$A$ $27\, mW$ laser beam has a cross-sectional area of $10\, mm^2$. The magnitude of the maximum electric field in this electromagnetic wave is given by: $........\, kV/m$. [Given permittivity of free space $\epsilon_0 = 9 \times 10^{-12}\, SI\, units$, speed of light $c = 3 \times 10^8\, m/s$]

$A$ plane $EM$ wave travelling in vacuum along $z$-direction is given by $\vec E = E_0 \sin(kz - \omega t) \hat i$ and $\vec B = B_0 \sin(kz - \omega t) \hat j$.
$(i)$ Evaluate $\int \vec E \cdot d\vec l$ over the rectangular loop $1234$ shown in the figure.
$(ii)$ Evaluate $\int \vec B \cdot d\vec s$ over the surface bounded by loop $1234$.
$(iii)$ Use $\int \vec E \cdot d\vec l = -\frac{d\phi_E}{dt}$ to prove $\frac{E_0}{B_0} = c$.
$(iv)$ By using a similar process and the equation $\int \vec B \cdot d\vec l = \mu_0 I + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$,prove that $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

An electromagnetic wave of frequency $1 \times 10^{14} \, Hz$ is propagating along the $z$-axis. The amplitude of the electric field is $4 \, V/m$. If $\varepsilon_0 = 8.8 \times 10^{-12} \, C^2/N \cdot m^2$,then the average energy density of the electric field will be:

The refractive index of the ionosphere is

For an $EM$ wave,the electric field intensity due to a bulb of $100\ W$ at a distance of $3\ m$ is $E$. What is the electric field intensity produced by radiations coming from a $50\ W$ bulb at the same distance?

$A$ 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 $400 \, \text{electrons}/cm^3$).