JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
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 \(9.73 \)
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 \(U_{a v}=\frac{P}{4 \pi r^{2} \times c}\) \(\frac{P}{4 \pi r^{2}}=\frac{1}{2} \varepsilon_{0} E^{2} \times c\)…
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