In the given electromagnetic wave \[ E_y = 600 \sin(\omega t - kx) \, \text{V/m}, \] intensity of the associated light beam is (in W/m$^2$); (Given $\epsilon_0 = 9 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2}$).
- 486
- 243
- 729
- 972
The Correct Option is A
Approach Solution - 1
To determine the intensity of the electromagnetic wave, we can use the formula for the intensity of an electromagnetic wave, which is given by:
\(I = \frac{1}{2} c \epsilon_0 E_m^2\)
where:
- \(c\) is the speed of light in vacuum, approximately \(3 \times 10^8 \, \text{m/s}\)
- \(\epsilon_0\) is the permittivity of free space, given as \(9 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2}\)
- \(E_m\) is the amplitude of the electric field, provided as \(600 \, \text{V/m}\)
Substituting these values into the formula, we get:
\(I = \frac{1}{2} \times 3 \times 10^8 \, \text{m/s} \times 9 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \times (600)^2 \, \text{V}^2/\text{m}^2\)
Simplifying further:
\(I = \frac{1}{2} \times 3 \times 9 \times 600^2 \times 10^8 \times 10^{-12}\) \(I = \frac{1}{2} \times 27 \times 360000 \times 10^{-4}\) \(I = \frac{1}{2} \times 9720000 \times 10^{-4}\) \(I = \frac{1}{2} \times 972 \, \text{W/m}^2\) \(I = 486 \, \text{W/m}^2\)
Thus, the intensity of the associated light beam is \(486 \, \text{W/m}^2\). Therefore, the correct answer is: 486
Approach Solution -2
The intensity \( I \) of an electromagnetic wave is given by:
\[I = \frac{1}{2} \varepsilon_0 E_0^2 c\]
where \( E_0 = 600 \, \text{Vm}^{-1} \) and \( c = 3 \times 10^8 \, \text{m/s} \).
Substitute the values:
\[I = \frac{1}{2} \times 9 \times 10^{-12} \times (600)^2 \times 3 \times 10^8\]
\[= \frac{9}{2} \times 36 \times 3 = 486 \, \text{W/m}^2\]
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