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A ray of light of frequency 5.09 x 10^14hertz is incident on a water-air interface as shown in the diagram below.

1) Calculate the angle of refraction of the light ray in air for the situation shown. [Show all work, including the equation and substitution with units.]

2) For the situation shown, calculate the speed of the light while in the water. [Show all work, including the equation and substitution with units.]

A Ray Of Light Of Frequency 509 X 1014hertz Is Incident On A Waterair Interface As Shown In The Diagram Below 1 Calculate The Angle Of Refraction Of The Light R class=

Sagot :

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The angle of refraction of the light ray in air is 58.75⁰.

The speed of the light while in the water is 2.25 x 10⁸ m/s.

Angle of refraction of the light ray

The angle of refraction can be determined by applying Snell's law as shown below;

n₁sinθ₁ = n₂sinθ₂

where;

  • n₁ is refractive index of water
  • n₂ is refractive index of air
  • θ₁ is the angle incidence
  • θ₂ is the angle of refraction

1.33 x sin(40) = 1 x sinθ₂

0.855 = sinθ₂

θ₂ = sin⁻¹(0.855)

θ₂ = 58.75⁰

Speed of the light in water

sinθ₁/v₁ = sinθ₂/v₂

where;

  • v₂ is speed of light in air

sin(40)/v₁ = sin(58.75)/(3 x 10⁸)

sin(40)/v₁ = 2.85 x 10⁻⁹

v₁ = sin(40) / (2.85 x 10⁻⁹)

v₁ = 2.25 x 10⁸ m/s

Learn more about angle of refraction here: https://brainly.com/question/15838784

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