A medium is divided into regions $${\rm I}$$ and $${\rm I}$$$${\rm I}$$ about $$x = 0$$ plane, as shown in the Fig. below. An electromagnetic wave with electric field $${\overrightarrow E _1} = 4{\widehat a_x} + 3{\widehat a_y} + 5{\widehat a_z}$$ is incident normally on the interface form region-$${\rm I}$$ . The electric field $${E_2}$$ in region-$${\rm I}$$$${\rm I}$$ at the interface is

A medium of relative permittivityb $${\varepsilon _r} = 2$$ forms an interface with free-space. A point source of electromagnetic energy is located in the medium at a depth of $$1$$ meter from the interface. Due to the total internal reflection, the transmitted beam has a circular cross-section over the interface. The area of the beam cross-section at the interface is given by

A plane electromagnetic wave propagating in free space in incident normally on a large slab of loss-less, non-magnetic, dielectric material with $$\varepsilon > {\varepsilon _0}$$. Maxima and minima are observed when the electric field is measured in front of the slab. The maximum electric field is found to be $$5$$ times the minimum field. The intrinsic impedance of the medium should be

Medium $$1$$ has the electrical permittivity $${\varepsilon _1} = 1.5\,\,{\varepsilon _0}\,\,\,F/m$$ and occupies the region to left of $$x = 0$$ plane. Medium $$2$$ has the electrical permittivity $${\varepsilon _2} = 2.5\,\,{\varepsilon _0}\,\,\,F/m$$ and occupies the region to the right of $$x = 0$$ plane. If $${E_1}$$ in medium $$1$$ is $${E_1} = \left( {2\,{u_x} - 3\,{u_y} + 1\,{u_z}} \right)$$ volt/m, then $${E_2}$$ in medium $$2$$ is