Modern Physics 7 Question 19
19. 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, $\left.c=3 \times 10^{8} m / s\right]$
(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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Solution:
- 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 $n=1$.
$$ \begin{array}{lll} \therefore & I & =\frac{1}{2} c \cdot \varepsilon _0 E^{2} \\ \text { Also, } & I & =\frac{P}{A} \end{array} $$
From Eq. (i) and (ii), we get
$$ \begin{aligned} \frac{1}{2} c \varepsilon _0 E^{2} & =\frac{P}{A} \text { or } E^{2}=\frac{2 P}{A c \varepsilon _0} \\ E & =\sqrt{\frac{2 \times 27 \times 10^{-3}}{10 \times 10^{-6} \times 3 \times 10^{8} \times 9 \times 10^{-12}}} \\ & \simeq 1.4 \times 10^{3} V / m=1.4 kV / m \end{aligned} $$
$$ \text { or } $$
20 Equation of an amplitude modulated wave is given by the relation,
$$ C _m=\left(A _c+A _m \sin \omega _m t\right) \cdot \sin \omega _c \cdot t $$
For the given graph, maximum amplitude,
$$ A _c+A _m=10 $$
and minimum amplitude, $A _c-A _m=8$
From Eqs. (ii) and (iii), we get
$$ \text { and } \quad A _m=1 V $$
$\because$ For angular frequency of message signal and carrier wave, we use a relation
$$ \omega _c=\frac{2 \pi}{T _c}=\frac{2 \pi}{8 \times 10^{-6}} $$
(as from given graph, $T _c=8 \times 10^{-6} s$ )
$$ =2.5 \pi \times 10^{5} s^{-1} $$
and
$$ \omega _m=\frac{2 \pi}{T _m}=\frac{2 \pi}{100 \times 10^{-6}} $$
(as from given graph, $T _m=100 \times 10^{-6} s$ )
$$ =2 \pi \times 10^{4} s^{-1} $$
When we put values of $A _c, A _m, \omega _c$ and $\omega _m$ in Eq. (i), we get
$$ C _m=\left[9+\sin \left(2 \pi \times 10^{4} t\right)\right] \sin \left(2.5 \pi \times 10^{5} t\right) V $$
