Question

A piece of glass of index of refraction 1.50 is coated with a thin layer of magnesium fluoride of index of refraction 1.38. It is illuminated with light of wavelength 680 nm. Determine the minimum thickness of the coating that will result in no reflection

Answer 1

Answer:

123.18 nm

Explanation:

Given:

• Refractive index of the glass=1.5
• Index of magnesium fluoride=1.38
• Wavelength of the light =680 nm

Let t be the thickness of the film. The relation between the wavelength of light  in film and thickness of the film  in order to have no reflection is given  by

$2t=\dfrac{\lambda_{film}}{2}\\t=\dfrac{\lambda_{film}}{4}$

Now According to question we have

$\lambda_{film}=\dfrac{\lambda_{film}}{n}$

where n is the refractive index of the film.

$\lambda_{film}=\dfrac{680}{1.38}\\\lambda_{film}=492.75\ nm$

Now thickness t is given by

$t=\dfrac{492.75}{4}\\t=123.18\ \rm nm$

Hence the thickness of the film is calculated in order to have no reflection.