Geometrical Optics — JEE Advanced · 2015 · 24 May 2015 · Paper 2 Question with Solution

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}

.

For two structures namely

S_{1}

with

n_{1} = 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)

(A)NA of $S_{1}$ immersed in water is the same as that of $S_{2}$ immersed in a liquid of refractive index $\frac{16}{3 15}$ ✓ Correct
(B)NA of $S_{1}$ immersed in liquid of refractive index $\frac{16}{15}$ is that as that of $S_{2}$ immersed in water
(C)NA of $S_{1}$ placed in air is the same as that of $S_{2}$ immersed in liquid of refractive index $\frac{4}{15}$ ✓ Correct
(D)NA of $S_{1}$ placed in air is the same as that of $S_{2}$ placed in water

Answer: (A), (C)

Solution

The numerical aperture (NA) of an optical fiber is given by:

N A = \frac{1}{n _{s}} n_{1}^{2} - n_{2}^{2}

where

n_{s}

is the refractive index of the surrounding medium.

**For

S_{1}

in air:**

N A = \frac{1}{1} \frac{45}{16} - \frac{9}{4} = \frac{3}{4}

**For

S_{1}

in water (

n_{s} = 4/3

):**

N A = \frac{3}{4} \frac{45}{16} - \frac{9}{4} = \frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16}

**For

S_{2}

in liquid with

n_{s} = \frac{16}{3 15}

:**

N A = \frac{3 15}{16} \frac{64}{25} - \frac{49}{25} = \frac{3 15}{16} \cdot \frac{15}{5} = \frac{9}{16}

This matches

S_{1}

in water, so (A) is correct.

**For

S_{2}

in liquid with

n_{s} = \frac{4}{15}

:**

N A = \frac{15}{4} \frac{64}{25} - \frac{49}{25} = \frac{15}{4} \cdot \frac{15}{5} = \frac{3}{4}

This matches

S_{1}

in air, so (C) is correct.

The answer is (A, C).

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Geometrical Optics

Solution

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