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6 August, 17:07

A beam of light is incident from air on the surface of a liquid. If the angle of incidence is 25.5° and the angle of refraction is 15.6°, find the critical angle for the liquid when surrounded by air.

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  1. 6 August, 17:32
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    Фc = 38.66°

    Explanation:

    from Snell's law:

    n1*sin (Ф1) = n2*sin (Ф2)

    for air, n1 = 1.0

    1.0*sin (Ф1) = n2*sin (Ф2)

    n2 = 1.0*sin (Ф1) / sin (Ф2)

    n2 = 1.0*sin (25.5) / sin (15.6) = 1.601

    For the critical angle: Ф2 = 90° and n2 = 1.601

    n2*sin (Фc) = n1*sin (Ф2) = n1

    sin (Фc) = n1 / (n2*sin (Ф2))

    sin (Фc) = (1.0) / (1.601*sin (90))

    sin (Фc) = 0.625

    Фc = 38.66°

    Therefore, the critical angle is 38.66°
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