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An EM wave from air enters a medium...

An EM wave from air enters a medium .The electric fields are ` vecE _1 = vecE_(01 ) hat x cos [ 2 pi v ((z)/(c )-t)] ` in air and ` vecE_2 = E_(02) hat x cos [ k ( 2z-c t ) ]` in medium , where the wave number k and frequency v refer to their values in air . the medium is non - magnetic . if ` epsi_(r_1) and epsi_(r_2)` refer to relative permittivities of air and medium respectively , which of the following options is correct ?

A

`(epsilon_(r_(1)))/(epsilon_(r_(1)))=(1)/(4)`

B

`(epsilon_(r_(1)))/(epsilon_(r_(1)))=(1)/(2)`

C

`(epsilon_(r_(1)))/(epsilon_(r_(1)))=4`

D

`(epsilon_(r_(1)))/(epsilon_(r_(1)))=2`

Text Solution

Verified by Experts

The correct Answer is:
A
` In air : omega = 2 pi f, k = (2 pi f)/( c)`
In medium `: omega = kc , k. = 2k `
`k = (omega )/(c) , k. = (omega)/(c)`
`:. c. = (c)/(2) rArr (1)/( sqrt( mu_(0) epsilon_(0) mu_(r_(2)) epsilon_(r_(2)))) = (1)/(2) (1)/( sqrt( mu_(0) epsilon_(0) mu_(r_(1)) epsilon_(r_(1)))) `
as `mu_(r_(1)) = mu_(r_(2))`
`(epsilon_(r_(1)))/(epsilon_(r_(2))) = (1)/(4)`
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