Above is a schematic of a fiber optic cable, which maintains total internal reflection throughout the length of the cable. The index of refraction of the cladding of a cable is generally 1.52,while the core has an index of refraction of 1.62Choosing from the list below, what is the smallest angle of incidence light could have to enter from the acceptance cone that would be able to andlsquo;escapeandrsquo; the fiber optic cable?a)60anddeg;b)30anddeg;c)45anddeg;d)80anddeg;Correct answer is option 'B'. Can you explain this answer?

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Above is a schematic of a fiber optic cable, which maintains total internal reflection throughout the length of the cable. The index of refraction of the cladding of a cable is generally 1.52, while the core has an index of refraction of 1.62 Choosing from the list below, what is the smallest angle of incidence light could have to enter from the acceptance cone that would be able to ‘escape’ the fiber optic cable?

a)
60°
b)
30°
c)
45°
d)
80°

Correct answer is option 'B'. Can you explain this answer?

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Above is a schematic of a fiber optic cable, which maintains total int...

There are multiple ways to come to an answer. You can use the critical angle formula sin θ_c = n₂/n₁ but this requires some math gymnastics.
One important realization to make is that the larger an angle of incidence is, the more likely it will be internally reflected. The angle of incidence is measured between the ray of light and the normal of the inside of the cable, therefore in the figure, the rays coming in from outside the acceptance cone would have the lowest angle of incidence.
Since we’re looking for the smallest angle of incidence possible to escape the cable, we should simply pick 30°
The harder way, if we use the formula, would give the result of sin θ_c = 1.52/1.62. We probably would need a calculator to solve this directly, but we should know the sine of 30°, 45°, and 60° , which are the available answer choices. Additionally, we know we want to find the smallest angle, so we can focus on 30°
sin (30°) = 1/2, therefore sin (30°) is much smaller than the sin (θ_c), which is all we need to know. Remember, if an angle is smaller than the critical angle, it will not be internally reflected! Since 30° is our smallest incident angle choice, even if the other angles can ‘escape’ the cable, 30° must be the correct answer.

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Above is a schematic of a fiber optic cable, which maintains total int...

There are multiple ways to come to an answer. You can use the critical angle formula sin θ_c = n₂/n₁ but this requires some math gymnastics.
One important realization to make is that the larger an angle of incidence is, the more likely it will be internally reflected. The angle of incidence is measured between the ray of light and the normal of the inside of the cable, therefore in the figure, the rays coming in from outside the acceptance cone would have the lowest angle of incidence.
Since we’re looking for the smallest angle of incidence possible to escape the cable, we should simply pick 30°
The harder way, if we use the formula, would give the result of sin θ_c = 1.52/1.62. We probably would need a calculator to solve this directly, but we should know the sine of 30°, 45°, and 60° , which are the available answer choices. Additionally, we know we want to find the smallest angle, so we can focus on 30°
sin (30°) = 1/2, therefore sin (30°) is much smaller than the sin (θ_c), which is all we need to know. Remember, if an angle is smaller than the critical angle, it will not be internally reflected! Since 30° is our smallest incident angle choice, even if the other angles can ‘escape’ the cable, 30° must be the correct answer.

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