Ask Me Help Desk - How do I work out this q. On interference

Quote:

Q. What is the thickness of a surface coating of MgF2 whose index of refraction is n=1.38 and is designed to eliminate reflected light at wavelengths centered at 400nm when incident normally on glass from which n=1.50 (assume n(air)=1.0003)?

A 96.18 nm
B 108.7 nm
C 72.49 nm
D 99.67 nm

The wavelength is 400 nm. The light is "incident normally", in other words, it comes in at a 90°0 angle.

In interference films, the light passes through the film and then back. The path difference, in order to get interference, must be half a wavelength. If the thickness of the film is 1/2λ, the total path will be 2 x (1/2λ) = 1λ and you'll get constructive interference. To get destructive interference, the thickness needs to be half that or λ/4.

You also need to take into account that the wavelength of light is not the same as it is in vacuum.

The pertinent equation is

$\lambda_{film}=\frac {\lambda_{vacuum}}{n}$ where n is the index of refraction.

thus

$t=\frac {m\lambda}{4n}$

where n is the index of refraction of the film, m is the order of refraction (we'll choose 1).

$\lambda_{film}=\frac {400}{4 \times 1.38}=72.46\,nm$

The index of refraction of glass comes into play because it must be greater than the index of refraction of the film. The index of refraction of air (less than that of the film) indicates that there will be reflection off the top surface. This has to be taken into account in figuring out the equation because there is a λ/2 phase shift whenever there is a reflection. In this case, there are two reflections -- one at the top surface and one at the glass, so they cancel each other (I think)