Question

What is the amplitude of the electric filed in a parallel beam of light intensity $\left(\frac{15}{\pi}\right) \frac{W}{m^{2}}$
$\left(\right.$ Assume $\left.\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \frac{ Nm ^{2}}{C^{2}}\right)$

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

Answer 1

Answer: (A)

Intensity of parallel beam of light is given as
$I=\frac{1}{2} \varepsilon_{0} E_{0}^{2} \ldots . .( i )$
where, $E_{0}=$ amplitude of electric field.
Here, $I=\frac{15}{\pi} \frac{ W }{ m ^{2}}, \frac{1}{4 \pi \varepsilon_{0}}$
$=9 \times 10^{9} \frac{ Nm ^{2}}{ C ^{2}}$
$c=3 \times 10^{8} ms ^{-1}$
From Eq. (i), we get
$E_{0}^{2}=\frac{2 \times I}{\varepsilon_{0} \times c}=\frac{2 \times 4 \pi \times I}{4 \pi \varepsilon_{0} \times c}$
$=\left(\frac{2 \times 4 \pi \times \frac{15}{\pi} \times 9 \times 10^{9}}{3 \times 10^{8}}\right)$
$\Rightarrow E_{0}{ }^{2}=3600 $
$\Rightarrow E_{0}=60 \,N / C$