JEE Advanced 2015 · Paper 2 · Physics Q19

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PhysicsMultiple correct+4 / -2Question 19 of 60

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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 $\sin i_m$ .

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

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)

NA of

S_1

immersed in water is the same as that of

S_2

immersed in a liquid of refractive index

\dfrac{16}{3\sqrt{15}}

NA of

S_1

immersed in liquid of refractive index

\dfrac6{\sqrt{15}}

is the same as that of

S_2

immersed in water

NA of

S_1

placed in air is the same as that of

S_2

immersed in liquid of refractive index

\dfrac4{\sqrt{15}}

NA of

S_1

placed in air is the same as that of

S_2

placed in water

Answer and worked solution

Official answer: A, C

Numerical aperture. Total internal reflection at the core-cladding boundary limits the entry angle:
$n_0\sin i_m=\sqrt{n_1^2-n_2^2}\implies NA=\frac{\sqrt{n_1^2-n_2^2}}{n_0}.$

The two fibres.

$S_1$ : $\sqrt{\frac{45}{16}-\frac{36}{16}}=\frac34$ .
$S_2$ : $\sqrt{\frac{64}{25}-\frac{49}{25}}=\frac{\sqrt{15}}5$ .

Check the options.

(A) $\dfrac{3/4}{4/3}=\dfrac9{16}$ and $\dfrac{\sqrt{15}/5}{16/(3\sqrt{15})}=\dfrac9{16}$ . Equal. ✓
(B) $\dfrac{\sqrt{15}}8$ against $\dfrac{3\sqrt{15}}{20}$ . ✗
(C) $\dfrac34$ and $\dfrac{\sqrt{15}/5}{4/\sqrt{15}}=\dfrac34$ . Equal. ✓
(D) $\dfrac34$ against $\dfrac{3\sqrt{15}}{20}$ . ✗

(A), (C)

Topics:Total internal reflection Refraction at plane surfaces

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JEE Advanced 2015 · Paper 2, Physics question 19