Question

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

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

Answer 1

Answer: (A)

Given that $B n=\left(4.0 \times 10^{-6} T\right) \sin \left[\left(1.57 \times 10^{7} m^{-1}\right) y+\omega t\right]$
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 $C^{2}=\frac{1}{\varepsilon_{0} H_{0}}$
$ \therefore l=\frac{1}{2} C^{3} \varepsilon_{0} B_{0}^{2}$
$ l=1.9\, k W / m^{2}$