A medium is divided into region I and II about x - 0 plane- as shown in the figure below- An electromagnetic wave with electric field E1-4-x-3-ay-5-z is incident normally on the interface from region-I- The electric field E2 in region-II at the interface is

Ans : (c) nbsp; nbsp; nbsp; nbsp;E_t_1=3â_y+5â_z=E_t_2 nbsp; nbsp; nbsp;D_n_1=D_n_2 nbsp;∈_2 E_n_2=∈_1E_n_1 nbsp; nbsp; E_n_2=3× 4 ...

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Byju's Answer

Standard VIII

Physics

Loudness and Pitch

A medium is d...

Question

A medium is divided into region I and II about x = 0 plane, as shown in the figure below. An electromagnetic wave with electric field ${\to E}_{1} = 4 {^a}_{x} +^3 a_{y} + 5 {^a}_{z}$ is incident normally on the interface from region-I. The electric field ${\to E}_{2}$ in region-II at the interface is

${\to E}_{2} = {\to E}_{1}$

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$4 {^a}_{x} + 0.75 {^a}_{y} - 1.25 {^a}_{z}$

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$3 {^a}_{x} + 3 {^a}_{y} + 5 {^a}_{z}$

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- 3 {^a}_{x} + 3 {^a}_{y} + 5 {^a}_{z}

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Solution

The correct option is C
$3 {^a}_{x} + 3 {^a}_{y} + 5 {^a}_{z}$

Ans : (c)
${\to E}_{t_{1}} = 3 {^a}_{y} + 5 {^a}_{z} = {\to E}_{t_{2}}$

${\to D}_{n_{1}} = {\to D}_{n_{2}}$
$\in_{2} {\to E}_{n_{2}} = \in_{1} {\to E}_{n_{1}}$

${\to E}_{n_{2}} = \frac{3 \times 4 {^a}_{x}}{4} = 3 {^a}_{x}$

$∴ {\to E}_{2} = {\to E}_{t_{2}} + {\to E}_{n_{2}}$

$= 3 {^a}_{y} + 5 {^a}_{z} + 3 {^a}_{x}$

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A medium is divided into region I and II about x = 0 plane, as shown in the figure below. An electromagnetic wave with electric field →E1=4^ax+^3ay+5^az is incident normally on the interface from region-I. The electric field →E2 in region-II at the interface is

The correct option is C 3^ax+3^ay+5^az Ans : (c) →Et1=3^ay+5^az=→Et2 →Dn1=→Dn2 ∈2→En2=∈1→En1 →En2=3×4^ax4=3^ax ∴→E2=→Et2+→En2 =3^ay+5^az+3^ax