COMEDK · Physics · 24. Electromagnetic Waves
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 \(\left[\begin{array}{l}\text { Permittivity of space, } \varepsilon_0=9 \times 10^{-12} \\ \text { Speed of light, } c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\end{array}\right]\)
- A \(1.22 \mathrm{kV} / \mathrm{m}\)
- B \(12 \mathrm{kV} / \mathrm{m}\)
- C \(10 \mathrm{kV} / \mathrm{m}\)
- D \(201 \mathrm{kV} / \mathrm{m}\)
Answer & Solution
Correct Answer
(A) \(1.22 \mathrm{kV} / \mathrm{m}\)
Step-by-step Solution
Detailed explanation
\(I = \dfrac{P}{A} = \dfrac{30 \times 10^{-3}}{15 \times 10^{-6}} = 2 \times 10^3\,\text{W/m}^2\) \(I = \dfrac{1}{2}c\varepsilon_0 E_0^2 \Rightarrow E_0^2 = \dfrac{2I}{c\varepsilon_0}\)…
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