1.

The frequency of na electromagnetic wave in free space is 2 MHz. When it passes through a region of relative permittivity epsilon_(r )=4.0, then its wavelength …… and frequency ……

Answer»

becomes double, becomes half.
becomes double, remains constant.
becomes half, becomes double.
becomes half, remains constant.

SOLUTION :`("velocity of light in vacuum c")/("velocity of light in medium V")=sqrt((mu_(0)EPSILON)/(mu_(0)epsilon_(0)))` (accepting `mu_(0)` is not CHANGING)
`therefore (c )/(v)=sqrt((epsilon)/(epsilon_(0)))=sqrt(epsilon_(r ))=sqrt(4)=2 ""`…(1)
If medium is CHANGED then also frequency of light f remains constant.
`therefore c = f lambda, "" v = f lambda.`
`therefore (c )/(v)=(lambda)/(lambda.)=2 ""`(From EQUATION (1))
`therefore lambda.=(lambda)/(2)`
As frequency f remains constant and wavelength becomes half.


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