Q14Engineering Physics

Question

A ray of light entres from air into fiber. The refractive index of air is one. The fiber has a core of refractive index $1.5$ and cladding of refractive index $1.48$ . Find the critical angle, the fractional refractive index, acceptance angle and numerical aperture.

Answer 1

An optical fiber relies entirely on the principle of Total Internal Reflection (TIR) to transmit data via light pulses over vast distances. It consists of a dense central glass core (refractive index $n_{1}$ ) surrounded by a slightly less dense glass cladding (refractive index $n_{2}$ ). The parameters that define how much light the fiber can capture and guide are rigorously defined by basic geometric optics.

1. Numerical Aperture (NA)

The Numerical Aperture is a dimensionless number that quantitatively represents the light-gathering capability of the optical fiber. A larger NA means the fiber can accept light from a wider range of incident angles. It is strictly dependent on the refractive indices of the core and the cladding. The formula is:

NA = n_{1}^{2} - n_{2}^{2}

Given values: Core index $n_{1} = 1.5$ , Cladding index $n_{2} = 1.48$ .

NA = (1.5)^{2} - (1.48)^{2} = 2.25 - 2.1904 = 0.0596 \approx 0.244

2. Relative Refractive Index Difference ( $Δ$ )

The relative index difference ( $Δ$ ) is a fractional measure of how much the core's refractive index exceeds that of the cladding. It is a critical parameter in fiber manufacturing because it directly dictates the number of modes (paths) the light can take. A very small $Δ$ is required to create high-bandwidth, single-mode fibers. The formula is:

Δ = \frac{n _{1} - n _{2}}{n _{1}}

Using the given values:

Δ = \frac{1.5 - 1.48}{1.5} = \frac{0.02}{1.5} \approx 0.01333

Expressed as a percentage, $Δ = 1.33%$ . This low value is typical for standard telecommunications fibers to minimize modal dispersion (signal distortion).

3. Acceptance Angle ( $θ_{a}$ )

The acceptance angle is the maximum angle (measured from the fiber axis) at which incident light entering the fiber face will strike the core-cladding interface at an angle strictly greater than the critical angle, ensuring it undergoes Total Internal Reflection and gets successfully guided down the fiber. Any light entering at a steeper angle will refract out into the cladding and be lost. The acceptance angle is directly related to the NA by:

θ_{a} = sin^{- 1} (NA)

Assuming the light enters from air ( $n_{0} = 1$ ):

θ_{a} = sin^{- 1} (0.244) \approx 14.1 2^{\circ}

This defines an "acceptance cone" at the entrance of the fiber with a total angle of $28.2 4^{\circ}$ . Only light fired inside this specific cone will be transmitted.

Q14Engineering Physics

Question

Answer

1. Numerical Aperture (NA)

2. Relative Refractive Index Difference (Δ)

3. Acceptance Angle (θa​)

2. Relative Refractive Index Difference ( $Δ$ )

3. Acceptance Angle ( $θ_{a}$ )