Question

A monochromatic beam of light has a frequency \(\nu=\frac{3}{2\pi}\times10^{12}\,\mathrm{Hz}\) and is propagating along the direction \(\frac{\hat i+\hat j}{\sqrt2}\). It is polarized along the \(\hat k\) direction. The acceptable form for the magnetic field is :

• A. \(\frac{E_0}{C}\frac{(\hat i-\hat j)}{\sqrt2}\cos\left[10^4\frac{(\hat i+\hat j)}{\sqrt2}\cdot\vec r-(3\times10^{12})t\right]\)
• B. \(\frac{E_0}{C}\frac{(\hat i-\hat j)}{\sqrt2}\cos\left[10^4\frac{(\hat i-\hat j)}{\sqrt2}\cdot\vec r-(3\times10^{12})t\right]\)
• C. \(\frac{E_0}{C}\hat k\cos\left[10^4\frac{(\hat i+\hat j)}{\sqrt2}\cdot\vec r+(3\times10^{12})t\right]\)
• D. \(\frac{E_0}{C}\frac{(\hat i+\hat j+\hat k)}{\sqrt3}\cos\left[10^4\frac{(\hat i+\hat j)}{\sqrt2}\cdot\vec r+(3\times10^{12})t\right]\)

Hint:

\(\vec E\), \(\vec B\) and the propagation direction are mutually perpendicular with \(\vec E\times\vec B\) along propagation; also \(k=\omega/c\) and the phase must be \(\vec k\cdot\vec r-\omega t\) for forward propagation.

Answer 1

The Correct Option is A

\(\omega=2\pi\nu=3\times10^{12}\ \mathrm{rad/s}\), so \(k=\frac{\omega}{c}=\frac{3\times10^{12}}{3\times10^{8}}=10^4\ \mathrm{m^{-1}}\).
The wave travels along \(\hat n=\frac{\hat i+\hat j}{\sqrt2}\), so the phase is \(10^4\,\hat n\cdot\vec r-(3\times10^{12})t\) (minus sign for propagation along \(+\hat n\)). This rules out options (3) and (4), which have a plus sign.
\(\vec B\) is along \(\hat n\times\hat E\): \(\frac{\hat i+\hat j}{\sqrt2}\times\hat k=\frac{\hat i\times\hat k+\hat j\times\hat k}{\sqrt2}=\frac{-\hat j+\hat i}{\sqrt2}=\frac{\hat i-\hat j}{\sqrt2}\).
Amplitude \(B_0=\frac{E_0}{c}\).
Option (2) has the wrong direction vector \(\hat i-\hat j\) in the phase, so only option (1) is consistent.
Answer: (1) \(\frac{E_0}{C}\frac{(\hat i-\hat j)}{\sqrt2}\cos\left[10^4\frac{(\hat i+\hat j)}{\sqrt2}\cdot\vec r-(3\times10^{12})t\right]\)