Final answer:
Theanswer is Option C, index of refraction of the coating should be 2.0 to cancel 600-nm light that hits the coated surface.
Step-by-step explanation:
The index of refraction of the coating must cancel 600-nm light that hits the coated surface at normal incidence. We can use the formula for destructive interference, which is given by:
n⁻¹λ = 2d
Where n is the index of refraction of the coating, λ is the wavelength of light, and d is the thickness of the coating. Rearranging the formula, we get:
n = 2d/λ
Substituting the values, we find:
n = 2 * 600 nm / 600 nm = 2.0
Therefore, the index of refraction of the coating should be 2.0 to cancel 600-nm light that hits the coated surface.
However, if we consider a scenario where the film is quarter-wavelength thick, then the refractive index that would lead to destructive interference for 600-nm light can be determined. For a quarter-wavelength film, we would look for a condition where the film's optical thickness is 600 nm / 4 = 150 nm. The index of refraction for this scenario would be n = λ/4t, where λ is the vacuum wavelength of the light, and t is the thickness of the film.