A laser beam of power 27 mW has a cross-sectional area of 10 mm2 . The magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space ε0 ≈ 9 × 10-12 C 2 N -1 m -2, speed of light c =3 × 108 m s -1 ]

Question

A laser beam of power 27 mW has a cross-sectional area of 10 mm2 . The magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space ε0 ≈ 9 × 10-12 C 2 N -1 m -2, speed of light c =3 × 108 m s -1 ] (A) 1kVm- 1 (B) 1.4 kV m- 1 (C) 0.7kVm- 1 (D) 2kVm- 1

Tardigrade Admin · Accepted Answer

I=(P/A)=(1/2)ϵ 0E02c ∴ E0=√(2 P/ϵ 0 cA)=√(2 × 27 × 10- 3/9 × 10- 12 × 3 × 108 × 10 × 10- 6) E0≈1.4kVm- 1

$1 kV m^{- 1}$

$1.4 kV m^{- 1}$

$0.7 kV m^{- 1}$

$2 kV m^{- 1}$

$I = \frac{P}{A} = \frac{1}{2} ϵ_{0} E_{0}^{2} c$
$∴ E_{0} = \frac{2 P}{ϵ _{0} c A} = \frac{2 \times 27 \times 1 0 ^{- 3}}{9 \times 1 0 ^{- 12} \times 3 \times 1 0 ^{8} \times 10 \times 1 0 ^{- 6}}$
$E_{0} \approx 1.4 kV m^{- 1}$

Q. A laser beam of power 27mW has a cross-sectional area of 10mm2 . The magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space ε0​≈9×10−12C2N−1m−2, speed of light c=3×108ms−1 ]

1kVm−1

1.4kVm−1

0.7kVm−1

2kVm−1

I=AP​=21​ϵ0​E02​c ∴E0​=ϵ0​cA2P​​=9×10−12×3×108×10×10−62×27×10−3​​ E0​≈1.4kVm−1

$1 kV m^{- 1}$

$1.4 kV m^{- 1}$

$0.7 kV m^{- 1}$

$2 kV m^{- 1}$

$I = \frac{P}{A} = \frac{1}{2} ϵ_{0} E_{0}^{2} c$
$∴ E_{0} = \frac{2 P}{ϵ _{0} c A} = \frac{2 \times 27 \times 1 0 ^{- 3}}{9 \times 1 0 ^{- 12} \times 3 \times 1 0 ^{8} \times 10 \times 1 0 ^{- 6}}$
$E_{0} \approx 1.4 kV m^{- 1}$