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
299.7k+ 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
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

Understanding Uniform Acceleration in Physics

Effective Nuclear Charge for JEE

Hybridisation in Chemistry – Concept, Types & Applications

Isoelectronic Species: Definition, Examples & Importance

