In Fig. 35-45, a broad beam of light of wavelength 683 nm is sent directly downward through the top plate of a pair of glass plates. The plates are 120 mm long, touch at the left end, and are separated by $\mathbf{48}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathit{m}$ at the right end. The air between the plates acts as a thin film. How many bright fringes will be seen by an observer looking down through the top plate?

Wavelength $\lambda =683\text{nm}$

Separated by $L_R=48.0\mu m$

The phenomenon of several light waves interfering with one another under specific conditions causes the combined amplitudes of the waves to either grow or decrease is known as interference of light.

The thickness of the $L_R$ at the right end for bright fringes is given by following expression.

$L_R=\frac{m\lambda }{2n_2}$

Here, m is the number of fringes, $\lambda$is the wavelength, and $n_2$ is the index of refraction of the medium between the wedges.

The index of refraction $n_2$ is equal to 1 since air is the medium between the glass plates.

Rearrange the above equation $L_R=\frac{m\lambda }{2n_2}$ to m.

$m=\frac{2L_R{n}_2}{\lambda }$

Substitute 683 nm for $\lambda$, 1 for $n_2$, and $48.0\mu m$ for $L_R$in the above equation to solve for m.

$$\begin{gathered} m=\frac{2\left(48.0\mu \text{m}\right)\left(\frac{1\text{m}}{{10}^6\mu \text{m}}\right)\left(1\right)}{683\text{nm}\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)} \\ =\frac{2\left(48\times {10}^{-6}\text{m}\right)}{683\times {10}^{-9}\text{m}} \\ =\frac{96\times {10}^{-6}\text{m}}{683\times {10}^{-9}\text{m}} \\ =140 \end{gathered}$$

Thus, the number of bright fringes observed is 140.