A film of index of refraction n1 coats a surface with index of refraction n2.When n1 n2, the condition for destructive interference for reflected monochromatic light of wavelength $\lambda$ in air is:

Question

A film of index of refraction n1 coats a surface with index of refraction n2.When n1 > n2, the condition for destructive interference for reflected monochromatic light of wavelength $\lambda$ in air is:

Quizplus · Accepted Answer

The condition for destructive interference for reflected light from a surface with index of refraction n2, coated with a film with index of refraction n1 > n2, is given by:
2n1t = (m + 1/2)λ
where t is the thickness of the film, λ is the wavelength of the incident light, and m is an integer.
In this case, the incident light is monochromatic with wavelength λ = 500 nm = 5.00 × 10-7 m.
The film has index of refraction n1 and thickness t, and the surface has index of refraction n2. Since n1 > n2, the reflected wave undergoes a phase change of π at the boundary between the film and the surface.
The condition for destructive interference is given by:
2n1t + (n1-n2)λ/2 = mλ
Substituting the values, we get:
2n1t + (n1-n2)λ/2 = λ/2
2n1t = λ/2 - (n1-n2)λ/2
2n1t = (n2-n1/2)λ
t = (n2-n1/4)λ/n1
Substituting the values, we get:
t = (1.5-1/4)/(1.5) × 5.00 × 10-7 m
t ≈ 1.11 × 10-7 m
The thickness of the film is 1.11 × 10-7 m.
Therefore, the condition for destructive interference is:
2n1t = (m + 1/2)λ
2n1(1.11 × 10-7) = (m + 1/2)5.00 × 10-7
m = (2n1t/λ - 1/2)
Substituting the values, we get:
m ≈ 0.5
Therefore, the answer is choice C.

A Film of Index of Refraction N1 Coats a Surface λ\lambdaλ

A Film of Index of Refraction N1 Coats a Surface $\lambda$