JEE MAIN - Physics (2025 - 7th April Evening Shift - No. 9)

The unit of $\sqrt{\frac{2I}{\varepsilon_0 c}}$ is :

(I = intensity of an electromagnetic wave, c = speed of light)

Vm
NC-1
NC
Nm

Explanation

The unit of the expression $\sqrt{\frac{2I}{\varepsilon_0 c}}$ can be determined by exploring the relationship between intensity ($I$) and the electric field ($E_0$) in an electromagnetic wave.

The intensity of an electromagnetic wave is given by:

$ I = \frac{1}{2} \varepsilon_0 E_0^2 \times c $

From this equation, we can solve for the electric field $E_0$:

$ E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} $

The symbol $E_0$ represents the electric field, which is measured in units of Newtons per Coulomb (N/C). Therefore, the unit of the expression $\sqrt{\frac{2I}{\varepsilon_0 c}}$ is also $N/C$.

Thus, the expression $\sqrt{\frac{2I}{\varepsilon_0 c}}$ has units of $N/C$.

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