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Contests/Gnit Sunday Open 002 (GSO002)/Problem 11 Interference Fringe Shift from Buoyancy and Elastic Support
Problem 11

Interference Fringe Shift from Buoyancy and Elastic Support

Finished
300 ptsLv.6 IntermediateWaves
2026/03/22 19:30〜2026/03/22 21:00
Author: admin

Problem Statement

A glass plate B with a flat horizontal upper surface is fixed in place. On top of it lies a uniform rectangular glass plate A of length LLL. The left end of plate A (at x=0x = 0x=0) is in contact with the upper surface of plate B, and this joint acts like a smooth hinge that can rotate — plate A can tilt by a small angle in the vertical plane.

Between the lower surface of the right end of plate A (at x=Lx = Lx=L) and the upper surface of plate B, a small cylindrical elastic spacer of natural length l0l_0l0​ and unknown spring constant is inserted, supporting plate A. The spacer deformation follows Hooke's law; its mass and volume are negligible.

Let the mass of plate A be MMM, its density ρ\rhoρ, and the magnitude of gravitational acceleration be ggg.

Between plates A and B, a very thin wedge-shaped gap is formed.

[Experiment 1] The entire apparatus is placed in air. Monochromatic light of wavelength λ\lambdaλ is shone vertically downward onto the apparatus, and bright and dark interference fringes are observed due to interference between reflections from the lower surface of plate A and the upper surface of plate B.

The spacing between adjacent bright fringes is measured to be Δx1\Delta x_1Δx1​.

The refractive index of air is 111; the density and buoyancy of air are negligible.

[Experiment 2] The entire apparatus is fully submerged in a transparent unknown liquid of density ρ0\rho_0ρ0​ and refractive index nnn, filling the wedge-shaped gap with the liquid as well.

After sufficient time and the plate A reaches equilibrium, monochromatic light of wavelength λ\lambdaλ is again shone vertically downward. The spacing between adjacent bright fringes is now Δx2\Delta x_2Δx2​.

Assume the refractive indices of plates A and B are both greater than nnn. The spring constant of the spacer does not change due to the liquid.

From these results, find the refractive index nnn of the liquid.

Constraints

  • Mass of plate A: M=0.40 kgM = 0.40 \mathrm{\ kg}M=0.40 kg
  • Density of plate A: ρ=2.0×103 kg/m3\rho = 2.0 \times 10^3 \mathrm{\ kg/m^3}ρ=2.0×103 kg/m3
  • Length of plate A: L=0.20 mL = 0.20 \mathrm{\ m}L=0.20 m
  • Natural length of elastic spacer: l0=1.0×10−4 ml_0 = 1.0 \times 10^{-4} \mathrm{\ m}l0​=1.0×10−4 m
  • Gravitational acceleration: g=10 m/s2g = 10 \mathrm{\ m/s^2}g=10 m/s2
  • Wavelength of monochromatic light: λ=6.0×10−7 m\lambda = 6.0 \times 10^{-7} \mathrm{\ m}λ=6.0×10−7 m
  • Density of liquid: ρ0=1.0×103 kg/m3\rho_0 = 1.0 \times 10^3 \mathrm{\ kg/m^3}ρ0​=1.0×103 kg/m3
  • Fringe spacing in air: Δx1=0.75×10−3 m\Delta x_1 = 0.75 \times 10^{-3} \mathrm{\ m}Δx1​=0.75×10−3 m
  • Fringe spacing in liquid: Δx2=0.50×10−3 m\Delta x_2 = 0.50 \times 10^{-3} \mathrm{\ m}Δx2​=0.50×10−3 m

Input Format

The refractive index nnn is expressed as an irreducible fraction AB\frac{A}{B}BA​ with A,BA, BA,B coprime positive integers. Compute 100A+B100A + B100A+B and give the answer as a positive integer.

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