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Sophie [7]
2 years ago
11

Monochromatic light of wavelength λ=136.8μ m is shone at normal incidence through a thin film of thickness t resting atop a full

y 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?
Physics
1 answer:
inn [45]2 years ago
7 0

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:

Thickness=\frac {\lambda}{4\times n}

Where, n is the refractive index of the film

{\lambda} is the wavelength

So, thickness is:

Thickness=\frac {136.8\ \mu\ m}{4\times 19}

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

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