- JEE Mains 2018
- Hard
- MCQ
A plane polarized monochromatic \(EM\) wave is travelling a vacuum along \(z\) direction such that at \(t\, = t_1\) it is found that the electric field is zero at a spatial point \(z_1\) . The next zero that occurs in its neighbourhood is at \(z_2\). The frequency of the electromagnetic wave is:
Answer & Solution
(A) \(\frac{{3 \times {{10}^8}}}{{\left| {{z_2} - {z_1}} \right|}}\)
Step-by-step Solution
Detailed explanation
Given, at \(t=t_{1}, z=z_{1}, E=0\)
the next zero that occurs in it's neighborhood is at \(z_{2},\) the frequency of the electromagnetic wave at \(t_{2}\)
\(e^{i\left(k z_{1}-\omega t_{1}\right)}=e^{i\left(k z_{2}-\omega x_{2}\right)}\)
\(k z_{1}-\omega t_{1}=k z_{2}-\omega t_{2}\)
\(\left(t_{2}-t_{1}\right) \omega=k\left(z-z_{1}\right)\)
where \(k=\frac{2 \pi}{\lambda}=2 \pi v\)
\(\left(t_{2}-t_{1}\right)=\frac{2 \pi}{\lambda \times 2 \pi v}\left(z_{2}-z_{1}\right)\)
\(=\frac{1}{x \times v}\left(z_{2}-z_{1}\right)\)
\(\Rightarrow \lambda \times v=\frac{\left(z_{2}-z_{1}\right)}{\left(t_{2}-t_{1}\right)}=C\)
\(\left(t_{2}-t_{1}\right)=\frac{\left(z_{2}-z_{1}\right)}{C}\)
Frequency is \(f \propto \frac{1}{t}\) then \(\frac{1}{\left(t_{2}-t_{1}\right)}=\frac{C}{\left(z_{2}-z_{1}\right)}\)
\(\therefore \text { Frequency, } f=\frac{3 \times 10^{8}}{\left(z_{2}-z_{1}\right)}\)
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