GATE 2018 IN Q.55: solution and answer

GATE 2018 · INQ.55 · 2 marks · NAT

A multi-mode optical fiber with a large core diameter has a core refractive index $n_{1} = 1.5$ and cladding refractive index $n_{2} = 1.4142$ . The maximum value of $θ_{A}$ (in degrees) for which the incident light from air will be guided in the optical fiber is $\pm$ ______.

GATE 2018 IN question 55: Longitudinal section of a step-index fiber: a horizontal core strip labelled 'Core, n_1 = 1.5' lies between two cladding… — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

ANSWER

The maximum acceptance angle is $θ_{A} = sin^{- 1} n_{1}^{2} - n_{2}^{2} = sin^{- 1} (0.5) \approx 3 0^{\circ}$ , so light is guided for launch angles within $\pm 3 0^{\circ}$ .

Official answer: 29 to 31Source: Official answer key, GATE 2018 IN (IIT Guwahati)

What the fibre needs: the ray is guided only if it meets the core–cladding boundary at an angle (measured from the normal to that boundary) of at least the critical angle $θ_{c}$ , so that it is totally internally reflected. The largest launch angle $θ_{A}$ in air that still meets this condition is the acceptance angle.
Critical angle at the core–cladding boundary: $sin θ_{c} = \frac{n _{2}}{n _{1}}$ . The fibre axis runs parallel to the boundary, so at this limit the ray inside the core makes an angle $θ_{r} = 9 0^{\circ} - θ_{c}$ with the axis.
Snell's law at the air–core end face ( $n_{0} = 1$ for air): $n_{0} sin θ_{A} = n_{1} sin θ_{r} = n_{1} cos θ_{c} = n_{1} 1 - \frac{n _{2}^{2}}{n _{1}^{2}} = n_{1}^{2} - n_{2}^{2}$ This quantity is the numerical aperture, $NA = n_{1}^{2} - n_{2}^{2}$ .
Substitute the values: $n_{1}^{2} = 1. 5^{2} = 2.25, n_{2}^{2} = 1.414 2^{2} = 1.99996$ $NA = 2.25 - 1.99996 = 0.25004 \approx 0.5000$ (Note that $1.4142 \approx 2$ , so $n_{2}^{2} \approx 2$ .)
Acceptance angle: $θ_{A} = sin^{- 1} (NA) = sin^{- 1} (0.5000) \approx 3 0^{\circ}$ (Keeping every digit gives $30.00 3^{\circ}$ .)
Result: light from air is guided for launch angles up to $\pm 3 0^{\circ}$ from the fibre axis.

KEY IDEA

For a fibre launched from air, the acceptance angle satisfies $sin θ_{A} = NA = n_{1}^{2} - n_{2}^{2}$ . This comes from total internal reflection at the core–cladding boundary ( $sin θ_{c} = n_{2} / n_{1}$ ) together with Snell's law at the end face. Here $n_{2} \approx 2$ , so $NA = 2.25 - 2 = 0.5$ .

WATCH OUT FOR

Answering with the critical angle $θ_{c} = sin^{- 1} (n_{2} / n_{1}) = sin^{- 1} (0.9428) \approx 70. 5^{\circ}$ , or its complement inside the core, $9 0^{\circ} - θ_{c} \approx 19. 5^{\circ}$ , instead of the angle in air. The core-side angle has to be carried through Snell's law at the end face, $sin θ_{A} = n_{1} cos θ_{c}$ . Another slip is to take $NA = n_{1} - n_{2}$ instead of $n_{1}^{2} - n_{2}^{2}$ .

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 6 Waveguides & fibres questions, with notes

GATE 2018 IN Q.55: solution and answer

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