The dispersion of a medium for wavelength $\lambda$ is D. Then the dispersion for the wavelength 2$\lambda$ will be:
(A) (D/8)
(B) (D/4)
(C) (D/2)
(D) D
Answer
310.5k+ views
Hint We know that dispersion is a term which is used to describe the situation in which the phase velocity of a wave depends on its frequency. Wave with a higher frequency is diffracted more than those which have a lower frequency.
Complete step by step answer
As we know that, Cauchy’s Dispersion formula is:
$\mu = A + \dfrac{B}{{{\lambda ^2}}}$
And the dispersion is given as:
$D = - \dfrac{{d\mu }}{{d\lambda }}$
Therefore, from the above 2 equations we can say that:
$D = - ( - 2{\lambda ^{-3}})B = \dfrac{{2B}}{{{\lambda ^3}}}$
$D\propto \dfrac{1}{{{\lambda ^3}}}$
This is implied by the expression that:
$D\propto \dfrac{1}{{{\lambda ^3}}}$
Hence, we can say that:
$\dfrac{{{D^/}}}{D} = {\left( {\dfrac{\lambda }{{{\lambda ^/}}}} \right)^3}$
As the ${\lambda ^/} = 2\lambda$
Therefore, the value of ${D^/} = D/8$.
Note: To avoid any confusion while solving such problems we should know that dispersion is defined as the property where the light is spread out of the according to its colour as it passes through an object. For instance, when we shine a while light into a, all of the different colours of light are bent different amounts, so they are spread out and make a rainbow.
Complete step by step answer
As we know that, Cauchy’s Dispersion formula is:
$\mu = A + \dfrac{B}{{{\lambda ^2}}}$
And the dispersion is given as:
$D = - \dfrac{{d\mu }}{{d\lambda }}$
Therefore, from the above 2 equations we can say that:
$D = - ( - 2{\lambda ^{-3}})B = \dfrac{{2B}}{{{\lambda ^3}}}$
$D\propto \dfrac{1}{{{\lambda ^3}}}$
This is implied by the expression that:
$D\propto \dfrac{1}{{{\lambda ^3}}}$
Hence, we can say that:
$\dfrac{{{D^/}}}{D} = {\left( {\dfrac{\lambda }{{{\lambda ^/}}}} \right)^3}$
As the ${\lambda ^/} = 2\lambda$
Therefore, the value of ${D^/} = D/8$.
Note: To avoid any confusion while solving such problems we should know that dispersion is defined as the property where the light is spread out of the according to its colour as it passes through an object. For instance, when we shine a while light into a, all of the different colours of light are bent different amounts, so they are spread out and make a rainbow.
Recently Updated Pages
JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Maths Question Paper with Answer Key

Apparent Frequency Explained: Formula, Uses & Examples

JEE Main 2023 (February 1st Shift 2) Chemistry Question Paper with Answer Key

Hydrogen and Its Type Important Concepts and Tips for JEE Exam Preparation

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

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Hybridisation in Chemistry – Concept, Types & Applications

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

Class 12 Physics Formulas: Complete Chapter-wise PDF Guide

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2026-27 PDF Download (Login Required)

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

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Understanding Uniform Acceleration in Physics

