Light guidance in an optical fibre 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 im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im. For two structures namely S1 with n1 = \sqrt 45 \over 4 and n2 = 3 \over 2 , and S2 with n1 = 8 \over 5 and n2 = 7 \over 5 and taking the refractive index of water to be 4 \over 3 and that to air to be 1, the correct options is/are :

NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index 16 \over 3\sqrt 15, NA of S1 placed in air is the same as that S2 immersed in liquid of refractive index 4 \over \sqrt 15

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

NA of S1 immersed in liquid of refractive index 6 \over \sqrt 15 is the same as that of S2 immersed in water

NA of S1 placed in air is the same as that S2 immersed in liquid of refractive index 4 \over \sqrt 15

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

Geometrical Optics PYQ — JEE Advanced 2015 · Shift 2 · Q57 with Solution

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Correct answer: A, C

Formula for numerical aperture in an external medium

For an optical fibre with core refractive index $n_1$ and cladding refractive index $n_2$ , if the surrounding external medium has refractive index $n_0$ , then

$n_0 \sin i_m = \sqrt{n_1^2-n_2^2}$

Hence the numerical aperture as defined in the question is

$\text{NA} = \sin i_m = \frac{\sqrt{n_1^2-n_2^2}}{n_0}$

So we first compute $\sqrt{n_1^2-n_2^2}$ for both structures.

For structure $S_1$

Given:

$n_1=\frac{\sqrt{45}}{4}, \qquad n_2=\frac{3}{2}$

Then

$n_1^2 = \frac{45}{16}, \qquad n_2^2 = \frac{9}{4}=\frac{36}{16}$

So,

$n_1^2-n_2^2 = \frac{45-36}{16} = \frac{9}{16}$

Therefore,

$\sqrt{n_1^2-n_2^2} = \frac{3}{4}$

Thus for $S_1$ ,

$\text{NA}(S_1)=\frac{3}{4n_0}$

For structure $S_2$

Given:

$n_1=\frac{8}{5}, \qquad n_2=\frac{7}{5}$

Then

$n_1^2 = \frac{64}{25}, \qquad n_2^2 = \frac{49}{25}$

So,

$n_1^2-n_2^2 = \frac{64-49}{25} = \frac{15}{25} = \frac{3}{5}$

Therefore,

$\sqrt{n_1^2-n_2^2} = \sqrt{\frac{3}{5}} = \frac{\sqrt{15}}{5}$

Thus for $S_2$ ,

$\text{NA}(S_2)=\frac{\sqrt{15}}{5n_0}$

Check option A

$S_1$ immersed in water: $n_0=\frac{4}{3}$

$\text{NA} = \frac{3}{4\cdot (4/3)} = \frac{3}{16/3} = \frac{9}{16}$

$S_2$ immersed in liquid of refractive index $\frac{16}{3\sqrt{15}}$ :

$\text{NA} = \frac{\sqrt{15}}{5\cdot \frac{16}{3\sqrt{15}}}$

= \sqrt{15}\cdot \frac{3\sqrt{15}}{80} = \frac{45}{80} = \frac{9}{16}$$ Both are equal, so **A is correct**. --- 5. **Check option B** - $S_1$ immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$: $$\text{NA} = \frac{3}{4\cdot \frac{6}{\sqrt{15}}} = \frac{3\sqrt{15}}{24} = \frac{\sqrt{15}}{8}$$ - $S_2$ immersed in water: $n_0=\frac{4}{3}$ $$\text{NA} = \frac{\sqrt{15}}{5\cdot (4/3)} = \frac{\sqrt{15}}{20/3} = \frac{3\sqrt{15}}{20}$$ Now, $$\frac{\sqrt{15}}{8} \ne \frac{3\sqrt{15}}{20}$$ So **B is incorrect**. --- 6. **Check option C** - $S_1$ in air: $n_0=1$ $$\text{NA} = \frac{3}{4}$$ - $S_2$ immersed in liquid of refractive index $\frac{4}{\sqrt{15}}$: $$\text{NA} = \frac{\sqrt{15}}{5\cdot \frac{4}{\sqrt{15}}} = \frac{\sqrt{15}}{20/\sqrt{15}} = \frac{15}{20} = \frac{3}{4}$$ Both are equal, so **C is correct**. --- 7. **Check option D** - $S_1$ in air: $$\text{NA} = \frac{3}{4}$$ - $S_2$ in water: $$\text{NA} = \frac{3\sqrt{15}}{20}$$ Since $$\frac{3}{4} \ne \frac{3\sqrt{15}}{20}$$ So **D is incorrect**. --- 8. **Final answer** The correct options are: $$\boxed{A,\ C}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: A, C

Formula for numerical aperture in an external medium

For an optical fibre with core refractive index $n_1$ and cladding refractive index $n_2$ , if the surrounding external medium has refractive index $n_0$ , then

$n_0 \sin i_m = \sqrt{n_1^2-n_2^2}$

Hence the numerical aperture as defined in the question is

$\text{NA} = \sin i_m = \frac{\sqrt{n_1^2-n_2^2}}{n_0}$

So we first compute $\sqrt{n_1^2-n_2^2}$ for both structures.

For structure $S_1$

Given:

$n_1=\frac{\sqrt{45}}{4}, \qquad n_2=\frac{3}{2}$

Then

$n_1^2 = \frac{45}{16}, \qquad n_2^2 = \frac{9}{4}=\frac{36}{16}$

So,

$n_1^2-n_2^2 = \frac{45-36}{16} = \frac{9}{16}$

Therefore,

$\sqrt{n_1^2-n_2^2} = \frac{3}{4}$

Thus for $S_1$ ,

$\text{NA}(S_1)=\frac{3}{4n_0}$

For structure $S_2$

Given:

$n_1=\frac{8}{5}, \qquad n_2=\frac{7}{5}$

Then

$n_1^2 = \frac{64}{25}, \qquad n_2^2 = \frac{49}{25}$

So,

$n_1^2-n_2^2 = \frac{64-49}{25} = \frac{15}{25} = \frac{3}{5}$

Therefore,

$\sqrt{n_1^2-n_2^2} = \sqrt{\frac{3}{5}} = \frac{\sqrt{15}}{5}$

Thus for $S_2$ ,

$\text{NA}(S_2)=\frac{\sqrt{15}}{5n_0}$

Check option A

$S_1$ immersed in water: $n_0=\frac{4}{3}$

$\text{NA} = \frac{3}{4\cdot (4/3)} = \frac{3}{16/3} = \frac{9}{16}$

$S_2$ immersed in liquid of refractive index $\frac{16}{3\sqrt{15}}$ :

$\text{NA} = \frac{\sqrt{15}}{5\cdot \frac{16}{3\sqrt{15}}}$

= \sqrt{15}\cdot \frac{3\sqrt{15}}{80} = \frac{45}{80} = \frac{9}{16}$$ Both are equal, so **A is correct**. --- 5. **Check option B** - $S_1$ immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$: $$\text{NA} = \frac{3}{4\cdot \frac{6}{\sqrt{15}}} = \frac{3\sqrt{15}}{24} = \frac{\sqrt{15}}{8}$$ - $S_2$ immersed in water: $n_0=\frac{4}{3}$ $$\text{NA} = \frac{\sqrt{15}}{5\cdot (4/3)} = \frac{\sqrt{15}}{20/3} = \frac{3\sqrt{15}}{20}$$ Now, $$\frac{\sqrt{15}}{8} \ne \frac{3\sqrt{15}}{20}$$ So **B is incorrect**. --- 6. **Check option C** - $S_1$ in air: $n_0=1$ $$\text{NA} = \frac{3}{4}$$ - $S_2$ immersed in liquid of refractive index $\frac{4}{\sqrt{15}}$: $$\text{NA} = \frac{\sqrt{15}}{5\cdot \frac{4}{\sqrt{15}}} = \frac{\sqrt{15}}{20/\sqrt{15}} = \frac{15}{20} = \frac{3}{4}$$ Both are equal, so **C is correct**. --- 7. **Check option D** - $S_1$ in air: $$\text{NA} = \frac{3}{4}$$ - $S_2$ in water: $$\text{NA} = \frac{3\sqrt{15}}{20}$$ Since $$\frac{3}{4} \ne \frac{3\sqrt{15}}{20}$$ So **D is incorrect**. --- 8. **Final answer** The correct options are: $$\boxed{A,\ C}$$ This matches the stored correct answer.

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