Antireflection coatings can be used on the inner surfaces of eyeglasses to reduce the reflection of stray light into the eye, thus reducing eyestrain. part a) a 90-nm-thick coating is applied to the lens. what must be the coating's index of refraction to be most effective at 480 nm? assume that the coating’s index of refraction is less than that of the lens. n = .part b) if the index of refraction of the coating is 1.38, what thickness should the coating be so as to be most effective at 480 ? the thinnest possible coating is best. t = nm .

a) n = 1.33, b)  t = 87 10⁻⁹ m

Explanation:

Part A.

The thin anti-reflection film should create destructive interference for the desired wavelength.

      2t sin θ = (m + ½) λₙ

The 2 goes out of the path of the beam inside the film, lam is the wavelength modulated by the refractive index of the film

      λₙ =  λ₀ / n

      2t = (m + ½) λ₀ / n

      n = (m+ ½) λ₀/2t

Suppose we are in the first inference m = 0

     n = (½ 480 10-9)/ 2 90

     n = 1.33

Part B

We do not know the refractive index of the glass explicitly, but in general it is of the order of 1.5, so since the index of the film is lower, there is no phase change.

      2t = (m + ½) λ₀ / n

      t = 1/4 480 10⁻⁹ / 1.38

      t = 87 10⁻⁹ m

v=ir = 3x15 =45 v

conservation of mass

mass before something happens = mass afterwards