Light guidance in an optical fiber can be understood by considering a structure comprising a thin solid glass cylinder of refractive index n 1 surrounded by a medium of lower refractive index n 2

Question

Light guidance in an optical fiber can be understood by considering a structure comprising a 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 an angle of incidence $i$ less than a particular value $i_{m}$ are confined in the core. The Numerical Aperture (NA) of the structure is defined as $sin i_{m}$ .

For two structures, namely S₁ with $n_{1} = \frac{4}{5}$ and $n_{2} = \frac{3}{2}$ , and S₂ with $n_{1} = \frac{8}{5}$ and $n_{2} = \frac{7}{5}$ , and taking the refractive index of water to be $\frac{4}{3}$ and that of air to be 1, the correct option(s) is/are:

TestHub · Accepted Answer

For critical incidence, sin C = n 1 ​ n 2 ​ ​ s i n r s i n i ​ = n 0 ​ n 1 ​ ​ ⇒ sin i = n 0 ​ n 1 ​ ​ ⋅ sin r N A = sin i = n 0 ​ n 1 ​ ​ cos r = n 0 ​ n 1 ​ ​ ⋅ n 1 ​ n 7 2 ​ − n 2 2 ​ ​ ​ = n 0 ​ n 1 2 ​ − n 2 2 ​ ​ ​

Physics - Geometrical Optics Question with Solution | TestHub

Options:(select one or more)

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