Question

A thin flake of mica (n = 1.58) is used to cover one slit of a double-slit interference arrangement. The central point on the viewing screen is now occupied by what had been the seventh bright side fringe (m = 7). If $\mathit{\lambda }\mathbf{=}\mathbf{550}\mathbf{}\mathbf{\text{nm}}$ , what is the thickness of the mica?

Short answer

The thickness of mica is $6.64\mu \text{m}$.

Step-by-step solution

Identification of given data

The index of refraction for mica is $n=1.58$

The order of seventh fringe is $m=7$

The wavelength of light is $\lambda =550\text{nm}$

The thickness of the mica is found by using the formula for thin film interference for bright fringes.

Determination of thickness of the mica

The thickness of the mica is given as:

$t=\frac{m\lambda }{n-1}$

Substitute all the values in equation.

$\begin{array}{rcl}t& =& \frac{\left(7\right)\left(550\text{nm}\right)}{1.58-1}\\ t& =& 6.64\times {10}^{-6}\text{m}\\ t& =& \left(6.64\times {10}^{-6}\text{m}\right)\left(\frac{1\mu \text{m}}{{10}^{-6}\text{m}}\right)\\ t& =& 6.64\mu \text{m}\end{array}$

Therefore, the thickness of mica is $6.64\mu \text{m}$.