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An oil film floats on a water surface. The indices of refraction for water and oil, respectively, are 1.333 and 1.466. If a ray of light is incident on the air-to-oil surface at an angle of 37.0º with the normal, what is the angle of the refracted ray in the water?

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I18ntech

According to the Snell's law,

[tex] \frac{n_{1}}{n_{2}}= \frac{Sinθ₂}{Sinθ₁}[/tex]

Here, n₁= Refractive index of oil=1.466

n₂=Refractive index of water=1.333

θ₁=Angle of incident= 37.0°

θ₂= Angle of refraction

Substituting the value,

[tex] \frac{1.466}{1.33} =\frac{Sinθ₂}{Sin37.0°} [/tex]

Solving this we get, θ₂ =26.8°

∴ The angle of refracted ray in water is 26.8°.

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