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Electromagnetic Waves

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Solve this problem - Electromagnetic Waves - JEE Main

An EM wave from air enters a medium.The electric fields are

$\vec{E_{1}}= E_{01} \hat{x} cos [ 2\pi v\left ( \frac{z}{c} \ -t )\right ]$ in air and $\vec{E_{2}}= E_{02} \hat{x} cos [ k (2z-ct) ]$ in medium, where the wave number k and frequency ν refer to their values in air. The medium is non-magnetic. If _andrefer to relative permittivities of air and medium respectively, which of the following options is correct ?

Option 1)
Option 2)
$\frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=4$
Option 3)
$\frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=2$
Option 4)
$\frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=\frac{1}{4}$

Answers (2)

198 Views

N neha

D Divya Saini

As we learnt that

Speed of light formula in vacuum -

$c=\frac{1}{\sqrt{\mu_{o}\epsilon _{o}}}$

c=2.99793 X $10^{8}$ m/s

- wherein

c = Speed of light in vacuum

$\mu_{o}$ = Permeability of vacuum

$\epsilon _{o}$ = Permittivity of vacuum

$\frac{V}{C}=\frac{1}{2}$ (where v= speed in medium and c=speed of light in air)

For non magnetic medium;

$\frac{V}{C}=\sqrt{\frac{\epsilon 1}{\epsilon 2}}=\frac{1}{2}\: or\: {\frac{\epsilon 1}{\epsilon 2}} =\frac{1}{4}$

Option 1)

This is incorrect

Option 2)

$\frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=4$

This is incorrect

Option 3)

$\frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=2$

This is incorrect

Option 4)

$\frac{\epsilon _{r_{1}}}{\epsilon _{r_{2}}}=\frac{1}{4}$

This is correct

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