Solution to 1992 Problem 96

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.
\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).

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