 # A light wave is incident normally on a glass slab of refractive index 1.5. Question:

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 $30 \mathrm{~V} / \mathrm{m}$, then the amplitude of the electric field for the wave propogating in the glass medium will be:

1. (1) $30 \mathrm{~V} / \mathrm{m}$

2. (2) $10 \mathrm{~V} / \mathrm{m}$

3. (3) $24 \mathrm{~V} / \mathrm{m}$

4. (4) $6 \mathrm{~V} / \mathrm{m}$

Correct Option: , 3

Solution:

(3) As $4 \%$ of light gets reflected, so only $(100-4=96 \%)$ of light comes after refraction so,

$P_{\text {refracted }}=\frac{96}{100} P_{I}$

$\Rightarrow \mathrm{K}_{2} \mathrm{~A}_{\mathrm{t}}^{2}=\frac{96}{100} \mathrm{~K}_{1} \mathrm{~A}_{\mathrm{i}}^{2}$

$\Rightarrow \mathrm{r}_{2} \mathrm{~A}_{\mathrm{t}}^{2}=\frac{96}{100} \mathrm{r}_{1} \mathrm{~A}_{\mathrm{i}}^{2}$

$\Rightarrow \mathrm{A}_{\mathrm{t}}^{2}=\frac{96}{100} \times \frac{1}{\frac{3}{2}} \times(30)^{2}$

$A_{t} \sqrt{\frac{64}{100} \times(30)^{2}}=24$