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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?

Answer :

Answer:

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:

[tex]Thickness=\frac {\lambda}{4\times n}[/tex]

Where, n is the refractive index of the film

[tex]{\lambda}[/tex] is the wavelength

So, thickness is:

[tex]Thickness=\frac {136.8\ \mu\ m}{4\times 19}[/tex]

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

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