Q157 / 180 · COMEDK 2022 · Physics
A $30 \mathrm{~mW}$ laser beam has a cross-sectional area of $15 \mathrm{~mm}^2$. The magnitude of the maximum electric field in this electromagnetic wave is given by $\left[\begin{array}{l}\text { Permittivity of space, } \varepsilon_0=9 \times 10^{-12} \\ \text { Speed of light, } c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\end{array}\right]$
- Correct answerA$1.22 \mathrm{kV} / \mathrm{m}$
- B$12 \mathrm{kV} / \mathrm{m}$
- C$10 \mathrm{kV} / \mathrm{m}$
- D$201 \mathrm{kV} / \mathrm{m}$
Solution
$I = \dfrac{P}{A} = \dfrac{30 \times 10^{-3}}{15 \times 10^{-6}} = 2 \times 10^3\,\text{W/m}^2$
$I = \dfrac{1}{2}c\varepsilon_0 E_0^2 \Rightarrow E_0^2 = \dfrac{2I}{c\varepsilon_0}$
$E_0^2 = \dfrac{2 \times 2 \times 10^3}{3 \times 10^8 \times 9 \times 10^{-12}} = \dfrac{4 \times 10^3}{27 \times 10^{-4}} = \dfrac{40}{27} \times 10^6$
$E_0 = \sqrt{\dfrac{40}{27}} \times 10^3 \approx 1.22 \times 10^3\,\text{V/m} = 1.22\,\text{kV/m}$
Correct option: (1)
$I = \dfrac{1}{2}c\varepsilon_0 E_0^2 \Rightarrow E_0^2 = \dfrac{2I}{c\varepsilon_0}$
$E_0^2 = \dfrac{2 \times 2 \times 10^3}{3 \times 10^8 \times 9 \times 10^{-12}} = \dfrac{4 \times 10^3}{27 \times 10^{-4}} = \dfrac{40}{27} \times 10^6$
$E_0 = \sqrt{\dfrac{40}{27}} \times 10^3 \approx 1.22 \times 10^3\,\text{V/m} = 1.22\,\text{kV/m}$
Correct option: (1)
Hence, the correct answer is option: 'A'.