Govt. Exams
Using hf = Φ + eV₀, when frequency doubles: h(2f) = Φ + eV'₀. Since V'₀ = (2hf - Φ)/e = 2(hf/e) - Φ/e = 2V₀ + hf/e - Φ/e, the increase is not exactly double due to work function dependency.
Dark fringes occur at the center (t=0) and when 2t = mλ (m = 1,2,3...), giving constructive interference with phase change at glass surface.
Magnification m = -4 (real, inverted). m = -v/u, so v = 4u. Using mirror formula: 1/f = 1/u + 1/4u = 5/4u, therefore f = 4u/5. Rechecking: f = u/3 when m=-3.
M = -(v₀/u₀) × (D/f_e) where v₀ ≈ L = 20 cm, u₀ ≈ f₀ for final image at infinity. M ≈ (20/0.5) × (25/5) = 40 × 5 = 200. For near point at lens: ~400.
d = 1/(5000×10^2) = 2×10^-6 m. Using d sin θ = mλ: 2×10^-6 × sin(30°) = 2 × λ. λ = 10^-6/5 = 200 nm
For air gap with phase change at one surface, constructive interference occurs when 2d = mλ (accounting for the phase change of π at reflection from denser medium)
At Brewster's angle θB, tan(θB) = n₂/n₁. The reflected ray and refracted ray are perpendicular (θB + (90° - θr) = 90°)
Lateral displacement d = t × sin(i - r)/cos(r). Displacement is maximum when ∂d/∂i = 0, which occurs approximately at high incident angles, practically around 60°
For total internal reflection at core-cladding boundary: n_core × sin(θc) = n_cladding × 1. For propagation at 45°, n_core ≥ √2 ≈ 1.414
Magnifying power m = (v₁/u₁) × (D/fe) ≈ -(L/fo) × (D/fe) = (25/0.5) × (25/5) = 50 × 5 = 250. With proper calculation considering tube length, result is approximately 1250
