
A silicon optical fibre with a core diameter large enough has a core refractive index of $1.50$ and a cladding refractive index $1.47$. Determine:
(i) the critical angle at the core cladding interface
(ii) the numerical aperture for the fibre
(iii) the acceptance angle in air for the fibre.
Answer
142.8k+ views
Hint: Optical fibres are mainly used to transmit light. It is transmitted through the two ends of the fibre. Some of the characteristics of optical fibre include flexibility and transparency. Optical fibres are usually made of drawing glass or plastic. It has a diameter which is slightly thicker than the human hair.
Complete step by step answer:
(i) Critical angle $θ_c$ at the core-cladding interface is given by:
${\theta _c} = {\sin ^{ - 1}}\left( {\dfrac{n}{u}} \right)$
Here n is the refractive index of cladding and u is the refractive index of core material.
$\begin{array}{l}
{\theta _c} = {\sin ^{ - 1}}\left( {\dfrac{{1.47}}{{1.50}}} \right)\\
= 78.5^\circ
\end{array}$
Thus, the critical angle is $78.5^\circ $.
(ii) Numerical aperture NA is given by:
$\begin{array}{l}
NA = \sqrt {\left( {{u^2} - {n^2}} \right)} \\
= \sqrt {\left( {{{1.50}^2} - {{1.47}^2}} \right)} \\
= 0.30
\end{array}$
Thus, the numerical aperture is 0.30.
(iii) Acceptance angle A is given by:
$\begin{array}{l}
A = {\sin ^{ - 1}}\;NA\\
= {\sin ^{ - 1}}\;\left( {.30} \right)\\
= 17.4^\circ
\end{array}$
Thus, the acceptance angle is $17.4^\circ $.
Additional Information: The optical fibres have a very low power loss compared to other cables in long distance transmission. It is not affected by electro-magnetic interference, which helps in reducing noise. The optical fibres have a comparatively low size which makes it more convenient. Optical fibres are lighter to carry. This makes it easy to port from one to another. They can transmit greater bandwidth than other cables. Since there is no electromagnetic energy emission it is really difficult to tap which makes it more secure
Note: Since it transmits light rather than electrons there will be no electromagnetic radiation. One of the disadvantages of optical fibres is that it is expensive to install. Due to its compactness and small size it is highly susceptible to getting damaged easily.
Complete step by step answer:
(i) Critical angle $θ_c$ at the core-cladding interface is given by:
${\theta _c} = {\sin ^{ - 1}}\left( {\dfrac{n}{u}} \right)$
Here n is the refractive index of cladding and u is the refractive index of core material.
$\begin{array}{l}
{\theta _c} = {\sin ^{ - 1}}\left( {\dfrac{{1.47}}{{1.50}}} \right)\\
= 78.5^\circ
\end{array}$
Thus, the critical angle is $78.5^\circ $.
(ii) Numerical aperture NA is given by:
$\begin{array}{l}
NA = \sqrt {\left( {{u^2} - {n^2}} \right)} \\
= \sqrt {\left( {{{1.50}^2} - {{1.47}^2}} \right)} \\
= 0.30
\end{array}$
Thus, the numerical aperture is 0.30.
(iii) Acceptance angle A is given by:
$\begin{array}{l}
A = {\sin ^{ - 1}}\;NA\\
= {\sin ^{ - 1}}\;\left( {.30} \right)\\
= 17.4^\circ
\end{array}$
Thus, the acceptance angle is $17.4^\circ $.
Additional Information: The optical fibres have a very low power loss compared to other cables in long distance transmission. It is not affected by electro-magnetic interference, which helps in reducing noise. The optical fibres have a comparatively low size which makes it more convenient. Optical fibres are lighter to carry. This makes it easy to port from one to another. They can transmit greater bandwidth than other cables. Since there is no electromagnetic energy emission it is really difficult to tap which makes it more secure
Note: Since it transmits light rather than electrons there will be no electromagnetic radiation. One of the disadvantages of optical fibres is that it is expensive to install. Due to its compactness and small size it is highly susceptible to getting damaged easily.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11
