About $20 \%$ of the power of a 100 W bulb is converted to visible radiation. Assuming that the radiation is emitted isotropically and neglecting reflection, the average intensity of visible radiation at a distance of 5 m is $\frac{\alpha}{25 \pi} \mathrm{~W} / \mathrm{m}^2$. The value of $\alpha$ is

An electromagnetic wave has its electric and magnetic fields given by

$$ \mathbf{E}(t)=\mathbf{E}_m \sin (k x-\omega t) ; \quad \mathbf{B}(t)=\mathbf{B}_m \sin (k x-\omega t) $$

If the direction of $\mathbf{E}_m$ and $\mathbf{B}_m$ are in the direction of $(\hat{\mathbf{i}}+\hat{\mathbf{j}})$ and $(\hat{\mathbf{i}}-\hat{\mathbf{j}})$ respectively, the unit vector that gives the direction of propagation of the wave is

A beam of white light is incident normally on a plane surface absorbing 70\% of the light and reflecting the rest. If the incident beam carries 10 W of power, the force exerted by it on the surface is

An electromagnetic wave is propagating in vacuum along $-\hat{\mathbf{j}}$ direction. The magnetic field of the wave is given by $\mathbf{B}=\left(2 \times 10^{-8}\right) \cos \left[\pi \times 10^{15}\left(t+\frac{y}{c}\right)\right] \hat{\mathbf{k}} \mathrm{T}$. The electric field $\mathbf{E}$ of this wave is ( $c \equiv$ speed of light)