JEE MAIN - Physics (2004 - No. 1)
An electromagnetic wave of frequency $$v=3.0$$ $$MHz$$ passes from vacuum into a dielectric medium with permittivity $$ \in = 4.0.$$ Then
wave length is halved and frequency remains unchanged
wave length is doubled and the frequency becomes half
wave length is doubled and the frequency remains unchanged
wave length and frequency both remain unchanged.
Explication
Frequency remains constant during refraction
$${v_{med}} = {1 \over {\sqrt {{\mu _0}{ \in _0} \times 4} }} = {c \over 2}$$
$${{{\lambda _{med}}} \over {{\lambda _{air}}}} = {{{v_{med}}} \over {{v_{air}}}} = {{c/2} \over c} = {1 \over 2}$$
$$\therefore$$ wavelength is halved and frequency remains unchanged
$${v_{med}} = {1 \over {\sqrt {{\mu _0}{ \in _0} \times 4} }} = {c \over 2}$$
$${{{\lambda _{med}}} \over {{\lambda _{air}}}} = {{{v_{med}}} \over {{v_{air}}}} = {{c/2} \over c} = {1 \over 2}$$
$$\therefore$$ wavelength is halved and frequency remains unchanged