SATHEE: Modern Physics 7 Question 19

Modern Physics 7 Question 19

19. A $27 m W$ laser beam has a cross-sectional area of $10 m m^{2}$ . The magnitude of the maximum electric field in this electromagnetic wave is given by

[Take, permittivity of space, $ε_{0} = 9 \times 10^{- 12}$ SI units and speed of light, $c = 3 \times 10^{8} m / s]$

(Main 2019, 11 Jan II)

(a) $1 k V / m$

(b) $0.7 k V / m$

(d) $1.4 k V / m$

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Solution:

Given,

Power of laser beam $(P) = 27 m W = 27 \times 10^{- 3} W$

Area of cros- section $(A) = 10 m m^{2} = 10 \times 10^{- 6} m^{2}$

Permittivity of free space $(ε_{0}) = 9 \times 10^{- 12}$ SI unit

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

Intensity of electromagnetic wave is given by the relation

$I = \frac{1}{2} n c ε_{0} E^{2}$

where, $n$ is refractive index, for air $n = 1$ .

$\begin{array}{lll} ∴ & I & = \frac{1}{2} c \cdot ε_{0} E^{2} \\ Also, & I & = \frac{P}{A} \end{array}$

From Eq. (i) and (ii), we get

$\begin{aligned} \frac{1}{2} c ε_{0} E^{2} & = \frac{P}{A} or E^{2} = \frac{2 P}{A c ε_{0}} \\ E & = \sqrt{\frac{2 \times 27 \times 10^{- 3}}{10 \times 10^{- 6} \times 3 \times 10^{8} \times 9 \times 10^{- 12}}} \\ ≃ 1.4 \times 10^{3} V / m = 1.4 k V / m \end{aligned}$

$or$

20 Equation of an amplitude modulated wave is given by the relation,

$C_{m} = (A_{c} + A_{m} \sin ω_{m} t) \cdot \sin ω_{c} \cdot t$

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

$and A_{m} = 1 V$

$∵$ For angular frequency of message signal and carrier wave, we use a relation

$ω_{c} = \frac{2 π}{T_{c}} = \frac{2 π}{8 \times 10^{- 6}}$

(as from given graph, $T_{c} = 8 \times 10^{- 6} s$ )

$= 2.5 π \times 10^{5} s^{- 1}$

and

$ω_{m} = \frac{2 π}{T_{m}} = \frac{2 π}{100 \times 10^{- 6}}$

(as from given graph, $T_{m} = 100 \times 10^{- 6} s$ )

$= 2 π \times 10^{4} s^{- 1}$

When we put values of $A_{c}, A_{m}, ω_{c}$ and $ω_{m}$ in Eq. (i), we get

$C_{m} = [9 + \sin (2 π \times 10^{4} t)] \sin (2.5 π \times 10^{5} t) V$

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Modern Physics 7 Question 19

19. A $27 m W$ laser beam has a cross-sectional area of $10 m m^{2}$ . The magnitude of the maximum electric field in this electromagnetic wave is given by

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Modern Physics 7 Question 19

19. A 27mW laser beam has a cross-sectional area of 10mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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19. A $27 m W$ laser beam has a cross-sectional area of $10 m m^{2}$ . The magnitude of the maximum electric field in this electromagnetic wave is given by