For two structures namely with and and with and and taking the refractive index of water to be and that of air to be 1, the correct option(s) is(are) [2015]

Question

Light guidance in an optical fibre can be understood by considering a structure comprising of a thin solid glass cylinder of refractive index $n_{1}$ surrounded by a medium of lower refractive index $n_{2}$ . The light guidance in the structure takes place due to successive total internal reflections at the interface of the media $n_{1}$ and $n_{2}$ , 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 $n_{1}$ . The numerical aperture (NA) of the structure is defined as $\sin i_{m} .$

Q. For two structures namely $S_{1}$ with $n_{1} = \frac{\sqrt{45}}{4}$ and $n_{2} = \frac{3}{2}$ and $S_{2}$ with $n_{1} = \frac{8}{5}$ and $n_{2} = \frac{7}{5}$ and taking the refractive index of water to be $\frac{4}{3}$ and that of air to be 1, the correct option(s) is(are) [2015]

Smart Achievers · Accepted Answer

(1, 3)

Using Snell's law at P; $n_{s} \sin i_{m} = n_{1} \sin (90 ° - C)$ ...(i)

$n_{s} = Refractive index of surrounding$

Also, $\sin C = \frac{n_{2}}{n_{1}}$ $∴$ $\cos C = \sqrt{1 - {(\frac{n_{2}}{n_{1}})}^{2}}$

Now from eq. (i),

Numerical aperture,

$N A = \sin i_{m} = \frac{n_{1}}{n_{s}} \cos C = \frac{n_{1}}{n_{s}} \sqrt{1 - \frac{n_{2}^{2}}{n_{1}^{2}}}$

$∴$ $N A = \frac{\sqrt{n_{1}^{2} - n_{2}^{2}}}{n_{s}}$

For $S_{1}$ (in air)

$N A = \sqrt{\frac{45}{16} - \frac{9}{4}} = \frac{3}{4}$

For $S_{1}$ (in $n_{s} = \frac{6}{\sqrt{15}}$ )

$N A = \frac{\sqrt{15}}{6} \sqrt{\frac{45}{16} - \frac{9}{4}}$

$= \frac{3 \sqrt{15}}{24} = \frac{3}{4} \frac{\sqrt{15}}{5}$

For $S_{2}$ (in air)

$N A = \sqrt{\frac{64}{25} - \frac{49}{25}} = \frac{\sqrt{15}}{5}$

For $S_{2}$ (in $n_{s} = \frac{16}{3 \sqrt{15}}$ )

$N A = \frac{3 \sqrt{15}}{16} \sqrt{\frac{64}{25} - \frac{49}{25}} = \frac{9}{16}$

For $S_{1}$ (in water)

$N A = \frac{3}{4} \sqrt{\frac{45}{16} - \frac{9}{4}} = \frac{9}{16}$

For $S_{2}$ (in water)

$N A = \frac{3}{4} \sqrt{\frac{64}{25} - \frac{49}{25}} = \frac{3 \sqrt{15}}{20}$

For $S_{2}$ (in $n_{s} = \frac{4}{\sqrt{15}}$ )

$N A = \frac{\sqrt{15}}{4} \sqrt{\frac{64}{25} - \frac{49}{25}} = \frac{3}{4}$

Solution

1 NA of $S_{1}$ immersed in water is the same as that of $S_{2}$ immersed in a liquid of refractive index $\frac{16}{3 \sqrt{15}}$

2 NA of $S_{1}$ immersed in a liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $S_{2}$ immersed in water

3 NA of $S_{1}$ placed in air is the same as that of $S_{2}$ immersed in a liquid of refractive index $\frac{4}{\sqrt{15}}$

4 NA of $S_{1}$ placed in air is the same as that of $S_{2}$ placed in water

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Solution

1 NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index 16315

2 NA of S1 immersed in a liquid of refractive index 615 is the same as that of S2 immersed in water

3 NA of S1 placed in air is the same as that of S2 immersed in a liquid of refractive index 415

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

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1 NA of $S_{1}$ immersed in water is the same as that of $S_{2}$ immersed in a liquid of refractive index $\frac{16}{3 \sqrt{15}}$

2 NA of $S_{1}$ immersed in a liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $S_{2}$ immersed in water

3 NA of $S_{1}$ placed in air is the same as that of $S_{2}$ immersed in a liquid of refractive index $\frac{4}{\sqrt{15}}$

4 NA of $S_{1}$ placed in air is the same as that of $S_{2}$ placed in water