Question

The amplitude of the electric field in a parallel beam of light of intensity $4·0W{m}^{-2}$ is __________.

• A. $54·87N{C}^{-1}$
• B. $19·4N{C}^{-1}$
• C. $9·76N{C}^{-1}$
• D. None of these

Answer 1

The correct option is A

$54·87N{C}^{-1}$

Step 1: Given data

Intensity of light=$4·0W{m}^{-2}$

Step2: Formula used

The intensity of the plane electromagnetic wave is given by the formula $I=\frac{1}{2}c{\epsilon }_{\circ }{E}^2$

Where $c=speedoflightandE=Amplitudeofelectricfield$

We know that the value of permittivity is ${\epsilon }_o=8.85\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2}$

Speed of light $c=3\times {10}^{8}m/s$

Step3: Compute the amplitude of the electric field

Substitute the known values in the formula

$I=\frac{1}{2}c{\epsilon }_{\circ }{E}^2$,

$$\begin{gathered} 4=\frac{1}{2}\times \left(3\times {10}^8\right)\times \left(8·85\times {10}^{-12}\right)\times E^2 \\ E=\sqrt{\frac{8}{\left(3\times {10}^8\right)\left(8·85\times {10}^{-12}\right)}} \\ E\simeq 54·87N{C}^{-1} \end{gathered}$$

So, the amplitude of the electric field in a parallel beam of light will be $54·87N{C}^{-1}$.

Hence, option A is the correct answer.