Question

The amplitude of the electric field associated with a light beam of intensity $\frac{15}{\pi}$ Wm$^{-2}$ is

• A. 120 NC$^{-1}$
• B. 15 NC$^{-1}$
• C. 60 NC$^{-1}$
• D. 30 NC$^{-1}$

Hint:

The formula for the intensity of an EM wave is $I = \frac{1}{2}\epsilon_0 c E_0^2$. For easier calculation, it's often helpful to use the relation $\epsilon_0 c = \frac{1}{4\pi k c} \cdot c = \frac{1}{4\pi k}$ where $k=9\times 10^9$, or to substitute for $\epsilon_0$ from $k = 1/(4\pi\epsilon_0)$.

Answer 1

The Correct Option is C

The intensity ($I$) of an electromagnetic wave is related to the amplitude of the electric field ($E_0$) by the formula:
$I = \frac{1}{2} \epsilon_0 c E_0^2$.
Where:
$\epsilon_0$ is the permittivity of free space, $\epsilon_0 \approx 8.85 \times 10^{-12}$ F/m.
$c$ is the speed of light in vacuum, $c = 3 \times 10^8$ m/s.
A useful relation to simplify calculations is $c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}$ and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$. This gives $\epsilon_0 = \frac{1}{4\pi \times 9 \times 10^9}$.
Let's rearrange the formula to solve for $E_0$:
$E_0^2 = \frac{2I}{\epsilon_0 c}$.
Substitute the values: $I = \frac{15}{\pi}$ W/m$^2$.
$E_0^2 = \frac{2(15/\pi)}{\epsilon_0 c} = \frac{30}{\pi \epsilon_0 c}$.
Substitute $\epsilon_0 = \frac{1}{4\pi \times 9 \times 10^9}$:
$E_0^2 = \frac{30}{\pi (\frac{1}{4\pi \times 9 \times 10^9}) c} = \frac{30 \times 4\pi \times 9 \times 10^9}{\pi c}$.
$E_0^2 = \frac{120 \times 9 \times 10^9}{c} = \frac{1080 \times 10^9}{3 \times 10^8} = 360 \times 10 = 3600$.
Now, take the square root to find the amplitude $E_0$:
$E_0 = \sqrt{3600} = 60$ V/m or N/C.