A 27, mW laser beam has a cross-sectional are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The intensity $I$ of an electromagnetic wave is given by the relation:$I = \frac{	ext{Power}}{	ext{Area}} = \frac{1}{2} \epsilon_0 E_0^2 c$Given

Vedclass · Accepted Answer

$1.4$

Vedclass · Answer

(D) The intensity $I$ of an electromagnetic wave is given by the relation:$I = \frac{	ext{Power}}{	ext{Area}} = \frac{1}{2} \epsilon_0 E_0^2 c$Given:Power $P = 27 	imes 10^{-3}\, W$Area $A = 10 	imes 10^{-6}\, m^2$$\epsilon_0 = 9 	imes 10^{-12}\, SI\, units$$c = 3 	imes 10^8\, m/s$Substituting the values:$\frac{27 	imes 10^{-3}}{10 	imes 10^{-6}} = \frac{1}{2} 	imes (9 	imes 10^{-12}) 	imes E_0^2 	imes (3 	imes 10^8)$$2700 = \frac{27 	imes 10^{-4}}{2} 	imes E_0^2$$E_0^2 = \frac{2700 	imes 2}{27 	imes 10^{-4}} = 200 	imes 10^4 = 2 	imes 10^6$$E_0 = \sqrt{2} 	imes 10^3\, V/m \approx 1.414 	imes 10^3\, V/m$Since $10^3\, V/m = 1\, kV/m$, we get $E_0 \approx 1.4\, kV/m$.

$A$ $27\, mW$ laser beam has a cross-sectional area of $10\, mm^2$. The magnitude of the maximum electric field in this electromagnetic wave is given by: $........\, kV/m$. [Given permittivity of free space $\epsilon_0 = 9 \times 10^{-12}\, SI\, units$, speed of light $c = 3 \times 10^8\, m/s$]

Similar Questions

In an electromagnetic wave,the electric field vector and magnetic field vector are given as $\vec{E} = E_{0} \hat{i}$ and $\vec{B} = B_{0} \hat{k}$ respectively. The direction of propagation of the electromagnetic wave is along:

In the propagation of electromagnetic waves,the angle between the direction of propagation and the plane of polarisation is (in $^{\circ}$)

The electric field and magnetic field components of an electromagnetic wave going through vacuum are described by
$E_x = E_0 \sin(kz - \omega t)$
$B_y = B_0 \sin(kz - \omega t)$
Then the correct relation between $E_0$ and $B_0$ is given by

The solar radiation is:

If the rms value of the electric field of electromagnetic waves at a distance of $3 \ m$ from a point source is $3 \ N C^{-1}$,then the power of the source is (in $W$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module