As shown in the figure, light from a medium with a refractive index of n 1 is emitted onto a thin film with a thickness of d and a refractive index of n2, and reflected at the interface of the two layers respectively. The optical path difference of two beams of light A and B reflected in a medium with refractive index of n 1 is δ = 2D. Generally, an antireflection film is a transparent crystal film plated on the interface between air and glass, so N 1 = 1 (air refractive index), n2 is the refractive index of the film, and n3 is the refractive index of the glass, so the optical path difference δ = 2d (because the A beam has a half-wave loss at the first interface and the B beam has a half-wave loss at the second interface, the total δ' = 0). Generally speaking, paraxial rays, that is, i 1 is very small and close to 0, so the optical path difference δ = 2n2d.

According to the interference principle of light:

(k = 0.1.2,3, ...), λ is the wavelength of light in air.

By selecting the thickness d of the coating and the refractive index (n2) of the medium, the optical path difference between the two reflected beams is (2k+ 1), and interference cancellation occurs, thus weakening the intensity of the reflected light and increasing the intensity of the transmitted light.

When k=0, d=

Because n2=(v is the propagation speed of light wave in thin film medium), the wavelength of light in thin film medium is =, that is, d=. Therefore, middle school textbooks emphasize that "when the film thickness is 1/4 of the wavelength of incident light in the film, the light reflected on both sides of the film ... cancels each other, greatly reducing the reflection loss of light and enhancing the intensity of projected light." However, this point was ignored by the National Science Comprehensive College Entrance Examination 16 in 2006. In the original question, "λ represents the wavelength of infrared rays" is better than "λ represents the wavelength of infrared rays in this film".

In addition to selecting the film thickness, the refractive index should also be selected. It can be proved that the refractive index n2 of thin film medium must be the middle term of the ratio of air refractive index n 1 to glass refractive index n3, that is, [1].

For example, if the refractive index of lens glass is n3= 1.50 and the refractive index of air is n 1= 1, the refractive index of its surface antireflection film is n2==, but no transparent material with such refractive index has been found yet. Among the existing optical thin film materials, magnesium fluoride (MgF2) has the smallest refractive index of 1.38. Therefore, this material was chosen in the process. If the wavelength of light wave is λ = 550 nm, its thickness is:

d=，

Then d≈ 10-7m.

Magnesium fluoride coating can reduce the reflectivity of light from 4% to 65438 0.8%. Of course, in order to improve the transmittance, a multi-film antireflection system can be used.

2. Anti-reflection film

In practical technical application, sometimes the purpose of coating is to increase the intensity of reflected light in a certain spectral region and improve reflectivity. This kind of film is called antireflection film (and reflective film). For example, the total reflector of He-Ne laser resonator is coated with 15 ~ 19 zinc sulfide-magnesium fluoride film system, which can make the reflectivity of 6328 angstrom wavelength as high as 99.6%.

In order to improve the reflectivity, it is required that the light waves reflected from both sides of the film have constructive interference, that is, the optical path difference δ = (2k) and 2n2d=(2k), so the film thickness is d = k (k = 0, 1, 2,3, ...).

The function of antireflection film and antireflection film is to change the energy distribution ratio of refracted light and reflected light. Therefore, by controlling the thickness of the film, it can be an antireflection film for some light and an antireflection film for other light. For example, sunglasses need more reflection for λ 1 = 550 nm and more transmission for λ 2 = 500 nm, which is on the glass surface (n3 = 1.5).

―――――――①

d =――②

Solution: k=5

Then from ① or ②: d=996nm.

That is to say, a magnesium fluoride film with a thickness of 996nm is plated on the glass surface to make it an antireflection film for 550nm light and an antireflection film for 500nm light.