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A fiber-optic rod consists of a central strand of material surrounded by an outer coating. the interior portion of the rod has an index of refraction of 1.55. if all rays striking the interior walls of the rod with incident angles greater than 59.5° are subject to total internal reflection, what is the index of refraction of the coating?

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Answer:

the index of refraction of the coating is 1.33

Step-by-step explanation:

Given the data in the question;

refraction index of interior portion of the rod η
_{interior = 1.55

angle of incidence θ
_i = 59.5°

From Snell's law, we know that;

η
_{interior × sinθ
_i = η
_{coating × sinθ
_r

where η
_{interior is the index of refraction of the rod ( material 1 )

θ
_i is the angle of incidence

η
_{coating is the index of refraction in outer coating ( material 2 )

θ
_r is the angle of refraction

so we substitute our values into the equation;

η
_{interior × sinθ
_i = η
_{coating × sinθ
_r

1.55 × sin( 59.5° ) = η
_{coating × sin( 90° )

1.55 × 0.861629 = η
_{coating × 1

1.3355 = η
_{coating × 1

η
_{coating = 1.33 { 2 decimal places }

Therefore, the index of refraction of the coating is 1.33

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