Monochromatic light of wavelength λ=136.8μ m is shone at normal incidence through a thin film of thickness t resting atop a fully reflective surface. the film has an index of refraction of nfilm=19. what is the smallest thickness t (in hm) the film could have to get constructive interference?

1.8 × 10⁻⁸ Hm

Explanation:

Given that:

The refractive index of the film = 19

The wavelength of the light = 136.8 μ m

The thickness can be calculated by using the formula shown below as:

Where, n is the refractive index of the film

is the wavelength

So, thickness is:

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

the answer is a. hope this .

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..