An electromagnetic wave of frequency `v=3.0 MHz` passes from vacuum into a dielectric medium with permittivity `epsilon=4.0`. Then

Explanation

The velocity of light in a medium of relative permittivity `epsilon` is `(c)/(sqrt(epsilon))`. Thus the change in wavelength of light `(` from its value in vacuum to its value in the medium `)` is
`Delta lambda=(c//sqrt(epsilon))/(v)-(c)/( v)=(c)/(v)((1)/sqrt(epsilon)-1)=-50M`.