Consider a ray incident on the end of a fiber-optic cable as illustrated in the figure below. In the experiment, you will explore the behavior of the fiber as a function of the angle of incidence a. When a ray strikes the end of the fiber it is bent toward the normal. By Snell's law, the angle inside the core is sin core (no/n₁) sin a. Once inside the core, the ray travels until it strikes the cladding of the fiber. The angle of incidence at the cladding i is the complementary angle of core. If i is a small angle, the ray will propagate into the cladding and be lost from the fiber. If i is large, however, the ray will be internally reflected and bounce down the fiber. By Snell's law, the critical angle for the ray to be internally reflected is sin icrit = n₂/n₁. In the experiment, you cannot directly measure angles inside the fiber but you can measure a. Using the facts that, (i) for complementary angles, sin core = cos i, (ii) the trigonometric identity cos i = √I sin² i, and (iii) the index of refraction of air is no = 1, it can be shown that sin acrit = √√²-n². This quantity is known as the numerical aperture (N.A.) of the fiber. For a cable with a core that has an index of refraction of 1.5 and a cladding with an index of refraction of 1.25, what do you expect the critical angle acrit to be (in degrees)? 40.8 Cladding → Core → ecore refracted reflected
Consider a ray incident on the end of a fiber-optic cable as illustrated in the figure below. In the experiment, you will explore the behavior of the fiber as a function of the angle of incidence a. When a ray strikes the end of the fiber it is bent toward the normal. By Snell's law, the angle inside the core is sin core (no/n₁) sin a. Once inside the core, the ray travels until it strikes the cladding of the fiber. The angle of incidence at the cladding i is the complementary angle of core. If i is a small angle, the ray will propagate into the cladding and be lost from the fiber. If i is large, however, the ray will be internally reflected and bounce down the fiber. By Snell's law, the critical angle for the ray to be internally reflected is sin icrit = n₂/n₁. In the experiment, you cannot directly measure angles inside the fiber but you can measure a. Using the facts that, (i) for complementary angles, sin core = cos i, (ii) the trigonometric identity cos i = √I sin² i, and (iii) the index of refraction of air is no = 1, it can be shown that sin acrit = √√²-n². This quantity is known as the numerical aperture (N.A.) of the fiber. For a cable with a core that has an index of refraction of 1.5 and a cladding with an index of refraction of 1.25, what do you expect the critical angle acrit to be (in degrees)? 40.8 Cladding → Core → ecore refracted reflected
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Transcribed Image Text:Consider a ray incident on the end of a fiber-optic cable as illustrated in the figure below. In the experiment, you will explore the behavior of the fiber as a function of the angle of incidence α. When a ray strikes the end of the fiber it is bent toward the normal. By Snell's law, the angle inside the core is
sin θ_core = (n₀/n₁) sin α.
Once inside the core, the ray travels until it strikes the cladding of the fiber. The angle of incidence at the cladding i is the complementary angle of θ_core. If i is a small angle, the ray will propagate into the cladding and be lost from the fiber. If i is large, however, the ray will be internally reflected and bounce down the fiber. By Snell's law, the critical angle for the ray to be internally reflected is
sin i_crit = n₂/n₁.
In the experiment, you cannot directly measure angles inside the fiber but you can measure α. Using the facts that, (i) for complementary angles,
sin θ_core = cos i,
(ii) the trigonometric identity
cos i = √(1 - sin² i),
and (iii) the index of refraction of air is n₀ = 1, it can be shown that
sin α_crit = √(n₁² - n₂²).
This quantity is known as the numerical aperture (N.A.) of the fiber. For a cable with a core that has an index of refraction of 1.5 and a cladding with an index of refraction of 1.25, what do you expect the critical angle α_crit to be (in degrees)?
40.8
**Diagram Explanation:**
The diagram shows the structure of a fiber-optic cable with two main parts: the core and the cladding. A light ray enters the core at an angle α. As it enters, it refracts and bends towards the normal, traveling through the core with an angle θ_core. The ray then reaches the boundary between the core and cladding, where it encounters the critical angle i. If this angle is greater than i_crit, the ray reflects internally and continues to bounce down the fiber. Otherwise, it refracts out into the cladding and is lost.
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