Modern Physics Ques 265
- 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
[Take, permittivity of space, $\varepsilon _0=9 \times 10^{-12}$ SI units and speed of light, $c=3 \times 10^{8} $ $m / s]$
(Main 2019, 11 Jan II)
(a) $1 $ $kV / m$
(b) $0.7 $ $kV / m$
(c) $2 $ $kV / m$
(d) $1.4 $ $kV / m$
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Answer:
Correct Answer: 265.(d)
Solution:
Formula:
Relation Between The Magnetic Field Vector And The Electric Field Vector:
- Given,
Power of laser beam $(P)=27 mW=27 \times 10^{-3} $ $W$
Area of cros- section $(A)=10 m m^{2}=10 \times 10^{-6} m^{2}$
Permittivity of free space $\left(\varepsilon _0\right)=9 \times 10^{-12}$ SI unit
Speed of light $(c)=3 \times 10^{8} $ $m / s$
Intensity of electromagnetic wave is given by the relation
$ I=\frac{1}{2} n c \varepsilon _0 E^{2} $
where, $n$ is refractive index, for air $n=1$.
$\therefore I =\frac{1}{2} c \cdot \varepsilon _0 E^{2}$ $\quad$ …….(i)
$\text { Also, } I =\frac{P}{A}$ $\quad$ …….(ii)
From Eq. (i) and (ii), we get
$ \begin{aligned} \frac{1}{2} c \varepsilon _0 E^{2} & =\frac{P}{A} \text { or } E^{2}=\frac{2 P}{A c \varepsilon _0} \\ E & =\sqrt{\frac{2 \times 27 \times 10^{-3}}{10 \times 10^{-6} \times 3 \times 10^{8} \times 9 \times 10^{-12}}} \\ & \simeq 1.4 \times 10^{3} V / m=1.4 kV / m \end{aligned} $
BETA
