InterviewSolution
Saved Bookmarks
| 1. |
The change in the wavelength of light when it travels from air to glass of refractive index 1.5 and the frequency of light `4 xx 10^(14) Hz` is |
|
Answer» Given : `mu_(g) = 1.5` `v = 4 xx 10^(14) Hz` `c = 3 xx 10^(8) m//s` `lambda_(g) = ?` As we know `c = v lambda_(a)` `:. lambda_(a) = c/v` `= (3 xx 10^(8))/(4 xx 10^(4)) = 7500 Å` `lambda_(g) = (c)/(v_(g)) = (vlambda_(a))/(vlambda_(g))` `lambda_(g) = (lambda_(a))/(mu_(g))` `= (7500)/(1.5)` `= 5000 Å` `:. lambda_(a) -lambda_(g) = 7500-5000` `= 2500 Å` Wave number in glass `barlambda_(g) = 1/(lambda_(g))` `barlambda_(g) = (1)/(5 xx 10^(-7))` ` :. lambda_(g) = 2 xx 10^(6)` per m |
|