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 $W \ m^{-2}$)

The pressure exerted by an electromagnetic wave of intensity $I \ (W/m^2)$ on a non-reflecting surface is [$c$ is the velocity of light].

For a plane electromagnetic wave,the magnetic field at a point $x$ and time $t$ is $\overrightarrow{ B }( x , t ) = [1.2 \times 10^{-7} \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \hat{ k }] \text{ T}$. The instantaneous electric field $\overrightarrow{ E }$ corresponding to $\overrightarrow{ B }$ is: (speed of light $c = 3 \times 10^{8} \text{ m/s}$)

$A$ point source of $100\,W$ emits light with $5\%$ efficiency. At a distance of $5\,m$ from the source,the intensity produced by the electric field component is:

The following travelling electromagnetic wave $E_x=0$,$E_y=E_0 \sin (kx + \omega t)$,$E_z=-2E_0 \sin (kx - \omega t)$ is:

The electric field associated with an electromagnetic $(EM)$ wave in vacuum is given by $\vec{E} = \hat{i} 40 \cos (kz - 6 \times 10^{8} t)$,where $E$,$z$,and $t$ are in $V/m$,$m$,and $s$ respectively. The value of the wave vector $k$ is .... $m^{-1}$.