
An optical communication system is operating at a wavelength of \[800nm\], its optical source frequency is
(A) \[3.8 \times {10^{14}}Hz\]
(B) \[3.8 \times {10^{12}}Hz\]
(C) \[3.8 \times {10^{10}}Hz\]
(D) \[3.8 \times {10^9}Hz\]
Answer
232.8k+ views
Hint Understand the meaning behind optical communication. Optical communication involves light rays for transferring of messages. Now, directly find the frequency by dividing the speed of light rays with the given wavelength.
Complete Step By Step Solution
Optical communication is a type of communication method, which uses light rays to carry information. Optical fibers are required to transmit light rays constantly throughout for a larger distance. Since the speed of light is greater than speed of sound or speed of radio waves, optical communication is said to be the fastest communication mode. It is also understandable that the construction of optical fibre cables are robust, which makes optical communication much more secure than others.
It is given that the optical communication system is operating at a given wavelength of \[800nm\]. We need to find out the frequency of the optical source. We know that , frequency is given by the formula,
\[f = \dfrac{c}{\lambda }\], where c is the speed of light and \[\lambda \]is the wavelength of the light ray passed.
Substituting the wavelength value in the above equation we get,
\[ \Rightarrow f = \dfrac{{3 \times {{10}^8}}}{{800 \times {{10}^{ - 9}}}}\]
On simplifying this equation we get,
\[ \Rightarrow f = \dfrac{{3 \times {{10}^{15}}}}{8}\]
\[ \Rightarrow f = 3.8 \times {10^{14}}Hz\]
Thus, option (A) is the right answer for the given question.
Note
An optical wave is a type of electromagnetic wave that has the ability to propagate in free space and also can be guided with dielectric. The optical waves undergo the concept of total internal reflection inside an optical fiber cable between the core and the cladding part of the cable. The optical communication has a wide range of electromagnetic spectrum, enabling it to be deployed in large distance communication.
Complete Step By Step Solution
Optical communication is a type of communication method, which uses light rays to carry information. Optical fibers are required to transmit light rays constantly throughout for a larger distance. Since the speed of light is greater than speed of sound or speed of radio waves, optical communication is said to be the fastest communication mode. It is also understandable that the construction of optical fibre cables are robust, which makes optical communication much more secure than others.
It is given that the optical communication system is operating at a given wavelength of \[800nm\]. We need to find out the frequency of the optical source. We know that , frequency is given by the formula,
\[f = \dfrac{c}{\lambda }\], where c is the speed of light and \[\lambda \]is the wavelength of the light ray passed.
Substituting the wavelength value in the above equation we get,
\[ \Rightarrow f = \dfrac{{3 \times {{10}^8}}}{{800 \times {{10}^{ - 9}}}}\]
On simplifying this equation we get,
\[ \Rightarrow f = \dfrac{{3 \times {{10}^{15}}}}{8}\]
\[ \Rightarrow f = 3.8 \times {10^{14}}Hz\]
Thus, option (A) is the right answer for the given question.
Note
An optical wave is a type of electromagnetic wave that has the ability to propagate in free space and also can be guided with dielectric. The optical waves undergo the concept of total internal reflection inside an optical fiber cable between the core and the cladding part of the cable. The optical communication has a wide range of electromagnetic spectrum, enabling it to be deployed in large distance communication.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding the Electric Field of a Uniformly Charged Ring

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

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

