Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 201815 AprilEvening ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Evening Shift
Mode
Online
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2018 · 15 April · Evening Shift · Online · Physics · Electromagnetic Waves

A plane polarized monochromatic EM wave is travelling a vacuum...

A plane polarized monochromatic EM wave is travelling a vacuum along zz direction such that at t=t1\mathrm{t}=\mathrm{t}_1 it is found that the electric field is zero at a spatial point z1z_1. The next zero that occurs in its neighbourhood is at z2z_2. The frequency of the electromagnetic wave is:

Options

  1. A3×108∣z2−z1∣\frac{3 \times 10^8}{\left|z_2-z_1\right|}
  2. B6×108∣z2−z1∣\frac{6 \times 10^8}{\left|\mathrm{z}_2-\mathrm{z}_1\right|}
  3. C1.5×108∣z2−z1∣\frac{1.5 \times 10^8}{\left|\mathrm{z}_2-\mathrm{z}_1\right|}
  4. D1t1+∣z2−z1∣3×108\frac{1}{t_1+\frac{\left|z_2-z_1\right|}{3 \times 10_8}}

Correct answer

A. 3×108∣z2−z1∣\frac{3 \times 10^8}{\left|z_2-z_1\right|}

Step-by-step solution

Using E=E0−ei(kz−ωt)E=E_0-e^i(k z-\omega t) Given, at t=t1,z=z1,E=0t=t_1, z=z_1, E=0 the next zero that occurs in it's neighborhood is at z2z_2, the frequency of the electromagnetic wave at t2t_2ei(kz1−\omegat1)=ei(kz2−\omegat2)kz1−\omegat1=kz2−\omegat2(t2−t1)ω=k(z−z1) where k=2πλ=2\piv(t2−t1)=2πλ×2πv(z2−z1)\begin{aligned}&e^{i\left(k z_1-\omegat_1\right)}=e^{i\left(k z_2-\omegat_2\right)} \\ &k z_1-\omegat_1=k z_2-\omegat_2 \\ &\left(t_2-t_1\right) \omega=k\left(z-z_1\right) \\ &\text{ where }k=\frac{2 \pi}{\lambda}=2 \piv \\ &\left(t_2-t_1\right)=\frac{2 \pi}{\lambda \times 2 \pi v}\left(z_2-z_1\right) \end{aligned}=1x×v(z2−z1)⇒λ\timesv=(z2−z1)(t2−t1)=C(t2−t1)=(z2−z1)C Frequency is f∝1t then 1(t2−t1)=C(z2−z1)∴ Frequency, f=3×108(z2−z1)\begin{aligned}&\quad=\frac{1}{x \times v}\left(z_2-z_1\right) \\ &\Rightarrow \lambda \timesv=\frac{\left(z_2-z_1\right)}{\left(t_2-t_1\right)}=C \\ &\quad\left(t_2-t_1\right)=\frac{\left(z_2-z_1\right)}{C}\\ &\text{ Frequency is }f \propto \frac{1}{t} \text{ then } \frac{1}{\left(t_2-t_1\right)}=\frac{C}{\left(z_2-z_1\right)}\\ &\therefore \text{ Frequency, }f=\frac{3 \times 10^8}{\left(z_2-z_1\right)} \end{aligned}$
Practice Electromagnetic Waves on Quantrex Academy →

More from Electromagnetic Waves

The electric field associated with an electromagnetic wave travelling in vacuum is given… JEE Advanced 2026A monochromatic source of light operating at 15 kW emits 2.5× 10^ 22 photons/s. The… JEE Main 2026For an electromagnetic wave propagating through vacuum, k , E and ω represent propagation… JEE Main 2026A point light source emits E.M. waves in free space. A detector, placed at a distance of… JEE Main 2026A displacement current of 4.0 A can be set up in the space between two parallel plates of… JEE Main 2026A magnetic field vector in an electromagnetic wave is represented by B =B_0 (2π vt-(2π… JEE Main 2026An electromagnetic wave travels in free space along the x -direction. At a particular… JEE Main 2026An electromagnetic wave travelling in x -direction is described by field equation E_y =… JEE Main 2026 Full Electromagnetic Waves list All JEE Main PYQs