If a $10 \ W$ bulb emits electromagnetic waves uniformly in all directions,then the intensity of light at a distance $0.5 \ m$ from the source is nearly (in $W \ m^{-2}$)

Suppose that the electric field of an electromagnetic wave in vacuum is $E = (3.1 \text{ NC}^{-1}) \cos [(1.8 \text{ rad m}^{-1}) y + (5.4 \times 10^6 \text{ rad s}^{-1}) t] \hat{i}$. What is the wavelength $\lambda$ (in $\text{ m}$)?

The electric and magnetic fields associated with an electromagnetic wave propagating along the $Z$-axis can be represented by:

Statement-$I$ :- During propagation of electromagnetic wave,$\vec{E}$,$\vec{B}$ and direction of propagation are perpendicular to each other.
Statement-$II$ :- During propagation of electromagnetic wave,the energy density due to electric and magnetic fields are equal.

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 $W \ m^{-2}$)

If the relative permeability and dielectric constant of a medium are $\mu_r$ and $K$ respectively,then the refractive index $n$ of the medium is: