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JEE Advanced 2015Paper 2
Exam
JEE Advanced
Year
2015
Paper
Paper 2
Subject
Physics
Chapter
Ray Optics

JEE Advanced 2015 · Paper 2 · Physics · Ray Optics

Paragraph: Light guidance in an optical fiber can be understood by...

Paragraph: Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1n_{1} surrounded by a medium of lower refractive index n2n_{2}. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1n_{1} and n2n_{2} as shown in the figure. All rays with the angle of incidence ii less than a particular value imi_{m} are confined in the medium of refractive index n1n_{1}. The numerical aperture (NA) of the structure is defined as sin⁡im\sin i_{m}. Paragraph: Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid

Question: For two structures namely S1S_{1} with n1=45/4n_{1}=\sqrt{45} / 4 and n2=3/2n_{2}=3 / 2, and S2S_{2} with n1=8/5n_{1}=8 / 5 and n2=7/5n_{2}=7 / 5 and taking the refractive index of water to be 4/34 / 3 and that of air to be 11 , the correct option(s) is(are)

Options

  1. ANA of S1{S}_{1} immersed in water is the same as that of S2{S}_{2} immersed in a liquid of refractive index 16315\dfrac{16}{3\sqrt{15}}
  2. BNA of S1{S}_{1} immersed in liquid of refractive index 615\dfrac{6}{\sqrt{15}} is the same as that of S2{S}_{2} immersed in water
  3. CNA of S1{S}_{1} placed in air is the same as that of S2{S}_{2} immersed in liquid of refractive index 415\dfrac{4}{\sqrt{15}}
  4. DNA of S1{S}_{1} placed in air is the same as that of S2{S}_{2} placed in water

Correct answer

A. NA of S1{S}_{1} immersed in water is the same as that of S2{S}_{2} immersed in a liquid of refractive index 16315\dfrac{16}{3\sqrt{15}}

Step-by-step solution

Let the whole structure is placed in a medium of refractive index {n}^{′} , then
{n}^{&prime;}sin i = {n}_{1}cos\left(90 - \theta\right)<br><br >{n}^{′}sin i = {n}_{1}cos\theta.....(i)<br>Herefor.....(i)<br >Here for{i}_{m}; \theta = C and sinC=n2n1sin C =\dfrac{{n}_{2}}{{n}_{1}}
From equation (i), {n}^{&prime;}sin {i}_{m} = {n}_{1}\sqrt{\dfrac{{1 - n}_{2}^{2}}{{n}_{1}^{2}}} = \sqrt{{n}_{1}^{2} - {n}_{2}^{2}}
⇒ sin {i}_{m} = \dfrac{\sqrt{{n}_{1}^{2} - {n}_{2}^{2}}}{{n}^{&prime;}}
Now, for (i) (NA)s1=344516−94=34×34=916{{\left(NA\right)}_{s}}_{1} = \dfrac{3}{4}\sqrt{\dfrac{45}{16} - \dfrac{9}{4}} = \dfrac{3}{4} \times \dfrac{3}{4} = \dfrac{9}{16}
(NA)s2=315166425−4925=315161515=916{\left(NA\right)}_{{s}_{2}} = \dfrac{3\sqrt{15}}{16}\sqrt{\dfrac{64}{25} - \dfrac{49}{25}} = \dfrac{3\sqrt{15}}{16}\dfrac{1}{5}\sqrt{15} = \dfrac{9}{16}
For (ii) (NA)s1=156×34=158{{\left(NA\right)}_{s}}_{1} = \dfrac{\sqrt{15}}{6} \times \dfrac{3}{4} = \dfrac{\sqrt{15}}{8}
(NA)s2=34=155{{\left(NA\right)}_{s}}_{2} = \dfrac{3}{4} = \dfrac{\sqrt{15}}{5} Not equal
For (iii) (NA)s1=1×34=34{{\left(NA\right)}_{s}}_{1} = 1 \times \dfrac{3}{4} = \dfrac{3}{4}
(NA)s2=154×155=154×5=34{{\left(NA\right)}_{s}}_{2} = \dfrac{\sqrt{15}}{4} \times \dfrac{\sqrt{15}}{5} = \dfrac{15}{4 \times 5} = \dfrac{3}{4}
For (iv) (NA)s1=34{\left(NA\right)}_{{s}_{1}} = \dfrac{3}{4}
(NA)s2=34155{{\left(NA\right)}_{s}}_{2} = \dfrac{3}{4}\dfrac{\sqrt{15}}{5} Not equal
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