$A$ point source of electromagnetic radiation has an average power output of $960 \,W$. The peak value of the electric field at a distance $400 \,cm$ from the source is (in $\,Vm^{-1}$)

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?

If an optical medium possesses a relative permeability of $\frac{10}{\pi}$ and relative permittivity of $\frac{1}{0.0885}$,then the velocity of light is greater in vacuum than that in this medium by . . . . . . times. $\left(\mu_0=4 \pi \times 10^{-7} \ H/m, \varepsilon_0=8.85 \times 10^{-12} \ F/m, c=3 \times 10^8 \ m/s\right)$

If the magnetic field of a plane electromagnetic wave is given by (The speed of light $c = 3 \times 10^8 \, m/s$):
$B = 100 \times 10^{-6} \sin \left[ 2\pi \times 2 \times 10^{15} \left( t - \frac{x}{c} \right) \right]$
Then the maximum electric field associated with it is:

$A$ plane electromagnetic wave of wave intensity $6 \ W/m^2$ strikes a small mirror of area $40 \ cm^2$,held perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second will be:

$A$ laser beam has intensity $2.1 \times 10^{15} \ W/m^2$. The amplitude of the magnetic field in the beam is approximately: (in $T$)