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You are designing a thin transparent reflective coating for the front surface of a sheet of glass. The index of refraction of the glass is 1.52, and when it is in use the coated glass has air on both sides. Because the coating is expensive, you want to use a layer that has the minimum thickness possible, which you determine to be 104 nm. What should the index of refraction of the coating be if it must cancel 550 nm light that hits the coated surface at normal incidence?

Respuesta :

Answer:

The index of refraction of the coating  is 1.32.

Explanation:

Given that,

Index of refraction of the glass = 1.52

Thickness = 104 nm

Length = 550 nm

We need to calculate the index of refraction of the coating

Using formula of index

[tex]n = \dfrac{L}{4t}[/tex]

Where, L = length

t = thickness

Put the value into the formula

[tex]n=\dfrac{550}{4\times104}[/tex]

[tex]n=1.32[/tex]

Hence, The index of refraction of the coating  is 1.32.

onyebuchinnaji

The index of refraction of the coating that hits the coated surface at normal incidence is 1.32.

The given parameters;

  • refractive index of the coating = 1.52
  • thickness of the coating, t = 104 nm
  • length of light that must be cancelled, L = 550 nm

The index of refraction of the coating that hits the coated surface at normal incidence is calculated as follows;

[tex]n = \frac{L}{4t} \\\\n = \frac{550}{4 \times 104} \\\\n = 1.32[/tex]

Thus, the index of refraction of the coating that hits the coated surface at normal incidence is 1.32.

Learn more here:https://brainly.com/question/17163951

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