$40$ fringes corresponds to a phase difference of $80 \pi$ . This must be the phase difference between light that travels $10 \mbox{ cm}$ in vacuum and light that travels $10 \mbox{ cm}$ in the gas. Recall that the wavelength of light is reduced a factor of the index of refraction when the light travels through a transparant medium.
Therefore,
$\begin{align*}\frac{2 \pi}{\lambda/n} \cdot 10 \cdot 10^{-2} \mbox{ m} - \frac{2 \pi}{\lambda} \cdot 10 \cdot 10^{-2} \mbox{ ...$
Solving for $n$ , we find that
$\begin{align*}n = \frac{40 \lambda + 10 \cdot 10^{-2} \mbox{ m}}{10 \cdot 10^{-2} \mbox{ m}} = \boxed{1.0002}\end{align*}$
Therefore, the correct answer is (C).

Solution to 1992 Problem 96