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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$
$30$
$10$
$24$
$6$
Solution
$I=\frac{1}{2} \varepsilon_{0} E_{0}^{2} C$ ….. $(i)$
$0.96\, \mathrm{I}=\frac{1}{2} \mathrm{E}_{0}^{2} \mathrm{V}$ ….. $(ii)$
$\Rightarrow \quad 0.96=\left(\frac{E_{0}^{\prime}}{E_{0}}\right)^{2} \frac{\varepsilon}{\varepsilon_{0}} \frac{V}{C}$
$0.96 = {\left( {\frac{{E_0^\prime }}{{{E_0}}}} \right)^2}\frac{\varepsilon }{{{\varepsilon _0}}}\frac{1}{{1.5}}$
${0.96=\left(\frac{E_{0}^{\prime}}{E_{0}}\right)^{2} \varepsilon_{r} \frac{1}{1.5}}$ ….. $(iii)$
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 $(iii)$
$0.96=\left(\frac{E_{0}^{\prime}}{E_{0}}\right)^{2}(1.5)^{2}\left(\frac{1}{1.5}\right)$
$\Rightarrow E_{\circ}^{\prime}=24 \mathrm{V} / \mathrm{m}$
