GSO001 Problem 16
Problem Statement
Fringe Shift in a Michelson Interferometer Combined with Mechanics
Problem Statement
A Michelson interferometer uses a monochromatic light source with vacuum wavelength λ. Light from the source is split into two paths by a half-mirror (beam splitter). Path 1 has a fixed mirror M1, and Path 2 has a movable mirror M2 placed on a smooth horizontal floor with mass M. The movable mirror M2 is connected to a light spring with spring constant k and can oscillate horizontally.
Furthermore, the movable mirror M2 and the space it moves in are completely immersed in a long narrow tank filled with a transparent liquid of absolute refractive index n. (The half-mirror and fixed mirror M1 are in air, with refractive index 1.)
In the initial state, the movable mirror M2 is at rest near the natural length of the spring, and the detector observes interference fringes of a certain brightness.
A bullet of mass m flies horizontally at speed v0 and embeds completely into the back of the movable mirror M2 (perfectly inelastic collision). The collision time is extremely short, so immediately after the collision, M2 (with the bullet) acquires velocity V and begins oscillating in the liquid. The resistance of the liquid is negligible.
When the movable mirror M2 oscillates, the optical path length of Path 2 changes continuously, causing the interference fringes at the detector to change periodically (fringe shift). Find the maximum frequency fmax (number of fringe changes per second) observed at the detector during this oscillation.
Assume the speed of light is incomparably larger than the speeds of the bullet and mirror, so Doppler wavelength changes are negligible. Only changes in optical path difference need to be considered.
Constraints
- Vacuum wavelength of light source: λ=5.60×10−7 m
- Absolute refractive index of liquid: n=1.25
- Mass of movable mirror: M=0.480 kg
- Spring constant: k=200 N/m
- Mass of bullet: m=0.020 kg
- Initial speed of bullet: v0=175 m/s
Input Format
Compute fmax in units of MHz (106 Hz) and give the answer as a positive integer equal to the value multiplied by 100.
Solution
1. Maximum Speed After Collision
By conservation of momentum during the extremely short collision:
mv0=(M+m)V V=M+mmv0=0.480+0.0200.020×175=0.5003.50=7.00 m/sImmediately after the collision, the combined mass (M+m) is at the center of oscillation (natural length position). In simple harmonic motion, the maximum speed occurs when passing through the center, so:
vmax=7.00 m/s(Note: The spring constant k determines the amplitude and period, but is not needed to find the maximum speed.)
2. Relationship Between Optical Path Change and Fringe Frequency
Let the distance from the half-mirror to M2 in the liquid be L2. Since Path 2 travels through liquid (refractive index n), the optical path length is 2nL2.
The optical path difference:
Δ=2nL2−2L1The fringe order N=Δ/λ changes as L2 changes. The fringe frequency f is the absolute value of the time derivative of N:
f=dtdN=λ2ndtdL2=λ2n∣v∣3. Numerical Calculation
The frequency f is maximum when the mirror speed ∣v∣=vmax=7.00 m/s:
fmax=λ2nvmax=5.60×10−72×1.25×7.00=5.60×10−717.5=3.125×107 HzConverting to MHz:
fmax=31.25 MHzThe required answer is 31.25×100=3125.
Final Answer: 3125