Math Problem Statement

Find NA and acceptance angle of a fiber with core index 1.4 and Δ = 0.02.

Solution

To calculate the Numerical Aperture (NA) and the Acceptance Angle of the fiber, we can use the following formulas:

1. Numerical Aperture (NA):

NA=n12n22\text{NA} = \sqrt{n_1^2 - n_2^2} Where:

  • n1n_1 = Core refractive index.
  • n2n_2 = Cladding refractive index.

The relation between n1n_1 and n2n_2 is given by the Relative Index Difference (Δ): Δ=n12n222n12\Delta = \frac{n_1^2 - n_2^2}{2n_1^2} From this, we can solve for n2n_2: n2=n112Δn_2 = n_1 \sqrt{1 - 2\Delta}

2. Acceptance Angle (θₐ):

The acceptance angle (in air) is related to the Numerical Aperture: sinθa=NA\sin \theta_a = \text{NA} Then, θa=arcsin(NA)\theta_a = \arcsin(\text{NA})

Given:

  • Core refractive index n1=1.4n_1 = 1.4
  • Δ=0.02\Delta = 0.02

Let's compute the Numerical Aperture and the Acceptance Angle.### Results:

  1. Numerical Aperture (NA): 0.28
  2. Acceptance Angle (θₐ): 16.26°

Would you like more details or have any further questions?

Here are 5 related questions for further understanding:

  1. How does increasing the core index affect the Numerical Aperture?
  2. What is the role of the cladding in fiber optics?
  3. How does the relative index difference Δ\Delta influence the fiber's performance?
  4. Can Numerical Aperture be greater than 1? Why or why not?
  5. How does the acceptance angle impact light propagation within the fiber?

Tip: A larger Numerical Aperture allows the fiber to capture light at wider angles, improving light collection efficiency.

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Math Problem Analysis

Mathematical Concepts

Numerical Aperture (NA)
Acceptance Angle
Relative Index Difference (Δ)
Optical Fiber Theory

Formulas

NA = sqrt(n1^2 - n2^2)
Δ = (n1^2 - n2^2) / (2n1^2)
n2 = n1 * sqrt(1 - 2Δ)
sin(θa) = NA
θa = arcsin(NA)

Theorems

Snell's Law
Fiber Optic Light Propagation

Suitable Grade Level

Undergraduate level (Physics/Optics)