A light wave of frequency `5xx10^14 Hz` enters a medium of refractive index `1.5`. In the velocity of the light wave is ….. And its wavelength is ……

Explanation

`._(2)^(1) mu = ("Velocity of light in medium" 1)/("Velocity of light in medium" 2)`
`implies (1)/(2pi) = (v lambda_(1))/(v lambda_(2)) = (lambda_(1))/(lambda_(2))`
In air, `c_(1) = v_(1) lambda_(1)`
`implies lambda_(1) = (c_(1))/(v_(1)) = (3 xx 10^(8))/(5 xx 10^(14)) = 0.6 xx 10^(-6) m`
From (i), we have
`lambda_(2) = (lambda_(1))/(mu) = (0.6 xx 10^(-6))/(1.5) = 0.4 xx 10^(-6) m`
Also, velocity of ligth in medium 2
`= ("Velocity is ligth in medium" 1)/(mu)`
`(3 xx 10^(8))/(1.5) = 2 xx 10^(8) m s^(-1)`