A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be:

We are told that 4 % of the incident light energy is reflected from the air-glass interface, and we also know that the incident electric-field amplitude is $$E_0 = 30\ \text{V m}^{-1}.$$

First, we recall the relation between intensity and amplitude. For an electromagnetic wave, the average intensity $$I$$ is proportional to the square of the peak electric-field amplitude:

$$I \propto E_0^{\,2}.$$

Because of this proportionality, the fraction of the reflected intensity is equal to the square of the fraction of the reflected amplitude. Mathematically, if $$R = 0.04$$ represents the reflected intensity fraction, then

$$\left(\dfrac{E_r}{E_0}\right)^2 = R = 0.04,$$

where $$E_r$$ is the reflected electric-field amplitude. Taking the square root on both sides gives

$$\dfrac{E_r}{E_0} = \sqrt{0.04} = 0.2.$$

Substituting the given incident amplitude $$E_0 = 30\ \text{V m}^{-1},$$ we obtain the reflected amplitude:

$$E_r = 0.2 \times 30\ \text{V m}^{-1} = 6\ \text{V m}^{-1}.$$

The transmitted (or propagated) wave is the remainder of the incident wave after the reflected portion has been subtracted. Hence, its amplitude $$E_t$$ is

$$E_t = E_0 - E_r = 30\ \text{V m}^{-1} - 6\ \text{V m}^{-1} = 24\ \text{V m}^{-1}.$$

Thus the electric-field amplitude of the wave that continues inside the glass slab is $$24\ \text{V m}^{-1}.$$

Hence, the correct answer is Option C.