The wavelenght of light from sodium source in vacuum is 5893Å. What are its (a) wavelenght, (b) speed and ( c) frequency when this light travels in water which has a refractive index of 1.33.

Explanation

The refractive index of vacuum, `n_(1) = 1`
The wavelenght in vaccum, `lambda_(1) = 5893 Å`
The speed in vacuum, `c = 3 xx 10^(8) ms^(-1)`
The refractive index of water `n_(2) = 1.33`
The wavelenght of light in water, `lambda_(2)`
The speed of light in water, `v_(2)`
(a) The equation relating the wavelenght and refractive index is, `(lambda_(1))/(lambda_(2)) = (n_(2))/(n_(1))`
Rewriting, `lambda_(2) = (n_(1))/(n_(2)) xx lambda_(1)`
Substituting the values, `lambda_(2) = (1)/(1.33) xx 5893 Å = 4431Å rArr lambda_(2) = 4431Å`
(b) The equation relating the speed and refractive index is, `(v_(1))/(v_(2)) = (n_(1))/(n_(2))`
Rewriting, `v_(2) = (n_(1))/(n_(2)) xx v_(1)`
Substituting the values, `v_(2) = (1)/(1.33) xx 3 xx 10^(8) = 2.256 xx 10^(8)`
`v_(2) = 2.256 xx 10^(8) ms^(-1)`
( c) Frequency of light in vacuum is, `v_(1) = (c )/(lambda_(1))`
Substituting the values, `v_(1) = (3 xx 10^(8))/(5893 xx 10^(-10)) = 5.091 xx 10^(14) Hz`
Frequency of light in water is, `v_(2) = (v)/(lambda_(2))`
Substituting the values, `v_(2)=(2.256xxx10^(8)ms^(-1))/(4431xx10^(-10))=5.091xx10^(14)Hz`
The results shown that the frequency remains smae in all media.