Question

A coating is being applied to reduce the reflectivity of a pane of glass to light with a wavelength of 522 nm incident near the normal to the pane. If the material has an index of refraction of 1.375, how thick should the coating be?

Answer 1

Answer:

  t = 94.91 nm

Explanation:

given,

wavelength of the light = 522 nm

refractive index of the material  = 1.375

we know the equation

       c = ν λ

where ν is the frequency of the wave

           c is the speed of light

   $\nu= \dfrac{c}{\nu\lambda}$

   $\nu = \dfrac{3\times 10^8}{522 \times 10^{-9}}$

       ν = 5.75 x 10¹⁴ Hz

the thickness of the coating will be calculated using

        $t = \dfrac{\lambda}{4\mu_{material}}$

        $t = \dfrac{522 \times 10^{-9}}{4\times 1.375}$

              t = 94.91 nm

the thickness of the coating will be equal to t = 94.91 nm