Question

"A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.33. Wavelengths of 479 nm and 798 nm and no wavelengths between are intensified in the reflected beam. The thickness of the film is"

Answer 1

t = 8.98 10⁻⁷ m

Step-by-step explanation:

This is an exercise in interference by reflection, let's analyze what happens on each surface of the film.

* When the light ray shifts from a medium with a lower refractive index to a medium with a higher refractive index, the reflected ray has a reflection of 180

* The beam when passing to the middle its wavelength changes

λ = λ₀ / n

if we take this into account, the constructive interference equation for normal incidence is

2t = (m + ½) λ₀ / n

let's apply this equation to our case

for λ₀ = 479 nm = 479 10⁻⁹ m

t = (m + ½) 479 10⁻⁹ / 1.33

(m + ½) = 1.33 t / 479 10⁻⁹

for λ₀ = 798 nm = 798 10⁻⁹ m

t = (m' + ½) 798 10⁻⁹ /1.33

(m' + ½) = 1.33 t / 798 10⁻⁹

as they tell us that no other constructive interference occurs between the two wavelengths, the order of interference must be consecutive, let's write the two equat⁻ions

(m + ½) = 1.33 t / 479 10⁻⁹

((m-1) + ½) = 1.33 t / 798 10⁻⁹

(m + ½) = 1.33 t / 798 10⁻⁹ +1

resolve

1.33 t / 479 10⁻⁹ = 1.33 t / 798 10⁻⁹ +1

1.33 t / 479 10⁻⁹ = (1.33t + 798 10⁻⁹) / 798 10⁻⁹

1.33t = (1 .33t + 798 10⁻⁹) 479/798

1.33t = (1 .33t + 798 10⁻⁹) 0.6

1.33 t = 0.7983 t + 477.6 10⁻⁹

t (1.33 - 0.7983) = 477.6 10⁻⁹

t = 477.6 10⁻⁹ /0.5315

t = 8.98 10⁻⁷ m