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11 November, 03:05

A green light is submerged 2.90 m beneath the surface of a liquid with an index of refraction 1.35. What is the radius of the circle from which light escapes from the liquid into the air above the surface?

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  1. 11 November, 03:11
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    radius is 3.197 m

    Explanation:

    Given data

    index of refraction = 1.35

    submerged h = 2.90 m

    to find out

    radius of the circle

    solution

    we know here that light will be escape only if angle of critical is less than angle of incidence

    so we can say

    sin (θ) i = 1 / index of refraction

    here angle (θ) i is critical angle so

    sin (θ) i = 1 / 1.35 = 0.7407

    (θ) i = 47.79 degree

    we consider here radius r of circle

    so

    tan (θ) i = R / h

    so

    R = 2.90 tan (47.79)

    R = 3.197 m

    so radius is 3.197 m
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