If $10 \%$ of the power of a $(100 \pi) \ W$ light bulb is converted to visible radiation,then the average intensity of visible radiation at a distance of $10 \ m$ is (in $W \ m^{-2}$)

An electromagnetic wave of intensity $50 \ W m^{-2}$ enters a medium of refractive index $n$ without any loss. The ratio of the magnitudes of electric fields and the ratio of the magnitudes of magnetic fields of the wave before and after entering the medium are respectively given by:

The intensity of the light from a bulb incident on a surface is $0.22 \, W/m^2$. The amplitude of the magnetic field in this light wave is . . . . . . $\times 10^{-9} \, T$. (Given: Permittivity of vacuum $\epsilon_0 = 8.85 \times 10^{-12} \, C^2 N^{-1} m^{-2}$,speed of light in vacuum $c = 3 \times 10^8 \, m/s$)

Electromagnetic waves are emitted from a point source. The output power of the source is $1500 \, W$. The maximum value of the electric field at a point $3 \, m$ away from the source is ........ $V \, m^{-1}$.

In a two-wire transmission line,the power flows .......

$A$ sky wave with a frequency $55 \, MHz$ is incident on the $D$-region of the Earth's atmosphere at $45^\circ$. The angle of refraction is ......... $^\circ$ (electron density for the $D$-region is $400 \, \text{electrons}/cm^3$).