Suppose that intensity of a laser is $\left(\frac{315}{\pi}\right) \mathrm{W} / \mathrm{m}^{2}$. The rms electric field, in units of $\mathrm{V} / \mathrm{m}$ associated with this source is close to the nearest integer is ____. $\left(\varepsilon_{0}=8.86 \times 10^{-12} \mathrm{C}^{2} \mathrm{Nm}^{-2} ; \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)$

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · EM Waves

Suppose that intensity of a laser is

(\frac{315}{π}) W / m^{2}

. The rms electric field, in units of

V / m

associated with this source is close to the nearest integer is ____.

(ε_{0} = 8.86 \times 1 0^{- 12} C^{2} Nm^{- 2}; c = 3 \times 1 0^{8} ms^{- 1})

Answer: 194

Solution

The time-averaged intensity of the wave is related to the rms field by

I = ε_{0} c E_{rms}^{2}

, so

E_{rms} = \frac{I}{ε _{0} c}

. With

I = \frac{315}{π} \approx 100.3 W/m^{2}

ε_{0} = 8.86 \times 1 0^{- 12}

and

c = 3 \times 1 0^{8}

, the denominator is

ε_{0} c = 2.658 \times 1 0^{- 3}

. Then

E_{rms} = \frac{100.3}{2.658 \times 1 0 ^{- 3}} = 3.77 \times 1 0^{4} \approx 194 V/m

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Integer answer

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