GSO002 Problem 11
Problem Statement
Interference Fringe Shift from Buoyancy and Elastic Support
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 L. The left end of plate A (at x=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=L) and the upper surface of plate B, a small cylindrical elastic spacer of natural length l0 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 M, its density ρ, and the magnitude of gravitational acceleration be g.
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 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.
The refractive index of air is 1; the density and buoyancy of air are negligible.
[Experiment 2] The entire apparatus is fully submerged in a transparent unknown liquid of density ρ0 and refractive index n, filling the wedge-shaped gap with the liquid as well.
After sufficient time and the plate A reaches equilibrium, monochromatic light of wavelength λ is again shone vertically downward. The spacing between adjacent bright fringes is now Δx2.
Assume the refractive indices of plates A and B are both greater than n. The spring constant of the spacer does not change due to the liquid.
From these results, find the refractive index n of the liquid.
Constraints
- Mass of plate A: M=0.40 kg
- Density of plate A: ρ=2.0×103 kg/m3
- Length of plate A: L=0.20 m
- Natural length of elastic spacer: l0=1.0×10−4 m
- Gravitational acceleration: g=10 m/s2
- Wavelength of monochromatic light: λ=6.0×10−7 m
- Density of liquid: ρ0=1.0×103 kg/m3
- Fringe spacing in air: Δx1=0.75×10−3 m
- Fringe spacing in liquid: Δx2=0.50×10−3 m
Input Format
The refractive index n is expressed as an irreducible fraction BA with A,B coprime positive integers. Compute 100A+B and give the answer as a positive integer.
Solution
1. Small Angle Approximation
The gap height at the right end (x=L) is D, and the tilt angle θ satisfies sinθ=D/L. Since D<l0=1.0×10−4 m, the angle θ<5.0×10−4 rad is extremely small, so cosθ≈1.
2. Analysis in Air (Experiment 1)
With the gap height at the right end D1, the fringe spacing formula for a wedge-shaped gap: Δx1=2D1Lλ Solving for D1: D1=2Δx1Lλ=2×0.75×10−30.20×6.0×10−7=8.0×10−5 m
Torque balance about the hinge (x=0). Plate weight W=Mg=4.0 N acts at the center of gravity (x=L/2). Spacer support force F1: F1=2W=2.0 N Spacer compression y1=l0−D1=1.0×10−4−8.0×10−5=2.0×10−5 m.
From Hooke's law F1=ky1: k=2.0×10−52.0=1.0×105 N/m
3. Analysis in Liquid (Experiment 2)
Buoyancy on plate A: Fb=ρ0ρMg=1.0×103×2.0×1030.40×10=2.0 N
Effective weight: W′=W−Fb=4.0−2.0=2.0 N
Spacer support force in liquid: F2=2W′=1.0 N
Spacer compression: y2=kF2=1.0×1051.0=1.0×10−5 m
Gap height at right end: D2=l0−y2=1.0×10−4−1.0×10−5=9.0×10−5 m
For a gap filled with liquid of refractive index n, the fringe spacing formula is: Δx2=2nD2Lλ Solving for n: n=2D2Δx2Lλ=2×9.0×10−5×0.50×10−30.20×6.0×10−7=9.0×10−81.2×10−7=34
4. Computing the Answer
n=34 is already an irreducible fraction, so A=4, B=3.
100A+B=100×4+3=403
Answer: 403