JEE Challenger
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PhyMedAdv 2015 • Paper 2

Effective Numerical Aperture of Two Optical Fibers Joined Longitudinally

Comprehension Passage

Light guidance within an optical fiber can be understood by considering a structure comprising a thin solid glass cylinder of refractive index n1n_1 surrounded by a cladding medium of lower refractive index n2n_2. Light propagation occurs due to successive total internal reflections at the core-cladding interface as illustrated in the figure. All rays entering with an incident angle ii less than a maximum limiting value imi_m are guided within the core of refractive index n1n_1. The numerical aperture (NA) of the structure is defined by sin⁡im\sin i_m.

Passage Diagram 1

Suppose two structures possessing the identical cross-sectional area, but with differing numerical apertures NA1NA_1 and NA2NA_2 (NA2<NA1NA_2 < NA_1), are joined longitudinally end-to-end. The numerical aperture of the combined fiber structure is

Options

A

NA1NA2NA1+NA2\frac{NA_1 NA_2}{NA_1 + NA_2}

B

NA1+NA2NA_1 + NA_2

C

NA1NA_1

D

NA2NA_2

Correct

Topics & Concepts

Ray Optics and Optical InstrumentsReflection and Refraction

Step-by-Step Solution

To find the numerical aperture of the combined optical fiber structure, let us recall the definition and working principle of an optical fiber:

  1. Condition for Total Internal Reflection (TIR): Consider a light ray entering the core of refractive index n1n_1 from air (n0=1n_0 = 1) at an angle of incidence ii. By Snell's law at the entrance face: 1⋅sin⁡i=n1sin⁡θ1 \cdot \sin i = n_1 \sin \theta where θ\theta is the angle of refraction inside the core with respect to the fiber axis.

    The ray strikes the core-cladding interface at an angle of incidence ϕ=90∘−θ\phi = 90^\circ - \theta. For total internal reflection to occur at the core-cladding interface (where the cladding has refractive index n2n_2 with n2<n1n_2 < n_1), the angle ϕ\phi must be greater than or equal to the critical angle θc\theta_c: sin⁡ϕ=sin⁡(90∘−θ)=cos⁡θ≥sin⁡θc=n2n1\sin \phi = \sin(90^\circ - \theta) = \cos \theta \ge \sin \theta_c = \frac{n_2}{n_1}

    Squaring both sides and using sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta: sin⁡θ≤1−(n2n1)2=n12−n22n1\sin\theta \le \sqrt{1 - \left(\frac{n_2}{n_1}\right)^2} = \frac{\sqrt{n_1^2 - n_2^2}}{n_1}

    Substituting this back into Snell's law at the entrance face: sin⁡i=n1sin⁡θ≤n12−n22\sin i = n_1 \sin \theta \le \sqrt{n_1^2 - n_2^2}

    Thus, the maximum angle of incidence imi_m is given by: sin⁡im=n12−n22=NA\sin i_m = \sqrt{n_1^2 - n_2^2} = \text{NA} where NA\text{NA} is the numerical aperture of the optical fiber.

  2. Propagation through the Combined Structure: When two optical fibers with numerical apertures NA1NA_1 and NA2NA_2 are joined longitudinally end-to-end, light must propagate successfully through both sections without leaking into the cladding.

    • For light to be guided through the first structure, the condition is: sin⁡i≤NA1\sin i \le NA_1
    • For light to also be guided through the second structure, the ray's angle with respect to the axis must be confined such that total internal reflection occurs in the second structure as well, which requires: sin⁡i≤NA2\sin i \le NA_2

    For light to be guided through the entire length of the combined fiber structure, it must satisfy both conditions simultaneously: sin⁡i≤min⁡(NA1,NA2)\sin i \le \min(NA_1, NA_2)

    Since it is given that NA2<NA1NA_2 < NA_1: sin⁡im=min⁡(NA1,NA2)=NA2\sin i_m = \min(NA_1, NA_2) = NA_2

Therefore, the numerical aperture of the combined fiber structure is NA2NA_2.

Correct Option: (D)

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