Question

A sheet of glass has an index of refraction n_g = 1.50. Assume that the index of refraction for air is n_a = 1.00.

a. Monochromatic light is incident on the glass sheet, as shown in the figure below, at an angle of incidence of 60°. On the figure, sketch the path the light takes the first time it strikes each of the two parallel surfaces. Calculate and label the size of each angle (in degrees) on the figure, including angles of incidence, reflection, and refraction at each of the two parallel surfaces shown.

b. Next a thin film of material is to be tested on the glass sheet for use in making reflective coatings. The film has an index of refraction n f below. It is observed that at a point where the light is incident on the film, light reflected from the surface = 1.38. White light is incident normal to the surface of the film as shown appears green (λ = 525 nm).

i. What is the frequency of the green light in air?

ii. What is the frequency of the green light in the film?

iii. What is the wavelength of the green light in the film?

iv. Calculate the minimum thickness of film that would produce this green reflection.

Answer 1

(a)

n_i sin θ_i = n_r sin θ_r … (1) sin (60) = 1.5 sin θ_g … θ_g = 35.3°

(b) i) c = f λ … 3x10⁸ = f (5.25x10^–7) … f = 5.71x10¹⁴ Hz.

ii) frequency does not change when it enters the film, same as air … f = 5.71x10¹⁴ Hz

iii) n₁ λ₁ = n₂ λ₂ … n_air λ_air = n_film λ_film … (1)(525) = (1.38) λ_film … λ_film = 380 nm.

iv) To see the green reflection, you would need to have constructive interference from the film. As the light travels from the air to film to glass, it undergoes a ½ λ phase shift at each boundary. These two phase shifts essentially cancel each other out to get zero phase shift from the flips. To produce constructive interference, a total of one wavelength needs to be covered from traveling in the film. This requires a film thickness of ½ of the wavelength in the film. t = ½ λ_film = ½ (380) = 190 nm.