JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
Suppose that intensity of a laser is \(\left(\frac{315}{\pi}\right)\, W / m ^{2} .\) The \(rms\) electric field, in units of \(V / m\) associated with this source is close to the nearest integer is \(\left(\epsilon_{0}=8.86 \times 10^{-12} C ^{2} Nm ^{-2} ; c =3 \times 10^{8} ms ^{-1}\right)\)
- A \(176\)
- B \(186\)
- C \(194\)
- D \(200\)
Answer & Solution
Correct Answer
(C) \(194\)
Step-by-step Solution
Detailed explanation
\(I =\epsilon_{0} E _{ rms }^{2} C\) \(E _{ rms }^{2}=\frac{ I }{\epsilon_{0} C }\) \(=\frac{315}{\pi \epsilon_{0}} \times \frac{1}{ C }\) \(=\frac{4 \times 315}{4 \pi \epsilon_{0}} \times \frac{1}{3 \times 10^{8}}\)…
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