Two rectangular glass plates $\mathbf{(}\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{)}$ are in contact along one edge (fig-35-45) and are separated along the opposite edge . Light with a wavelength of 600 nm is incident perpendicularly onto the top plate. The air between the plates acts as a thin film. Nine dark fringes and eight bright fringes are observed from above the top plate. If the distance between the two plates along the separated edges is increased by 600 nm, how many dark fringes will there then be across the top plate.

Order of bright fringe $m=8$

Wavelength of light in air $\lambda =600nm$

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

The wavelengths reflected from the top plate and are taken to be in phase. The condition for the bright fringes is

$2L=m\lambda$

Here, L thickness of the top plate, m order of bright fringe in an integer, and $\lambda$ is wavelength of light in air.

Rearrange for L,

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

Substitute 8 for m and 600 nm for $\lambda$.

$\begin{array}{rcl}L& =& \frac{\left(8\right)\times \left(\text{600}\text{nm}\right)}{2}\\ & =& 2400\text{nm}\end{array}$

Now the thickness is increased by 600 nm.

Therefore,

$\begin{array}{rcl}L\text{'}& =& 2400\text{nm + 600}\text{nm}\backslash \mathrm{hfill}\\ & =& 3000\text{nm}\end{array}$

Rearrange equation $2L=m\lambda$ for m,

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

Substitute 3000 nm for L and 600 nm for $\lambda$,

$\begin{array}{rcl}m& =& \frac{2\left(3000\text{nm}\right)}{600\text{nm}}\\ & =& 10\end{array}$

The number of dark fringes,

$10+1=11$

Therefore, the numbers of bright fringes are 10. The number of dark fringes will there be across the top plate is 11.