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.......\( V/m\)

\(0.96\, \mathrm{I}=\frac{1}{2} \mathrm{E}_{0}^{2} \mathrm{V}\)    ..... \((ii)\)

\(\Rightarrow \quad 0.96=\left(\frac{E_{0}^{\prime}}{E_{0}}\right)^{2} \frac{\varepsilon}{\varepsilon_{0}} \frac{V}{C}\)

\(0.96 = {\left( {\frac{{E_0^\prime }}{{{E_0}}}} \right)^2}\frac{\varepsilon }{{{\varepsilon _0}}}\frac{1}{{1.5}}\)

\({0.96=\left(\frac{E_{0}^{\prime}}{E_{0}}\right)^{2} \varepsilon_{r} \frac{1}{1.5}}\)   ..... \((iii)\)

and    \({\text{v}} = \frac{1}{{\sqrt {{\mu _{\text{o}}}{\varepsilon _{\text{o}}}{\mu _{\text{r}}}{\varepsilon _{\text{r}}}} }};\)  \({\text{v}} = \frac{{\text{C}}}{{\sqrt {{\mu _{\text{r}}}{\varepsilon _{\text{r}}}} }}\)

\(\sqrt {{\mu _r}{\varepsilon _r}}  = \frac{C}{v}\) ; \(\sqrt {{\varepsilon _r}}  = 1.5\,;\) \({\mu _r} \approx 1\) for transparent medium.

From equation \((iii)\)

\(0.96=\left(\frac{E_{0}^{\prime}}{E_{0}}\right)^{2}(1.5)^{2}\left(\frac{1}{1.5}\right)\)

\(\Rightarrow E_{\circ}^{\prime}=24 \mathrm{V} / \mathrm{m}\)