Physics - EM Waves Question with Solution | TestHub

Using $E=E_0-e^i(kz-\omega t)$ 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 $t_2$ $$\begin{array}{rl}& e^{i\left(k{z}_1-\omega t_1\right)}=e^{i\left(k{z}_2-\omega t_2\right)}\\ & k{z}_1-\omega t_1=k{z}_2-\omega t_2\\ & \left(t_2-t_1\right)\omega =k\left(z-z_1\right)\\ & \text{ where }k=\frac{2\pi }{\lambda }=2\pi v\\ & \left(t_2-t_1\right)=\frac{2\pi }{\lambda \times 2\pi v}\left(z_2-z_1\right)\end{array}$$ $$\begin{array}{rl}& =\frac{1}{x\times v}\left(z_2-z_1\right)\\ & \Rightarrow \lambda \times v=\frac{\left(z_2-z_1\right)}{\left(t_2-t_1\right)}=C\\ & \left(t_2-t_1\right)=\frac{\left(z_2-z_1\right)}{C}\\ & \text{ Frequency is }f\propto \frac{1}{t}\text{ then }\frac{1}{\left(t_2-t_1\right)}=\frac{C}{\left(z_2-z_1\right)}\\ & \therefore \text{ Frequency, }f=\frac{3\times 10^8}{\left(z_2-z_1\right)}\end{array}$$