Chapter 35: Q37P (page 1076)

The rhinestones in costume jewellery are glass with index of refraction $1.50$ . To make them more reflective, they are often coated with a layer of silicon monoxide of index of refraction $2.00$ .What is the minimum coating thickness needed to ensure that light of wavelength $560 nm$ and of perpendicular incidence will be reflected from the two surfaces of the coating with fully constructive interference?

Expert verified

The minimum thickness of coating with fully constructive interference is $70 nm$ .

Step by step solution

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 $λ = 560 nm$ .

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

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

$t = (m + \frac{1}{2}) \frac{λ}{2 n_{c}}$

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" $\begin{array}{rcl} t & = & (0 + \frac{1}{2}) \frac{(560 nm)}{2 (2)} \\ = & 70 nm \end{array}$

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

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