In Fig. 35-45, a broad beam of monochromatic light is directed perpendicularly through two glass plates that are held together at one end to create a wedge of air between them. An observer intercepting light reflected from the wedge of air, which acts as a thin film, sees 4001 dark fringes along the length of the wedge. When the air between the plates is evacuated, only 4000 dark fringes are seen. Calculate to six significant figures the index of refraction of air from these data.

Monochromatic lights are single-wavelength light, where mono refers to single, and chrome means color. Visible light of a narrow band of wavelengths is classified as monochromatic lights. It features a wavelength within a short wavelength range.

The expression for the minima condition for normal incidence in the case of thin films is,

$2L=m\frac{\lambda }{n}$

Here, L is thickness, m is order, $\lambda$ is wavelength, and n is refractive index of medium.

In case of 4001 dark fringe with index of right in air $n_{air}$, the condition of minima

$2L=\left(4001\right)\frac{\lambda }{n_{air}}$

In the case of 4000 dark fringe with index c vacuum 1.0 the condition for minima:

$2L=\left(4000\right)\frac{\lambda }{1.0}$

For both conditions the left side of the equation is 2L.

Thus the equation right hand side and solve for $n_{air}$.

$\begin{array}{rcl}\left(4001\right)\frac{\lambda }{n_{air}}& =& \left(4000\right)\frac{\lambda }{1.0}\\ n_{air}& =& 1.00025\end{array}$

Hence, the refractive index of air is $1.00025$.