Question

Light guidance in an optical fibre can be understood by considering a structure comprising of 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 $\text{i}$ less than a particular value ${\text{i}}_m$ are confined in the medium of refractive index $n_1$. The numerical aperture (NA) of the structure is defined as $\sin {\text{i}}_m$.
For two structures namely $S_1$ with $n_1=\sqrt{45}/4$ and $n_2=3/2$, and $S_2$ with $n_1=8/5$ and $n_2=7/5$ and
taking the refractive index of water to be $4/3$ and that of air to be $1$, the correct option(s) is(are)

• A. NA of $S_1$ immersed in water is the same as that of $S_2$ immersed in a liquid of refractive index $\frac{16}{3\sqrt{15}}$
• B. NA of $S_1$ immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $S_2$ immersed in water.
• C. NA of $S_1$ placed in air is the same as that of $S_2$ immersed in liquid of refractive index $\frac{4}{\sqrt{15}}$.
• D. NA of $S_1$ placed in air is the same as that of $S_2$ placed in water

Answer 1

Answer: (A), (C)

NA of $S_1$ placed in air is the same as that of $S_2$ immersed in liquid of refractive index $\frac{4}{\sqrt{15}}$.

$\theta \ge C$
$\Rightarrow {90}^{\circ }-r\ge C$
$\Rightarrow \sin ({90}^{\circ }-r)\ge \sin C$
$\Rightarrow \cos r\ge \sin C$----(1)




using $\frac{\sin i}{\sin r}=\frac{n_1}{n_m}$ and $\sin c=\frac{n_2}{n_1}$
From (1) ${\sin }^{2}r={\cos }^{2}C$

So we get, ${\sin }^{2}i=\frac{n_1^2-n_2^2}{n_m^2}$

So $NA=\sin i=\frac{\sqrt{n_1^2-n_2^2}}{n_m}$
For
$S_1⟹n_1=\frac{\sqrt{45}}{4};n_2=\frac{3}{2}$

$S_2⟹n_1=\frac{8}{5};n_2=\frac{7}{5}$

Option(A):
$N{A}_{S_1}=\frac{1}{\frac{4}{3}}\sqrt{\frac{45}{16}-\frac{9}{4}}=\frac{9}{16}$
$N{A}_{S_2}=\frac{1}{\frac{16}{3\sqrt{15}}}\sqrt{\frac{64}{25}-\frac{49}{25}}=\frac{9}{16}$

Option(B):
$N{A}_{S_1}=\frac{1}{\frac{\sqrt{15}}{6}}\sqrt{\frac{45}{16}-\frac{9}{4}}=\frac{9}{16}$
$N{A}_{S_2}=\frac{1}{\frac{\sqrt{15}}{6}}\sqrt{\frac{64}{25}-\frac{49}{25}}=\frac{\sqrt{15}}{5}$

Option(C):
$N{A}_{S_1}=\frac{3}{4}$
$N{A}_{S_2}=\frac{3}{4}$

Option(D):
$N{A}_{S_1}=\frac{3}{4}$
$N{A}_{S_2}=\frac{3\sqrt{15}}{20}$