GATE 2020 IN Q.46: solution and answer

GATE 2020 · INQ.46 · 2 marks · NAT

A laser beam of 10 mm beam diameter is focused onto an optical fibre using a thin biconvex lens as shown in the figure. The refractive index of the lens is 1.5. The refractive indices of the core and cladding of the fibre are 1.55 and 1.54 respectively. The minimum value of the focal length of the lens to attain the maximum coupling to the fibre (in mm, rounded off to one decimal place) is ____

GATE 2020 IN question 46: A collimated laser beam, drawn as a grey horizontal band, travels from left to right. — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

ANSWER

The minimum focal length is $f_{m i n} = 5 mm / tan θ_{a} = 5/0.17856 \approx 28.0 mm$ .

Official answer: 27.5 to 28.5Source: Official answer key, GATE 2020 IN (IIT Delhi)

Numerical aperture of the fibre. The medium around the fibre is not stated, so take it as air ( $n_{0} = 1$ ). With core index $n_{1} = 1.55$ and cladding index $n_{2} = 1.54$ : $NA = n_{0} sin θ_{a} = n_{1}^{2} - n_{2}^{2} = (1.55 - 1.54) (1.55 + 1.54) = (0.01) (3.09) = 0.0309 = 0.17578$
Acceptance half-angle. $θ_{a} = sin^{- 1} (0.17578) \approx 10.1 2^{\circ}$ . A ray that reaches the core face at a larger angle to the axis is not totally internally reflected at the core–cladding boundary, so it leaks out of the core and is lost.
Geometry of the focused beam. The collimated beam has diameter $10 mm$ , so its radius is $a = 5 mm$ . A thin lens brings a collimated beam to a focus at distance $f$ . The edge (marginal) ray therefore meets the axis at a half-angle $θ$ given by $tan θ = \frac{a}{f} = \frac{5}{f}$
Condition for maximum coupling. The whole beam enters the guided modes only if the marginal ray, which is the steepest ray, lies inside the acceptance cone: $θ \leq θ_{a}$ . Since $tan θ = 5/ f$ gets smaller as $f$ grows, this means $f \geq \frac{5}{tan θ _{a}}$ . The minimum focal length is the one that puts the marginal ray exactly at $θ_{a}$ .
Convert NA to a tangent. $tan θ_{a} = \frac{sin θ _{a}}{1 - sin ^{2} θ _{a}} = \frac{0.17578}{1 - 0.0309} = \frac{0.17578}{0.98443} = 0.17856$
Minimum focal length. $f_{m i n} = \frac{a}{tan θ _{a}} = \frac{5 mm}{0.17856} = 28.00 mm \approx 28.0 mm$
The lens index is not needed. The refractive index of the lens (1.5) would matter only for working out $f$ from the radii of curvature (lensmaker's equation). No radii are given and the question asks for $f$ itself, so the index plays no part.

KEY IDEA

For maximum coupling, every ray of the focused cone must lie inside the fibre's acceptance cone. The steepest ray is the marginal ray, at $tan θ = a / f$ where $a$ is the beam radius. Setting $θ = θ_{a}$ with $sin θ_{a} = NA = n_{1}^{2} - n_{2}^{2}$ gives the shortest focal length that still couples the whole beam: $f_{m i n} = a / tan θ_{a}$ .

WATCH OUT FOR

Using the full diameter (10 mm) where the geometry needs the radius (5 mm). That doubles the answer to about 56 mm. Other slips: forgetting the square root in $NA = n_{1}^{2} - n_{2}^{2}$ ; writing $NA = n_{1} - n_{2}$ ; treating the lens index 1.5 as part of the NA. Setting $sin θ_{a} = a / f$ instead of $tan θ_{a} = a / f$ gives about 28.4 mm, which happens to fall inside the key range, but the ray geometry is a tangent, not a sine.

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 6 Waveguides & fibres questions, with notes

GATE 2020 IN Q.46: solution and answer

Answer and solution

More Waveguides & fibres questions