Question

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n₁ surrounded by a medium of lower refractive index n₂. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n₁ and n₂ as shown in the figure. All rays with the angle of incidence i less than a particular value i_m are confined in the medium of refractive index n₁. The numerical aperture (NA) of the structure is defined as sin i_m.

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Linked Question 1

For two structures namely S₁ with n₁ = $\sqrt{45} / 4$ and n₂ = 3/2, and S₂ with n₁ = 8/5 and n₂ = 7/5 and taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is(are)

NA of S₁ placed in air is the same as that of S₂ placed in water.

NA of S₁ placed in air is the same as that of S₂ immersed in liquid of refractive index $\frac{4}{\sqrt{15}}$ .

NA of S₁ immersed in water is the same as that of S₂ immersed in a liquid of refractive index $\frac{16}{3 \sqrt{15}}$

NA of S₁ immersed in liquid of refractive index $\frac{16}{\sqrt{15}}$ is the same as that of S₂ immersed in water.

$sin C = \frac{µ_{r}}{µ_{d}} = \frac{n_{2}}{n_{1}}$

µ_s sin i = n sin r = n₁ sin (90 – Q)

µ_s sin i = n₁ cos Q = n₁ $\sqrt{1 - \frac{n_{2}^{2}}{n_{1}^{2}}} = \sqrt{n_{1}^{2} - n_{2}^{2}}$

sin i = $\frac{1}{μ_{s}} \sqrt{n_{1}^{2} - n_{2}^{2}}$

(A) $S_{1} : {sini}_{m_{1}} = \frac{3}{4} \sqrt{\frac{45}{16} - \frac{9}{4}} = \frac{3}{4} \sqrt{\frac{45 - 36}{16}} = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$

$S_{2} : {sini}_{m_{2}} = \frac{3 \sqrt{15}}{16} \sqrt{\frac{64}{25} - \frac{49}{25}} = \frac{3 \sqrt{15}}{16} \frac{\sqrt{15}}{5} = \frac{9}{16}$

(B) $S_{1} : {sini}_{m_{1}} = \frac{\sqrt{15}}{6} \sqrt{\frac{45}{16} - \frac{9}{4}} = \frac{\sqrt{15}}{6} \sqrt{\frac{45 - 36}{16}} = \frac{3}{4} \frac{\sqrt{15}}{6} = \frac{\sqrt{15}}{8}$

$S_{2} : {sini}_{m_{2}} = \frac{3}{4} \sqrt{\frac{64}{25} - \frac{49}{25}} = \frac{3}{4} \sqrt{\frac{15}{25}} = \frac{\sqrt{15} \times 3}{20}$

$S_{2} : {sini}_{m_{2}} = \frac{\sqrt{15}}{4} \sqrt{\frac{15}{25}} = \frac{15}{4 \times 5} = \frac{15}{20} = \frac{3}{4}$

(D) $S_{1} : {sini}_{m_{1}} = \frac{3}{4}$

$S_{2} : {sini}_{m_{2}} = \frac{\sqrt{15}}{5}$

Linked Question 2

If two structures of same cross-sectional area, but different numerical apertures NA₁ and NA₂ (NA₂ < NA₁) are joined longitudinally, the numerical aperture of the combined structure is

NA₁

$\frac{{NA}_{1} {NA}_{2}}{{NA}_{1} + {NA}_{2}}$

NA₁ + NA₂

NA₂