Consider a 400 nm thick film of water (n-1.33) on plastic (-1.9). visible light is incident. on the film firom air. (25 pts) sketch this situation and clearly indicate the rays which are interfering with one another. (5 pts) a b. find three wavelengths that will be reflected brightly from the film. (14 pts) c. find three wavelengths that will be reflected very dimly, if at all, from the film.

Explanation:

For constructive interference in thin films by reflected light , the condition is

Path difference =  2μd = odd multiple of λ / 2

Where μ is refractive index of water , d is thickness of water medium and λ

is wavelength of light

λ = 4μd or 4μd / 3 or 4μd / 5

= 4 x 1.33 x 400 nm = 2128 nm

or 2128 /3 = 709.33 nm

or 2128 /5 = 425.6 nm

For destructive interference ( dim reflection )

2μd = multiple of λ = λ or 2λ or 3λ

2 x 1.33 x 400 nm =1064

or 1064 x 2 = 2128 nm

or 1064 x 3 = 3192 nm.

the answer is false

explanation:

568.98

explanation:

its math