For the optical fiber shown in the figure, find the minimum angle of incidence (0,) that will result in total internal reflection if the refractive index for the cladding (n2) is 1.40 and the refractive index of the core is (A.) n = 1.57 and (B.) n = 1.71. Cone of acceptance Core Cladding (n.) A. В. 0, 2 Al

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Chapter1: Units, Trigonometry. And Vectors
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**Optical Fiber Problem Statement:**

For the optical fiber shown in the figure, find the minimum angle of incidence (\( \theta_i \)) that will result in total internal reflection if the refractive index for the cladding (\( n_2 \)) is 1.40. Determine the angle for the following refractive indices of the core:

- (A) \( n_1 = 1.57 \)
- (B) \( n_1 = 1.71 \)

**Diagram Explanation:**

The diagram illustrates the path of light through an optical fiber, highlighting the core and cladding. 

- The core, where total internal reflection occurs, has a refractive index of \( n_1 \).
- The surrounding cladding has a refractive index of \( n_2 = 1.40 \).
- Light is incident at an angle \( \theta_i \) represented by the angle between the incoming ray and the normal at the boundary.
- The light travels in a zigzag pattern within the core, reflecting off the boundary due to total internal reflection.
- The cone of acceptance illustrates the range of angles for which light will successfully propagate through the fiber.

**Blank Fields:**

- A. \( \theta_i \ge \) _______ °
- B. \( \theta_i \ge \) _______ °

The task is to calculate the minimum values of \( \theta_i \) for conditions A and B that allow total internal reflection.
Transcribed Image Text:**Optical Fiber Problem Statement:** For the optical fiber shown in the figure, find the minimum angle of incidence (\( \theta_i \)) that will result in total internal reflection if the refractive index for the cladding (\( n_2 \)) is 1.40. Determine the angle for the following refractive indices of the core: - (A) \( n_1 = 1.57 \) - (B) \( n_1 = 1.71 \) **Diagram Explanation:** The diagram illustrates the path of light through an optical fiber, highlighting the core and cladding. - The core, where total internal reflection occurs, has a refractive index of \( n_1 \). - The surrounding cladding has a refractive index of \( n_2 = 1.40 \). - Light is incident at an angle \( \theta_i \) represented by the angle between the incoming ray and the normal at the boundary. - The light travels in a zigzag pattern within the core, reflecting off the boundary due to total internal reflection. - The cone of acceptance illustrates the range of angles for which light will successfully propagate through the fiber. **Blank Fields:** - A. \( \theta_i \ge \) _______ ° - B. \( \theta_i \ge \) _______ ° The task is to calculate the minimum values of \( \theta_i \) for conditions A and B that allow total internal reflection.
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