QJEEM24Ph2ff8: JEE Main 2024 Physics - Class 12 JEEM24Ph2ff8 Solution: In the given electromagnetic wave , intensity of the associated …

To find the intensity of the given electromagnetic wave, we need to use the formula for the intensity of an electromagnetic wave:

$I = \frac{1}{2} \epsilon_0 c E_0^2$

where:

• $I$ is the intensity in $\mathrm{W} / \mathrm{m}^2$
• $\epsilon_0$ is the permittivity of free space, given as $9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$
• $c$ is the speed of light in vacuum, approximately $3 \times 10^8 \mathrm{~m/s}$
• $E_0$ is the peak electric field, given as $600 \mathrm{Vm}^{-1}$

Now, substitute these values into the formula:

$I = \frac{1}{2} \times 9 \times 10^{-12} \times 3 \times 10^8 \times (600)^2$

Simplify the expression step-by-step:

$I = \frac{1}{2} \times 9 \times 10^{-12} \times 3 \times 10^8 \times 360000$

First, calculate $9 \times 3 \times 360000$ :

$I = \frac{1}{2} \times 9.72 \times 10^{-4} \times 360000$

Combine 9 and 3 into 27, giving you:

$I = 13.5 \times 10^{-4} \times 360000$

Then, calculate the multiplication:

$I = 13.5 \times 36$

Finally, multiply the remaining values:

$I = 486 \times 10^{-4}$

The final value is:

The intensity, $I$ , is 486 $\mathrm{W} / \mathrm{m}^2$ . Therefore, the correct option is:

Option A: 486