Question 375: Ray Optics

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 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
.

Accepted Answer

$$
\sin i_m=n_1 \sin \left(90-	heta_cight)
$$
$$
\Rightarrow \sin i_m=n_1 \cos 	heta_c
$$
$$
\begin{aligned}
\Rightarrow N A & =n_1 \sqrt{1-\sin ^2 	heta_c} \\
& =n_1 \sqrt{1-\frac{n_2^2}{n_1^2}}=\sqrt{n_1^2-n_2^2}
\end{aligned}
$$
Substituting the values we get,
$$
N A_1=\frac{3}{4}
$$
$$
	ext { and } N A_2=\frac{\sqrt{15}}{5}=\sqrt{\frac{3}{4}}
$$
and $$
N A_2 < N A_1
$$
Therefore, the numerical aperture of combined structure is equal to the lesser of the two numerical aperture, which is $N A_2$.

Answer

$${{N{A_1}N{A_2}} \over {N{A_1} + N{A_2}}}$$

Answer

$$N{A_1} + N{A_2}$$

Answer

$$N{A_1}$$

Answer

$$N{A_2}$$

Ray Optics

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