A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the circle from wh

Question

A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the circle from which light escapes from the liquid into the air above the surface

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Bơ 4 years 2021-08-10T12:56:02+00:00 1 Answers 14 views 0

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  1. dangkhoa
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    2021-08-10T12:57:23+00:00
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    Answer:

    The radius is  r = 3.1905 \ m

    Explanation:

    From the question we are told that

            The  distance  beneath the liquid  is  d = 2.70 \ m

            The refractive index of the liquid is  n_i = 1.31

    Now the critical value is mathematically represented as

             \theta = sin ^{-1} [\frac{1}{n_i} ]

    substituting values

             \theta = sin ^{-1} [\frac{1}{131} ]

             \theta = 49.76^o

    Using SOHCAHTOA rule we have that

             tan \theta = \frac{ r}{d}

    =>     r = d * tan \theta

    substituting values  

            r = 2.7 * tan (49.76)

            r = 3.1905 \ m

             

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