Question

What is the amplitude of the electric filed in a parallel beam of light intensity $\left(\frac{15}{\pi }\right)\frac{W}{m^2}$
$($ Assume $\frac{1}{4\pi {\epsilon }_0}=9\times 10^9\frac{N{m}^2}{C^2})$

• A. $60N/C$
• B. $50N/C$
• C. $40N/C$
• D. $30N/C$

Answer 1

Answer: (A)

Intensity of parallel beam of light is given as
$I=\frac{1}{2}{\epsilon }_0{E}_0^2\dots ..(i)$
where, $E_0=$ amplitude of electric field.
Here, $I=\frac{15}{\pi }\frac{W}{m^2},\frac{1}{4\pi {\epsilon }_0}$
$=9\times 10^9\frac{N{m}^2}{C^2}$
$c=3\times 10^{8}m{s}^{-1}$
From Eq. (i), we get
$E_0^2=\frac{2\times I}{{\epsilon }_0\times c}=\frac{2\times 4\pi \times I}{4\pi {\epsilon }_0\times c}$
$=\left(\frac{2\times 4\pi \times \frac{15}{\pi }\times 9\times 10^9}{3\times 10^8}\right)$
$\Rightarrow E_0{}^2=3600$
$\Rightarrow E_0=60N/C$