Optics Ques 41
Passage Based Questions
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 $\sin i _m$.
- If two structures of same cross-sectional area, but different numerical apertures $N A _1$ and $N A _2\left(N A _2<N A _1\right)$ are joined longitudinally, the numerical aperture of the combined structure is
(2015 Adv.)
(a) $\frac{N A _1 N A _2}{N A _1+N A _2}$
(b) $N A _1+N A _2$
(c) $N A _1$
(d) $N A _2$
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Answer:
Correct Answer: 41.(d)
Solution:
Formula:
Laws of Refraction (at any Refracting Surface):
- (1) $\sin i _m=n _1 \sin \left(90^{\circ}-\theta _c\right)$
$\Rightarrow \quad \sin i _m=n _1 \cos \theta _c$
$\Rightarrow \quad N A=n _1 \sqrt{1-\sin ^{2} \theta _c}$
$ = \quad n _1 \sqrt{1-\frac{n _2^{2}}{n _1^{2}}}=\sqrt{n _1^{2}-n _2^{2}} $
Substituting the values we get,
$ \begin{aligned} & N A _1=\frac{3}{4} \text { and } N A _2=\frac{\sqrt{15}}{5}=\sqrt{\frac{3}{4}} \\ & N A _2<N A _1 \end{aligned} $
Therefore, the numerical aperture of combined structure is equal to the lesser of the two numerical aperture, which is $N A _2$.
BETA
