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8.Electromagnetic waves
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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
$200$
(JEE MAIN-2020)
Solution
$I =\epsilon_{0} E _{ rms }^{2} C$
$E _{ rms }^{2}=\frac{ I }{\epsilon_{0} C }$
$=\frac{315}{\pi \epsilon_{0}} \times \frac{1}{ C }$
$=\frac{4 \times 315}{4 \pi \epsilon_{0}} \times \frac{1}{3 \times 10^{8}}$
$=\frac{4 \times 315 \times 9 \times 10^{9}}{3 \times 10^{8}}$
$E _{ rms }^{2}=4 \times 315 \times 30$
$E _{ rms }=2 \sqrt{315 \times 30}$
$=194.42$
Standard 12
Physics
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