Question

A schematic of a Michelson interferometer, used for the measurement of refractive index of gas, is shown in the figure. The transparent chamber is filled with a gas of refractive index $n_g$, where $n_g \neq 1$, at atmospheric pressure. If a 532 nm laser beam produces 30 interference fringes on the screen, then the number of fringes produced by a 632.8 nm laser beam will be ________ (rounded off to one decimal place). Note: Assume that the effect of beamsplitter width is negligible. The setup is placed in air medium with refractive index equal to 1.

 

Hint:

The number of interference fringes produced in a Michelson interferometer is directly proportional to the optical path difference and inversely proportional to the wavelength of the light source.

Answer 1

Step 1: Determine the change in optical path difference ($\Delta L$).
The change in optical path length due to the gas in the 2 cm chamber (traversed twice) is $\Delta L = 2 \times 2 \times (n_g - 1) = 4(n_g - 1)$ cm.
Step 2: Relate the number of fringes to the change in optical path difference and wavelength for the first laser.
$\Delta N_1 = \frac{\Delta L}{\lambda_1} \implies 30 = \frac{4(n_g - 1)}{532 \times 10^{-9} \, \text{m}}$
$\Delta L = 30 \times 532 \times 10^{-9} \, \text{m} = 15960 \times 10^{-9} \, \text{m}$
Step 3: Calculate the number of fringes for the second laser.
$\Delta N_2 = \frac{\Delta L}{\lambda_2} = \frac{15960 \times 10^{-9} \, \text{m}}{632.8 \times 10^{-9} \, \text{m}} = \frac{15960}{632.8} \approx 25.22124$
Step 4: Round off the result to one decimal place.
$\Delta N_2 \approx 25.2$