Final answer:
The minimum thickness of the aluminum foil calculated using a Michelson interferometer and a He-Ne laser with a wavelength of 632.8 nm, with a fringe shift of 27, is 4.2764 micrometers.
Step-by-step explanation:
The minimum thickness of an aluminum anti-intrusion plate can be calculated using a Michelson interferometer and the properties of light interference.
When the movable mirror of the interferometer, which is mounted on a micrometer, clamps down on the aluminum foil, a difference of 27 fringes is observed in the interference pattern.
This fringe shift can be related to the thickness of the aluminum foil since each fringe represents a full wavelength shift of the light source, which in this case is a He-Ne laser with a wavelength of 632.8 nm (nanometers). The calculation follows this formula:
Thickness of the aluminum foil = (Number of fringes × Wavelength of light) / 2
Thickness = (27 fringes × 632.8 nm) / 2
Thickness = (27 × 632.8 nm) / 2
Thickness = 8,552.8 nm / 2
Thickness = 4,276.4 nm
Thickness = 4.2764 μm (micrometers)
Therefore, the minimum thickness of the aluminum foil measured using the Michelson interferometer and the given information is 4.2764 micrometers.