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Difficulty: MediumRefraction of Light, Total Internal Reflection, and Prisms

An optical fiber consists of a core with a refractive index of 1.501.50 surrounded by a cladding with a refractive index of 1.201.20. Which of the following correctly describes the direction a light ray must travel for total internal reflection to take place at the boundary, as well as the sine of the critical angle?

  1. From core to cladding with sin⁡C=0.80\sin C = 0.80Answer
  2. B
    From cladding to core with sin⁡C=0.80\sin C = 0.80
  3. C
    From core to cladding with sin⁡C=1.25\sin C = 1.25
  4. D
    From cladding to core with sin⁡C=1.25\sin C = 1.25

Answer

Light must travel from the core to the cladding with sin⁡C=0.80\sin C = 0.80
Total internal reflection takes place only when light travels from an optically denser medium (n1=1.50n_1 = 1.50) to an optically less dense medium (n2=1.20n_2 = 1.20). Applying Snell's law at the critical boundary gives sin⁡C=n2n1=1.201.50=0.80\sin C = \frac{n_2}{n_1} = \frac{1.20}{1.50} = 0.80.

Step-by-Step Solution

1
Determine the direction requirement for Total Internal Reflection (TIR)
Light must travel from the denser medium (core, n1=1.50n_1 = 1.50) toward the less dense medium (cladding, n2=1.20n_2 = 1.20).
TIR only occurs when light attempts to pass into a medium of lower optical density so that the refracted ray bends away from the normal.
2
Calculate the sine of the critical angle CC
\sin C = \frac{n_2}{n_1} = \frac{1.20}{1.50} = 0.80
By Snell's law at the critical angle, n1sin⁡C=n2sin⁡90∘  ⟹  sin⁡C=n2n1n_1 \sin C = n_2 \sin 90^\circ \implies \sin C = \frac{n_2}{n_1}.

Key Concept

Total Internal Reflection and Critical Angle
Estimated Time:1m 30s
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Topics

Refraction of Light, Total Internal Reflection, and Prisms

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