A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass i

Question

A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of \theta_{a} = 36.2°, the ray refracted into the water makes an angle of 49.8° with the normal to the interface. What is the smallest value of the incident angle \theta_{a} for which none of the ray refracts into the water?

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Kim Cúc 5 years 2021-07-24T03:43:56+00:00 2 Answers 14 views 0

Answers ( )

  1. Acacia
    0
    2021-07-24T03:45:04+00:00
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    Answer:

    Explanation:

    Using Snell’s law

    n₁ sinθ₁ = n₂ sinθ₂

    n₁ = refractive index of the first medium ( glass)

    n₂ = refractive index of the second medium ( water)

    θ₁ = 36.2°, θ₂ = 49.8° refractive index of water = 1.333

    n₁ sin 36.2 = 1.333 sin 49.8

    n₁ = 1.333 sin 49.8° / sin 36.2° = 1.724

    the smallest angle of incidence for which none of the ray refracts into the water is the critical angle which is just the angle in which there is an increase to the normal will cause total internal reflection

    n₁  sin citical angle = 1.333 sin 90° where the refracted angle for this is 90°

    sin critical angle = 1.3333 / 1.724 = 0.7732

    critical angle = sin⁻¹0.7732 = 50.64°

  2. thachthao
    0
    2021-07-24T03:45:39+00:00
    Reply

    Answer:

    The smallest value of the incident angle is 50.8°

    Explanation:

    According the law of refraction:

    n_{1} sin\theta _{1} =n_{2} sin\theta _{2}

    Where

    n₁ = index of refraction glass

    n₂ = index of refraction water = 1.333

    θ₁ = 36.2°

    θ₂ = 49.8°

    The index of refraction of the glass is:

    n_{1} =\frac{n_{2}sin\theta _{2} }{sin\theta _{1} } =\frac{1.333*sin49.8}{sin36.2} =1.72

    The smallest value of the incident angle is:

    \theta =sin^{-1} \frac{n_{2} }{n_{1} } =\frac{1.333}{1.72} =50.8

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