A light beam is traveling through an unknown substance. When it exits that substance and enters into air, the angle of reflection is 32.0°

Question

A light beam is traveling through an unknown substance. When it exits that substance and enters into air, the angle of reflection is 32.0° and the angle of refraction is 42.0°. What is the index of refraction of the substance?

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Hồng Cúc 5 years 2021-08-09T22:17:39+00:00 2 Answers 26 views 0

Answers ( )

  1. King
    0
    2021-08-09T22:18:59+00:00
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    Answer:

    0.79

    Explanation:

    Using Snell’s law, we have that:

    n(1) * sin θ1 = n(2) * sinθ2

    Where n(1) = refractive index of air = 1.0003

    θ1 = angle of incidence

    n(2) = refractive index of second substance

    θ2 = angle of refraction

    The angle of reflection through the unknown substance is the same as the angle of incidence of air. This means that θ1 = 32°

    => 1.0003 * sin32 = n(2) * sin42

    n(2) = (1.0003 * sin32) / sin42

    n(2) = 0.79

  2. Farah
    0
    2021-08-09T22:19:03+00:00
    Reply

    Answer:

    The index of refraction of the substance is 0.79

    Explanation:

    According to the second law of refraction which states that the ratio of the sine of angle of incidence (i) to the sine of angle of refraction (r) is a constant for a given pair of media. This constant is known as the refractive index (n). Mathematically;

    sin(i)/sin(r) = n

    If a light travelling through an unknown substance reflects at an angle of 32°, then the angle of incidence will also be 32° according to the second law of reflection which states that the angle of incidence is equal to the angle of reflection r’ i.e i = r’ = 32°

    Given angle of refraction r = 42°

    The index of refraction;

    n = sin32°/sin42°

    n = 0.5299/0.6691

    n = 0.79

    Therefore the index of refraction of the substance is 0.79

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