Electromagnetic Waves

JEE Main level, single correct, medium

116 of 165

Question 116 of 165

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:

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

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_2
ei(kz1−ωt1)=ei(kz2−ωt2)kz1−ωt1=kz2−ωt2(t2−t1)ω=k(z−z1) where k=2πλ=2πv(t2−t1)=2πλ×2πv(z2−z1)\begin{aligned} &e^{i\left(k z_1-\omega t_1\right)}=e^{i\left(k z_2-\omega t_2\right)} \\ &k z_1-\omega t_1=k z_2-\omega t_2 \\ &\left(t_2-t_1\right) \omega=k\left(z-z_1\right) \\ &\text { where } k=\frac{2 \pi}{\lambda}=2 \pi v \\ &\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×v=(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 \times v=\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}

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