A ligth of wavelength 6000 `Å` , in air, enters, a medium with refractive index 1.5. Inside the medium, its frequency is `"________________"` Hz and its wavelength is `"________________" Å`

Explanation

Frequency remains the same, i.e.,
`f=(c)/(lambda)=(3xx10^(3))/(6000xx10^(-10))=5xx10^(14)Hz`
`mu=(V_(1))/(V_(2))=(vlambda_(1))/(vlambda_(2))rArrlambda_(2)=(lambda_(1))/(mu)`
The wavelength gets modified to
`lambda_(2)=(lambda_(1))/(mu)=(6000Å)/(1.5)=4000Å`