Paragraph: Light guidance in an optical fiber can be understood by considering a structure comprising of…

Paragraph: Light guidance in an optical fiber 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 $\sin i_{m}$.

Question: 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)
  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 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 liquid of refractive index 415
  4. NA of S1 placed in air is the same as that of S2 placed in water

Solution

Let the whole structure is placed in a medium of refractive index n′ , then
n′sin⁡i=n1cos⁡90-θ
n′sin⁡i=n1 cosθ .....(i)
Here for im; θ=C and sin⁡C=n2n1 
From equation (i), n′sin⁡im=n1 1-n22n12=n1 2-n22
⇒sin⁡im=n12-n22n′
Now, for (i) NAs1=34 4516-94=34×34=916
NAs2=31516 6425-4925=315161515=916
For (ii) NAs1=156×34=158
NAs2=34=155 Not equal
For (iii) NAs1=1×34=34
NAs2=154×155=154×5=34
For (iv) NAs1=34
NAs2=34155 Not equal :

Asked in: JEE Advanced 2015 (Paper 2)

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