The stateof polarization of light with the electric field vector $$\overrightarrow {\bf{E}} = {\bf{\hat i}}{E_0}\cos \left( {kz - \omega t} \right) - {\bf{\hat j}}{E_0}\cos \left( {kz - \omega t} \right)$$ is
A. linearly polarized along Z-direction
B. linearly polarized at -45° to X-axis
C. circularly polarized
D. elliptically polarized with the major axis along X-axis
Answer: Option B
A. $$\frac{2}{{{\mu _0}}}\left( {x{\bf{\hat i}} + y{\bf{\hat j}}} \right)$$
B. $$ - \frac{2}{{{\mu _0}}}\left( {{\bf{\hat i}} + {\bf{\hat J}}} \right)$$
C. $$ - \frac{2}{{{\mu _0}}}\left( {{\bf{\hat i}} - {\bf{\hat j}}} \right)$$
D. $$\frac{2}{{{\mu _0}}}\left( {x{\bf{\hat i}} - y{\bf{\hat j}}} \right)$$
A. 0.033 μm
B. 0.330 μm
C. 3.300 μm
D. 33.000 μm
A. $${\bf{\hat z}}k$$
B. $${\bf{\hat x}}k\sin \alpha + {\bf{\hat y}}k\cos \alpha $$
C. $${\bf{\hat x}}k\cos \alpha + {\bf{\hat y}}k\cos \alpha $$
D. $$ - {\bf{\hat z}}k$$
A. vp = vg
B. vp = $${\text{v}}_{\text{g}}^{\frac{1}{2}}$$
C. vp vg = c2
D. vg = $${\text{v}}_{\text{p}}^{\frac{1}{2}}$$
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