JEE Main 2024 — Electromagnetic Waves Question with Solution

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

$$I=\frac{1}{2}{ϵ}_{0}c{E}_0^2$$

where:

• $$I$$ is the intensity in $$\mathrm{W}/{\mathrm{m}}^2$$
• $${ϵ}_0$$ 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 $$3\times 10^{8}m/s$$
• $$E_0$$ 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