Question

The magnetic component of a polarised wave of light is $B_x=\left(4.0\times 10^{-6}T\right)\sin \left[(1.57\times 10^7{m}^{-1}\right)y+\omega t]$ The intensity of light is

• A. $1.9kW/m^2$
• B. $3.8kW/m^2$
• C. $5.7kW/m^2$
• D. $7.6kW/m^2$

Answer 1

Answer: (A)

Given that $Bn=\left(4.0\times 10^{-6}T\right)\sin \left[\left(1.57\times 10^7{m}^{-1}\right)y+\omega t\right]$
This equation comparing with
$B=B_0\sin \left[\frac{2\pi V}{\lambda }+\omega t\right]$,
we get $B_o=4\times 10^{-6}T$
Now, from $l=\frac{C{B}_0^2}{2{\mu }_0}$
but $C^2=\frac{1}{{\epsilon }_0{H}_0}$
$\therefore l=\frac{1}{2}{C}^3{\epsilon }_0{B}_0^2$
$l=1.9kW/m^2$