1. Energy density in an electromagnetic wave

For an electromagnetic wave, the instantaneous energy densities are:

$u_E = \frac{1}{2}\varepsilon_0 E^2$

and

$u_B = \frac{1}{2\mu_0} B^2$

So the total instantaneous energy density is:

$u = u_E + u_B$

2. Relation between electric and magnetic parts

For an electromagnetic wave in vacuum,

$B = \frac{E}{c}$

Using

$c^2 = \frac{1}{\varepsilon_0 \mu_0}$

we get

$u_B = \frac{1}{2\mu_0} \left(\frac{E}{c}\right)^2 = \frac{1}{2\mu_0} \cdot \frac{E^2}{c^2}$

Since

$\frac{1}{c^2} = \varepsilon_0 \mu_0$

therefore,

$u_B = \frac{1}{2\mu_0} \cdot E^2 (\varepsilon_0 \mu_0) = \frac{1}{2}\varepsilon_0 E^2 = u_E$

Thus,

$u_E = u_B$

3. Average energy densities

Because the electric and magnetic energy densities are equal at every instant, their average values are also equal:

$\langle u_E \rangle = \langle u_B \rangle$

Hence total average energy density is

$\langle u \rangle = \langle u_E \rangle + \langle u_B \rangle = 2\langle u_E \rangle$

So the required ratio is

$\frac{\langle u_E \rangle}{\langle u \rangle} = \frac{\langle u_E \rangle}{2\langle u_E \rangle} = \frac{1}{2}$

4. Option check

• A: $1$ ❌
• B: $3$ ❌
• C: $2$ ❌
• D: $\frac{1}{2}$ ✅

Therefore, the correct answer is D.