JEE Advanced 2015 · Paper 2 · Physics Q20

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PARAGRAPH 2

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 $\sin i_m$ .

$An optical fibre: a core of index n₁ between cladding layers of index n₂ (n₁ > n₂). A ray from air enters the core's end face at angle i to the axis, refracts, and strikes the core-cladding interface at angle θ, then zig-zags along the core by total internal reflection.$

If two structures of same cross-sectional area, but different numerical apertures $NA_1$ and $NA_2$ $(NA_2<NA_1)$ are joined longitudinally, the numerical aperture of the combined structure is

\dfrac{NA_1NA_2}{NA_1+NA_2}

NA_1+NA_2

NA_1

NA_2

Answer and worked solution

Official answer: D

A ray must be guided in both sections. Rays accepted by the larger-NA section but outside the smaller cone escape in the other section. So the combination accepts only the smaller cone: $NA_2$ .

(D)

Topics:Total internal reflection

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JEE Advanced 2015 · Paper 2, Physics question 20