An optical fiber is a flexible, transparent fiber devised to transmit light between the two ends of the fiber. Complete transmission of light is achieved through total internal reflection. This problem aims to calculate the minimum index of refraction n of the optical fiber necessary to obtain total internal reflection for every possible incidence angle.
(a) Express sinθ, where the angle θ is defined in the figure above, in terms of the incidence angle α and the index of refraction n of the optical fiber. Evaluate this function for n=1.5 and α=70∘. Take the index of refraction of air to be 1.
(b) The condition on n for total internal reflection of all beams entering the fiber is achieved when θ=90∘ for all values of α. Determine the smallest value of n that satisfies that condition.
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(a) To express sinθ in terms of the incidence angle α and the index of refraction n, we can use Snell's law. Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction.
Mathematically, Snell's law can be written as:
n₁ * sin α = n₂ * sin θ
In this case, n₁ is the index of refraction of air (1) and n₂ is the index of refraction of the optical fiber (n). We want to solve this equation for sin θ.
Rearranging the equation, we get:
sin θ = (n₁ / n₂) * sin α
Substituting the values given, n₁=1, n₂=1.5, and α=70°, we can calculate sin θ:
sin θ = (1 / 1.5) * sin 70°
Using a calculator, we can find that sin θ ≈ 0.9117.
(b) To achieve total internal reflection for all possible incidence angles, we need θ to be equal to 90°. In this case, we want to find the minimum value of n that satisfies this condition.
Using the equation sin θ = (n₁ / n₂) * sin α, we can substitute sin θ = 1, sin α = 1 (since α can be any angle), and n₁ = 1 (index of refraction of air). Solving for n₂, we get:
1 = (1 / n₂) * 1
This simplifies to:
n₂ = 1
Therefore, the smallest value of n that satisfies the condition θ = 90° for all α is n = 1.