Verified Solution • 2025–26 Session moderate

QJEEAdv14Ph80d2: JEE Advanced 2014 Physics - Class 12 JEEAdv14Ph80d2 Solution: A light source, width emits two wavelengths 1 = 400 nm and 2 =…

A light source, width emits two wavelengths λ\lambda 1 = 400 nm and λ\lambda 2 = 600 nm, is used in a Young's double-slit experiment. If recorded fringe widths for λ\lambda 1 and λ\lambda 2 are β\beta 1 and β\beta 2 and the number of fringes for them within a distance y on one side of the central maximum are m1 and m2, respectively, then

Multiple Choice Options

  • A β\beta 2 > β\beta 1
  • B m1 > m2
  • C from the central maximum, 3rd maximum of λ\lambda 2 overlaps with 5th minimum of λ\lambda 1
  • D the angular separation of fringes of λ\lambda 1 is greater than λ\lambda 2

Step-by-Step Verified Solution

Fringe width β=λDd\beta = {{\lambda D} \over d} ; λ\lambda 1 = 400 nm; λ\lambda 2 = 400 nm

Since λ\lambda 2 > λ\lambda 1, β\beta 2 > β\beta 1

Number of fringes within a distance y is given by

m1=yβ1{m_1} = {y \over {{\beta _1}}}

m2=yβ2{m_2} = {y \over {{\beta _2}}}

Since, β\beta 2 > β\beta 1, we get m2 < m1.

Position of 3rd maxima of λ\lambda 2 :

y′=Dd(3λ2)=1800Ddy' = {D \over d}(3{\lambda _2}) = 1800{D \over d}

Position of 5th minima of λ\lambda 1 :

y′′=Dd(9λ12)=1800Ddy'' = {D \over d}\left( {{{9{\lambda _1}} \over 2}} \right) = 1800{D \over d}

Hence, y' = y''.

Angular fringe width is θ=βD=λd\theta = {\beta \over D} = {\lambda \over d}

Since, β\beta 2 > β\beta 1 ⇒\Rightarrow θ\theta 2 > θ\theta 1

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring