2, 3, … At these approximate values, sinβ ≈ ±1, so Iₙ/I₀ ≈ 1/βₙ². Hence I₁/I₀ ≈ 1/(3π/2)² = 4/(9π²), I₂/I₀ ≈ 1/(5π/2)² = 4/(25π²), I₃/I₀ ≈ 1/(7π/2)² = 4/(49π²). Therefore the intensities of successive maxima are nearly in the ratio 1 : 4/(9π²) : 4/(25π²) : 4/(49π²). The central maximum is the brightest; the secondary maxima fall off rapidly.

(c) Let the speed of each spaceship relative to the Earth observer be u. Since they approach each other, their velocities in the Earth frame are +u and −u along the line of motion. Using the relativistic velocity-addition formula, the speed of one spaceship relative to the other is w = (u + u)/(1 + u²/c²) = 2u/(1 + u²/c²). Given w = 0.7c, let x = u/c. Then 0.7 = 2x/(1 + x²). Thus 0.7(1 + x²) = 2x ⇒ 0.7x² − 2x + 0.7 = 0. Multiplying by 10, 7x² − 20x + 7 = 0. Solving, x = (20 ± √(400 − 196))/14 = (20 ± √204)/14 = (10 ± √51)/7. The positive sign gives x = (10 + √51)/7 ≈ 2.45 > 1, which is impossible. Hence x = (10 − √51)/7 ≈ 0.408. Therefore u ≈ 0.408c ≈ 1.22 × 10⁸ m s⁻¹. So, as measured by the stationary Earth observer, one spaceship moves with velocity +0.408c and the other with velocity −0.408c along their line of approach.