A beam of light starts at a point 8.0 cm beneath the surface of a liquid and strikes the surface 7.0 cm from the point directly above the li

Question

A beam of light starts at a point 8.0 cm beneath the surface of a liquid and strikes the surface 7.0 cm from the point directly above the light source. Total internal reflection occurs for the beam. What can be said about the index of refraction of the liquid?

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Dulcie 6 months 2021-07-26T08:56:08+00:00 1 Answers 6 views 0

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    2021-07-26T08:58:06+00:00

    Answer:

    Explanation:

    Given that,

    The light starts at a point 8cm beneath the surface

    It strikes the surface 7cm directly above the light

    The angle of incidence from the given distances is

    Using trigonometry

    Tan I = opp / adj

    tan I= (7.0 cm)/(8.0 cm) = 0.875

    I = arctan(0.875)

    I = 41.1859

    According to Snell’s law

    n(water) sin I = n(air) sin R

    Here R =90° for total reflection to occur

    nair = 1

    n(water) sin I = n(air) sinR

    n(water) = (1)sin 90

    n(water) = 1

    n(water) = 1/sin I

    n(water) =1/ sin(41.1859o)

    = 1.52

    The refractive index index of the liquid is 1.52

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