A yellow LED of power 0.2 W emits light uniformly around it. The amplitude of the electric field of light at a distance of 2 m from the LED is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1.73 V/m
Power of LED: $P = 0.2 \, \text{W}$
Distance: $r = 2 \, \text{m}$
Intensity at distance $r$:
$I = \frac{P}{4\pi r^2} = \frac{0.2}{4\pi (2^2)} = \frac{0.2}{16\pi}$
$I = \frac{0.0125}{\pi} \approx 3.98 \times 10^{-3} \, \text{W/m}^2$
Relation between intensity and electric field amplitude:
$I = \frac{1}{2} c \varepsilon_0 E_0^2$
⟹ $E_0 = \sqrt{\frac{2I}{c \varepsilon_0}}$
Using $c = 3 \times 10^8 \, \text{m/s}$, $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}$:
$E_0 = \sqrt{\frac{2 \times 3.98 \times 10^{-3}}{(3 \times 10^8)(8.85 \times 10^{-12})}}$
$E_0 = \sqrt{3.0} \approx 1.73 \, \text{V/m}$
Answer: $E_0 \approx 1.7 \, \text{V/m}$