Question

In a Michelson interferometer, \(100\) fringes cross the field of view when the movable mirror is moved through \(29.48\ \mu m\). The wavelength of light used is:

• A. \(5896\ \text{\AA}\)
• B. \(5896\ \text{nm}\)
• C. \(5896\ \text{mm}\)
• D. \(2048\ \text{\AA}\)

Hint:

In Michelson interferometer, mirror displacement \(d\) produces path difference \(2d\), so \(N\lambda=2d\).

Answer 1

The Correct Option is A

Concept:
In Michelson interferometer, when the mirror is moved by a distance \(d\), optical path difference changes by \(2d\). \[ N\lambda=2d \]

Step 1: Write the given values. \[ N=100 \] \[ d=29.48\ \mu m \]

Step 2: Apply Michelson formula. \[ N\lambda=2d \] \[ 100\lambda=2\times 29.48\ \mu m \] \[ 100\lambda=58.96\ \mu m \] \[ \lambda=0.5896\ \mu m \]

Step 3: Convert into Angstrom. \[ 1\ \mu m=10^4\ \text{\AA} \] \[ 0.5896\ \mu m=0.5896\times10^4\ \text{\AA} \] \[ \lambda=5896\ \text{\AA} \] \[ \therefore \text{Correct Answer is (A)} \]