Question

The rhinestones in costume jewellery are glass with index of refraction $\mathbf{1}.\mathbf{50}$. To make them more reflective, they are often coated with a layer of silicon monoxide of index of refraction $\mathbf{2}\mathbf{.}\mathbf{00}$.What is the minimum coating thickness needed to ensure that light of wavelength $\mathbf{560}\mathbf{}\mathbf{nm}$ and of perpendicular incidence will be reflected from the two surfaces of the coating with fully constructive interference?

Short answer

The minimum thickness of coating with fully constructive interference is $70\text{nm}$.

Step-by-step solution

Identification of given data

The index of refraction of silicon monoxide layer is$n_c=2.00$

The index of refraction for glass is $n_g=1.50$.

The wavelength of light is $\lambda =560\text{nm}$.

Understanding the concept

The index of refraction is the ratio of the speed of light in medium to speed of light in vacuum

Determination of minimum thickness of coating with fully constructive interference

The minimum thickness of coating with fully constructive interference is given as:

$t=\left(m+\frac{1}{2}\right)\frac{\lambda }{2n_c}$

Here, $m$is order of fringe and its value for minimum thickness for constructive interference is 0.

Substitute all the values in equation.

role="math" localid="1663135130327" $\begin{array}{rcl}t& =& \left(0+\frac{1}{2}\right)\frac{\left(560\text{nm}\right)}{2\left(2\right)}\\ & =& 70\text{nm}\\ & & \end{array}$

Therefore, the minimum thickness of coating with fully constructive interference is $70\text{nm}$.