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8.Electromagnetic waves
medium
A red $LED$ emits light at $0.1$ watt uniformly around it. The amplitude of the electric field of the light at a distance of $1\ m$ from the diode is....$ Vm^{-1}$
A
$2.45$
B
$5.48 $
C
$7.75$
D
$9.73 $
(JEE MAIN-2015)
Solution
Using $\mathrm{U}_{\mathrm{av}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2}$
But $U_{a v}=\frac{P}{4 \pi r^{2} \times c}$
$\frac{P}{4 \pi r^{2}}=\frac{1}{2} \varepsilon_{0} E^{2} \times c$
$\mathrm{E}_{0}^{2}=\frac{2 \mathrm{P}}{4 \pi \mathrm{r}^{2} \varepsilon_{0} \mathrm{c}}=\frac{2 \times 0.1 \times 9 \times 10^{9}}{1 \times 3 \times 10^{8}}$
$\mathrm{E}_{0}=\sqrt{6}=2.45 \,\mathrm{V} / \mathrm{m}$
Standard 12
Physics
