Question

When the movable arm of a Michelson interferometer in vacuum \( (n = 1) \) is moved by 325 \(\mu m\), the number of fringe crossings is 1000. The wavelength of the laser used in nanometers is \(\underline{\hspace{2cm}}\).

Hint:

In a Michelson interferometer, the number of fringe crossings is related to the distance moved and the wavelength of the light used.

Answer 1

Correct Answer: 650

The relationship between the number of fringe crossings \( N \), the distance moved \( d \), and the wavelength \( \lambda \) is given by: \[ N = \frac{2d}{\lambda} \] where \( d = 325 \, \mu m = 325 \times 10^{-6} \, m \). Given \( N = 1000 \), we can solve for \( \lambda \): \[ 1000 = \frac{2 \times 325 \times 10^{-6}}{\lambda} \] \[ \lambda = \frac{2 \times 325 \times 10^{-6}}{1000} = 650 \times 10^{-9} \, \text{m} = 650 \, \text{nm} \] Thus, the wavelength of the laser is \( 650 \, \text{nm} \).