91Ó°ÊÓ

Write the expression for the parallel of electric fields.

${\mathbf{E}}_{\mathbf{above}}\mathbf{−}{\mathbf{E}}_{\mathbf{below}}\mathbf{=}\mathbf{0}$ …… (1)

Here,${\mathrm{E}}_{\mathrm{above}}$ is the parallel component of electric field above the interface and${\mathrm{E}}_{\mathrm{below}}$ is the parallel component of electric filed below the interface.

Write the expression for perpendicular components.

${∈}_1{\mathrm{E}}_{\mathrm{above}}^{âŠ\yen }−{∈}_2{\mathrm{E}}_{\mathrm{below}}^{âŠ\yen }={\mathrm{σ}}_{\mathrm{f}}$ …… (2)

Here,${\mathrm{E}}_{\mathrm{above}}^{âŠ\yen }$ is the perpendicular component of electric field above the interface and${\mathrm{E}}_{\mathrm{below}}^{âŠ\yen }$ is the perpendicular component of electric field blow the interface.

Across the material interface, electrical fields satisfied boundary requirements. There is no free boundary charge here. Hence,

${\mathrm{σ}}_{\mathrm{f}}=0$

Now, applying boundary conditions,

Write the expression for parallel components of electric fields.

role="math" localid="1657601144914"

Substitute$E_1$for$E_{above}$and $E_2$for$E_{below}$in above equation.

Therefore,

$E_1=E_2$

Hence, the parallel component of electric field that is, $E$ is continuous.

Write the expression for perpendicular components.

${∈}_1{\mathrm{E}}_{\mathrm{above}}^{âŠ\yen }−{∈}_2{\mathrm{E}}_{\mathrm{below}}^{âŠ\yen }={\mathrm{σ}}_{\mathrm{f}}$

Substitute$E_1$ forrole="math" localid="1657601336234" $E_{above}$ and$E_2$ for$E_{below}$ and$0$ for${\mathrm{σ}}_f$ in above equation.

$\begin{array}{c}{∈}_1{E}_{above}^{âŠ\yen }−{∈}_2{E}_{below}^{âŠ\yen }={\mathrm{σ}}_f\\ {∈}_1{E}_1−{∈}_2{E}_2=0\end{array}$

$\begin{array}{c}{∈}_1{E}_1={∈}_2{E}_2\\ \frac{E_1}{E_2}=\frac{{∈}_2}{{∈}_1}\end{array}$

Write the expression for the tangent of angle ${\mathrm{θ}}_1$.

$\tan {\mathrm{θ}}_1=\frac{E_1}{E_1^{âŠ\yen }}$ …… (3)

Write the expression for the tangent of angle ${\mathrm{θ}}_2$.

$\tan {\mathrm{θ}}_2=\frac{E_2}{E_2^{âŠ\yen }}$ ……. (4)

Divide equation (4) by equation (3)

$\begin{array}{c}\frac{\tan {\mathrm{θ}}_2}{\tan {\mathrm{θ}}_1}=\frac{\left(\frac{E_2}{E_2^{âŠ\yen }}\right)}{\left(\frac{E_1}{E_1^{âŠ\yen }}\right)}\\ =\left(\frac{E_2}{E_1}\right)\left(\frac{E_1^{âŠ\yen }}{E_2^{âŠ\yen }}\right)\end{array}$

Substitute$E_1$ for$E_2$ and$\frac{{∈}_2}{{∈}_1}$ for$\left(\frac{E_1^{âŠ\yen }}{E_2^{âŠ\yen }}\right)$ in above equation

Therefore,

$\begin{array}{c}\frac{\tan {\mathrm{θ}}_2}{\tan {\mathrm{θ}}_1}=\left(\frac{E_2}{E_1}\right)\left(\frac{E_1^{âŠ\yen }}{E_2^{âŠ\yen }}\right)\\ =\left(\frac{E_1}{E_1}\right)\left(\frac{{∈}_2}{{∈}_1}\right)\\ =\frac{{∈}_2}{{∈}_1}\end{array}$

Hence, the relation is proved,$\frac{{\mathrm{³Ù²¹²Ôθ}}_2}{{\mathrm{³Ù²¹²Ôθ}}_1}=\frac{{∈}_2}{{∈}_1}$ .

If the medium 1 is air and medium 2 is dielectric then, ${∈}_2>{∈}_1$and filed lines bend away from the normal. Hence,

$\frac{\tan {\mathrm{θ}}_2}{\tan {\mathrm{θ}}_1}=\frac{{∈}_2}{{∈}_1}>1$

This is opposite of light rays. Thus, a convex lens would defocus the field line.

Hence, the convex lens of a dielectric material would defocus the electric field.