Easy INTEGER +4 / -1 PYQ · JEE Mains 2020

Suppose that intensity of a laser is ${{315} \over \pi }$ W/m2.
The rms electric field, in units of V/m associated with this source is close to the nearest integer is __________.

$\in$0 = 8.86 × 10–12 C2 Nm–2; c = 3 × 108 ms–1)

Answer (integer) 194

Solution

I = ${1 \over 2}$$\varepsilon$<sub>0</sub>$E_0^2$c <br><br>$\Rightarrow$ E<sub>0</sub> = $\sqrt {{{2I} \over {{\varepsilon _0}c}}}$ <br><br>$\therefore$ E<sub>rms</sub> = ${{{E_0}} \over {\sqrt 2 }}$ = $\sqrt {{I \over {{\varepsilon _0}c}}}$ <br><br>= $$\sqrt {{{{{315} \over \pi }} \over {8.86 \times {{10}^{ - 12}} \times 3 \times {{10}^8}}}} $$ <br><br>= 194

About this question

Subject: Physics · Chapter: Electromagnetic Waves · Topic: Properties of EM Waves

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