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A plastic film with index of refraction 1.70 is applied to the surface of a car window to increase the reflectivity and thus to keep the car's interior cooler. The window glass has index of refraction 1.52. Part A What minimum thickness is required if light of wavelength 550 nm in air reflected from the two sides of the film is to interfere constructively

Sagot :

Answer:

80.9 nm

Explanation:

The computation of the minimum thickness that needed in the case when the wave length is 550 nm

The bottom light wave should be twice to the thickness

= [tex]\lambda[/tex] ÷ 2

2t  =   [tex]\lambda[/tex] ÷ 2

[tex]\lambda[/tex] represent the light wave wavelength

As we know that

Wavelength of the light in air is

[tex]\lambda_0 = \lambda n[/tex]

Here n represent the refractive plastic index

now the minimum thickness is

= 550 nm ÷ 4( 1.70)

= 80.9 nm

okpalawalter8

We have that the  minimum thickness is mathematically given as

t = 80.9nm

From the question we are told

  • A plastic film with index of refraction 1.70 is applied to the surface of a car window
  • The window glass has index of refraction 1.52.
  • What minimum thickness is required if light of wavelength 550 nm in air reflected from the two sides of the film is to interfere constructively

Wave

Generally the equation for the constructive interference,  is mathematically given as

[tex]2t = ( m + 1/2) \lambda / n\\\\Therefore\\\\2t = \lambda / 2n\\\\t = \lambda / 4n\\\\t = \frac{550}{ 4 * 1.70}\\\\[/tex]

  • t = 80.9nm

For more information on Wave visit

https://brainly.com/question/3004869

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