Question

The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is ______.

Options

• `c : 1`

• `c^2 : 1`

• `1 : 1`

• `sqrt(c) : 1`

Answer 1

The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is 1 : 1.

Explanation:

Average energy by electric field E₀ is U_av

`U_(av) = 1/2 ε_0 E_0^2`

But E₀ = cB₀

(U_av) electric field = `1/2 ε_0 (cB_0)^2 = 1/2 ε_0c^2B_0^2`

= `1/2 ε_0* 1/(mu_0ε_0) (B_0)^2`  ......`[because c^2 = 1/(mu_0ε_0)]`

(U_av) electric field = `1/(2mu_0) B_0^2 (U_(av))_("magnetic field")`

Ratio = `((U_(av))_("electric field"))/((U_(av))_("magnetic field")) = 1/1`, i.e., 1:1

Hence the energy in electromagnetic wave is divided equally between electric field vector and magnetic field vector.

It means the ratio of contributions by the electric field and magnetic field components to the intensity of an electromagnetic wave is 1:1.

• When the incident EM wave is completely absorbed by a surface, it delivers energy u and momentum u/c to the surface.
• When a wave of energy u is totally reflected from the surface, the momentum delivered to surface is 2u/c.