The electric field of a plane electromagnetic wave is given by

$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos (\mathrm{kz}+\omega \mathrm{t})$ At $\mathrm{t}=0,$ a positively charged particle is at the point $(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(0,0, \frac{\pi}{\mathrm{k}}\right) .$ If its instantaneous velocity at $(t=0)$ is $v_{0} \hat{\mathrm{k}},$ the force acting on it due to the wave is

Nearly $10 \%$ of the power of a $110\,W$ light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of $1\, m$ from the bulb to a distance of $5\,m$ is $a \times 10^{-2}\,W / m ^{2}$. The value of ' $a$ ' will be.

A light wave is incident normally on a glass slab of refractive index $1.5$. If $4\%$ of light gets reflected and the amplitude of the electric field of the incident light is $30\, V/m$, then the amplitude of the electric field for the wave propagating in the glass medium will be.......$ V/m$

The electric field component of an electromagnetic wave in vaccum is given as $\vec E = 3\cos \,\left( {1.8y + 5.4 \times {{10}^8}\,t} \right)\hat i$ Its direction of propagation and wavelength is 

A laser beam can be focussed on an area equal to the square of its wavelength A $He-Ne$  laser radiates energy at the rate of $1\,mW$ and its wavelength is $632.8 \,nm$. The intensity of focussed beam will be

In an $E.M.$ wave the average energy density is associated with