JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2015
- Medium
- MCQ
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....\( Vm^{-1}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(2.45\)
Step-by-step Solution
Detailed explanation
Using \(\mathrm{U}_{\mathrm{av}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2}\)
But \(U_{a v}=\frac{P}{4 \pi r^{2} \times c}\)
\(\frac{P}{4 \pi r^{2}}=\frac{1}{2} \varepsilon_{0} E^{2} \times c\)
\(\mathrm{E}_{0}^{2}=\frac{2 \mathrm{P}}{4 \pi \mathrm{r}^{2} \varepsilon_{0} \mathrm{c}}=\frac{2 \times 0.1 \times 9 \times 10^{9}}{1 \times 3 \times 10^{8}}\)
\(\mathrm{E}_{0}=\sqrt{6}=2.45 \,\mathrm{V} / \mathrm{m}\)
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