Question

In S.I units, the electric field of an E.M wave in free space is deseribed by $E_y=400\sin \left[\pi \right({10}^{6}x-2ft\left)\right]$. $f$ is the frequency.

• A. Wave length of light is 2000 nm.
• B. Frequency is $1.5\times {10}^{14}$ Hz.
• C. This E.M wave is not visible for human eyes.
• D. All of the above

Answer 1

The correct option is D All of the above

$k=\pi \times {10}^6=2\frac{\pi }{\lambda }$
$\Rightarrow \lambda =\frac{2}{{10}^6}=\frac{2000}{{10}^9}=2000nm$ (infrared & not visible)
$f=\frac{C}{\lambda }=\frac{3\times {10}^8}{2000\times {10}^{-3}}=1.5\times {10}^5\times {10}^9=1.5\times {10}^{14}$
(all A, B & C correct)