You are driving in your car listening to musi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The propagation of radio waves into a tunnel depends on the phenomenon of diffraction. Diffraction is significant when the wavelength $(\lambda)$

Vedclass · Accepted Answer

Radio signal of frequency $f_2$

Vedclass · Answer

(B) The propagation of radio waves into a tunnel depends on the phenomenon of diffraction. Diffraction is significant when the wavelength $(\lambda)$ of the wave is comparable to the size of the aperture (diameter $D = 10 \, m$).The wavelength is given by $\lambda = c/f$,where $c \approx 3 	imes 10^8 \, m/s$.For $f_1 = 3 	imes 10^6 \, Hz$,$\lambda_1 = (3 	imes 10^8) / (3 	imes 10^6) = 100 \, m$.For $f_2 = 30 	imes 10^6 \, Hz$,$\lambda_2 = (3 	imes 10^8) / (30 	imes 10^6) = 10 \, m$.For $f_3 = 3 	imes 10^9 \, Hz$,$\lambda_3 = (3 	imes 10^8) / (3 	imes 10^9) = 0.1 \, m$.Since the diameter of the tunnel is $10 \, m$,the signal with $\lambda_2 = 10 \, m$ matches the aperture size,leading to effective diffraction and better penetration into the tunnel compared to the others.

Similar Questions

If $E$ and $H$ represent the intensity of the electric field and the magnetising field respectively,then the unit of $E/H$ will be:

The electric field in an electromagnetic wave is given by $E = 56.5 \sin \omega(t - x/c) \; NC^{-1}$. Find the intensity of the wave if it is propagating along the $x$-axis in free space. (Given $\varepsilon_{0} = 8.85 \times 10^{-12} \; C^{2} N^{-1} m^{-2}$ and $c = 3 \times 10^{8} \; m/s$)

The energy associated with the electric field is $(U_E)$ and with the magnetic field is $(U_B)$ for an electromagnetic wave in free space. Then

What is the displacement variable in a simple pendulum and the propagation of light?

Electromagnetic waves are incident normally on a perfectly reflecting surface having surface area $A$. If $I$ is the intensity of the incident electromagnetic radiation and $c$ is the speed of light in vacuum,the force exerted by the electromagnetic wave on the reflecting surface is

Vedclass Test Series

Exam Paper Generator

Online Exam Module